Chapter 6 Linear Inequalities

Given:

The inequality $30x < 200$.

To Find:

The solution set for the given inequality when:

(i) $x$ is a natural number.

(ii) $x$ is an integer.

Solution:

We are given the inequality:

$30x < 200$

To solve for $x$, we divide both sides of the inequality by 30. Since 30 is a positive number, the direction of the inequality sign remains unchanged.

$\frac{30x}{30} < \frac{200}{30}$

$x < \frac{\cancel{200}^{20}}{\cancel{30}_{3}}$

$x < \frac{20}{3}$

Converting the fraction to a decimal for easier comparison:

$x < 6.66...$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is a natural number:

Natural numbers are the set $\{1, 2, 3, 4, ...\}$.

We need to find the natural numbers $x$ that satisfy $x < 6.66...$.

The natural numbers less than $6.66...$ are $1, 2, 3, 4, 5, 6$.

Therefore, the solution set when $x$ is a natural number is $\{1, 2, 3, 4, 5, 6\}$.

(ii) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < 6.66...$.

The integers less than $6.66...$ are all integers less than or equal to 6.

Therefore, the solution set when $x$ is an integer is $\{..., -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$.

Given:

The inequality $5x - 3 < 3x + 1$.

To Find:

The solution set for the given inequality when:

(i) $x$ is an integer.

(ii) $x$ is a real number.

Solution:

We are given the inequality:

$5x - 3 < 3x + 1$

Subtract $3x$ from both sides of the inequality:

$5x - 3x - 3 < 3x - 3x + 1$

$2x - 3 < 1$

Add 3 to both sides of the inequality:

$2x - 3 + 3 < 1 + 3$

$2x < 4$

Divide both sides by 2. Since 2 is a positive number, the inequality sign remains unchanged.

$\frac{2x}{2} < \frac{4}{2}$

$x < 2$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < 2$.

The integers strictly less than 2 are all integers up to 1.

Therefore, the solution set when $x$ is an integer is $\{..., -3, -2, -1, 0, 1\}$.

(ii) When $x$ is a real number:

Real numbers include all rational and irrational numbers.

We need to find the real numbers $x$ that satisfy $x < 2$.

This includes all real numbers less than 2.

In interval notation, the solution set is $(-\infty, 2)$.

Therefore, the solution set when $x$ is a real number is $\{x \in \mathbb{R} \mid x < 2\}$ or $(-\infty, 2)$.

Given:

The inequality $4x + 3 < 6x + 7$.

To Find:

The solution set for the given inequality.

Solution:

We are given the inequality:

$4x + 3 < 6x + 7$

To solve for $x$, we want to gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $4x$ from both sides of the inequality:

$4x - 4x + 3 < 6x - 4x + 7$

$3 < 2x + 7$

Now, subtract 7 from both sides of the inequality:

$3 - 7 < 2x + 7 - 7$

$-4 < 2x$

Finally, divide both sides by 2. Since 2 is a positive number, the direction of the inequality sign remains unchanged.

$\frac{-4}{2} < \frac{2x}{2}$

$-2 < x$

This can also be written as $x > -2$.

Since the type of number $x$ represents (integer, natural, real) is not specified, we assume $x$ is a real number.

The solution set consists of all real numbers greater than -2.

In interval notation, the solution set is $(-2, \infty)$.

Therefore, the solution to the inequality $4x + 3 < 6x + 7$ is $x > -2$ or $(-2, \infty)$.

Given:

The inequality $\frac{5-2x}{3} \leq \frac{x}{6} - 5$.

To Find:

The solution set for the given inequality.

Solution:

We are given the inequality:

$\frac{5-2x}{3} \leq \frac{x}{6} - 5$

To eliminate the fractions, we can find the Least Common Multiple (LCM) of the denominators, which are 3 and 6. The LCM of 3 and 6 is 6.

Multiply both sides of the inequality by the LCM, 6. Since 6 is a positive number, the inequality sign remains unchanged.

$6 \times \left(\frac{5-2x}{3}\right) \leq 6 \times \left(\frac{x}{6} - 5\right)$

$\frac{\cancel{6}^2}{1} \times \frac{(5-2x)}{\cancel{3}_1} \leq \frac{\cancel{6}^1}{1} \times \frac{x}{\cancel{6}_1} - 6 \times 5$

$2(5-2x) \leq x - 30$

Expand the left side by distributing the 2:

$10 - 4x \leq x - 30$

Now, we rearrange the inequality to group the $x$ terms on one side and the constant terms on the other.

Add $4x$ to both sides:

$10 - 4x + 4x \leq x + 4x - 30$

$10 \leq 5x - 30$

Add 30 to both sides:

$10 + 30 \leq 5x - 30 + 30$

$40 \leq 5x$

Finally, divide both sides by 5. Since 5 is a positive number, the inequality sign remains unchanged.

$\frac{40}{5} \leq \frac{5x}{5}$

$8 \leq x$

This is equivalent to $x \geq 8$.

Assuming $x$ is a real number (as not specified otherwise), the solution set includes all real numbers greater than or equal to 8.

In interval notation, the solution set is $[8, \infty)$.

Therefore, the solution to the inequality $\frac{5-2x}{3} \leq \frac{x}{6} - 5$ is $x \geq 8$ or $[8, \infty)$.

Given:

The inequality $7x + 3 < 5x + 9$.

To Find:

The solution set for the given inequality and show the graph of the solutions on a number line.

Solution:

We are given the inequality:

$7x + 3 < 5x + 9$

Subtract $5x$ from both sides of the inequality:

$7x - 5x + 3 < 5x - 5x + 9$

$2x + 3 < 9$

Subtract 3 from both sides of the inequality:

$2x + 3 - 3 < 9 - 3$

$2x < 6$

Divide both sides by 2. Since 2 is a positive number, the inequality sign remains unchanged.

$\frac{2x}{2} < \frac{6}{2}$

$x < 3$

Assuming $x$ is a real number (as not specified otherwise), the solution set consists of all real numbers strictly less than 3.

Therefore, the solution to the inequality $7x + 3 < 5x + 9$ is $x < 3$.

In interval notation, the solution set is $(-\infty, 3)$.

Graph of the solution on the number line:

To represent the solution $x < 3$ on a number line:

1. Draw a horizontal line representing the real numbers.

2. Mark the point representing the number 3 on the line.

3. Since the inequality is strict ($x < 3$), the number 3 is not included in the solution set. We indicate this by drawing an open circle at 3.

4. Shade the portion of the number line to the left of the open circle at 3. This shaded region represents all real numbers less than 3.

5. Draw an arrow pointing to the left from the shaded region to indicate that the solution extends indefinitely towards negative infinity.

Given:

The inequality $\frac{3x-4}{2} \geq \frac{x+1}{4} - 1$.

To Find:

The solution set for the given inequality and show the graph of the solutions on a number line.

Solution:

We are given the inequality:

$\frac{3x-4}{2} \geq \frac{x+1}{4} - 1$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators 2 and 4, which is 4.

Multiply both sides of the inequality by 4. Since 4 is a positive number, the inequality sign remains unchanged.

$4 \times \left(\frac{3x-4}{2}\right) \geq 4 \times \left(\frac{x+1}{4} - 1\right)$

$\frac{\cancel{4}^2}{1} \times \frac{(3x-4)}{\cancel{2}_1} \geq \frac{\cancel{4}^1}{1} \times \frac{(x+1)}{\cancel{4}_1} - 4 \times 1$

$2(3x-4) \geq (x+1) - 4$

Distribute the 2 on the left side and simplify the right side:

$6x - 8 \geq x - 3$

Now, rearrange the inequality to group the $x$ terms on one side and the constant terms on the other.

Subtract $x$ from both sides:

$6x - x - 8 \geq x - x - 3$

$5x - 8 \geq -3$

Add 8 to both sides:

$5x - 8 + 8 \geq -3 + 8$

$5x \geq 5$

Divide both sides by 5. Since 5 is a positive number, the inequality sign remains unchanged.

$\frac{5x}{5} \geq \frac{5}{5}$

$x \geq 1$

Assuming $x$ is a real number (as not specified otherwise), the solution set consists of all real numbers greater than or equal to 1.

Therefore, the solution to the inequality $\frac{3x-4}{2} \geq \frac{x+1}{4} - 1$ is $x \geq 1$.

In interval notation, the solution set is $[1, \infty)$.

Graph of the solution on the number line:

To represent the solution $x \geq 1$ on a number line:

1. Draw a horizontal line representing the real numbers.

2. Mark the point representing the number 1 on the line.

3. Since the inequality is $x \geq 1$ (greater than or equal to), the number 1 is included in the solution set. We indicate this by drawing a closed circle (or solid dot) at 1.

4. Shade the portion of the number line to the right of the closed circle at 1. This shaded region represents all real numbers greater than or equal to 1.

5. Draw an arrow pointing to the right from the shaded region to indicate that the solution extends indefinitely towards positive infinity.

Given:

Marks obtained in the first terminal examination = 62.

Marks obtained in the second terminal examination = 48.

The student needs an average of at least 60 marks in three examinations (first terminal, second terminal, and annual).

To Find:

The minimum marks the student should get in the annual examination.

Solution:

Let $x$ be the marks obtained by the student in the annual examination.

The marks in the three examinations are 62, 48, and $x$.

The total marks obtained in the three examinations = $62 + 48 + x = 110 + x$.

The number of examinations is 3.

The average marks for the three examinations is given by:

Average = $\frac{\text{Total marks}}{\text{Number of examinations}} = \frac{110 + x}{3}$

According to the problem, the student must have an average of at least 60 marks. "At least 60" means the average must be greater than or equal to 60.

So, we can set up the inequality:

$\frac{110 + x}{3} \geq 60$

To solve for $x$, we first multiply both sides of the inequality by 3. Since 3 is a positive number, the inequality sign remains the same.

$3 \times \left(\frac{110 + x}{3}\right) \geq 3 \times 60$

$110 + x \geq 180$

Now, subtract 110 from both sides of the inequality:

$110 + x - 110 \geq 180 - 110$

$x \geq 70$

This inequality means that the marks obtained in the annual examination ($x$) must be greater than or equal to 70.

Therefore, the minimum marks the student should get in the annual examination to achieve an average of at least 60 marks is 70.

Graphical Representation of the Solution:

The solution $x \geq 70$ can be represented on a number line, showing all possible scores the student can get to meet the average requirement.

Given:

We are looking for pairs of consecutive odd natural numbers.

Condition 1: Both numbers in the pair must be larger than 10.

Condition 2: The sum of the two numbers in the pair must be less than 40.

To Find:

All pairs of consecutive odd natural numbers that satisfy the given conditions.

Solution:

Let the smaller of the two consecutive odd natural numbers be $x$.

Since the numbers are consecutive odd natural numbers, the next odd natural number will be $x+2$.

According to the first condition, both numbers must be larger than 10.

$x > 10$

$x+2 > 10$ (This condition is automatically satisfied if $x > 10$)

Since $x$ must be an odd natural number greater than 10, the smallest possible value for $x$ is 11.

According to the second condition, the sum of the two numbers must be less than 40.

$x + (x+2) < 40$

Simplify the inequality:

$2x + 2 < 40$

Subtract 2 from both sides:

$2x < 40 - 2$

$2x < 38$

Divide both sides by 2:

$\frac{2x}{2} < \frac{38}{2}$

$x < 19$

So, we are looking for odd natural numbers $x$ that satisfy both conditions: $x > 10$ and $x < 19$.

Combining the conditions, we need odd natural numbers $x$ such that $10 < x < 19$.

The odd natural numbers between 10 and 19 are 11, 13, 15, and 17.

Now, we find the pairs $(x, x+2)$ for each possible value of $x$:

1. If $x = 11$, the pair is $(11, 11+2) = (11, 13)$.

Check: Both are odd, both > 10. Sum = $11 + 13 = 24 < 40$. This pair satisfies the conditions.

2. If $x = 13$, the pair is $(13, 13+2) = (13, 15)$.

Check: Both are odd, both > 10. Sum = $13 + 15 = 28 < 40$. This pair satisfies the conditions.

3. If $x = 15$, the pair is $(15, 15+2) = (15, 17)$.

Check: Both are odd, both > 10. Sum = $15 + 17 = 32 < 40$. This pair satisfies the conditions.

4. If $x = 17$, the pair is $(17, 17+2) = (17, 19)$.

Check: Both are odd, both > 10. Sum = $17 + 19 = 36 < 40$. This pair satisfies the conditions.

If $x = 19$, the condition $x < 19$ is not satisfied, so we stop here.

The pairs of consecutive odd natural numbers satisfying the given conditions are (11, 13), (13, 15), (15, 17), and (17, 19).

Therefore, the required pairs are (11, 13), (13, 15), (15, 17), (17, 19).

Given:

The inequality $24x < 100$.

To Find:

The solution set for the given inequality when:

(i) $x$ is a natural number.

(ii) $x$ is an integer.

Solution:

We are given the inequality:

$24x < 100$

To solve for $x$, we divide both sides of the inequality by 24. Since 24 is a positive number, the direction of the inequality sign remains unchanged.

$\frac{24x}{24} < \frac{100}{24}$

$x < \frac{\cancel{100}^{25}}{\cancel{24}_{6}}$

$x < \frac{25}{6}$

Converting the fraction to a decimal for easier comparison:

$\frac{25}{6} = 4 \frac{1}{6} \approx 4.167$

So the inequality is $x < 4.167$.

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is a natural number:

Natural numbers are the set $\{1, 2, 3, 4, 5, ...\}$.

We need to find the natural numbers $x$ that satisfy $x < 4.167$.

The natural numbers less than $4.167$ are $1, 2, 3, 4$.

Therefore, the solution set when $x$ is a natural number is $\{1, 2, 3, 4\}$.

(ii) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < 4.167$.

The integers less than $4.167$ are all integers less than or equal to 4.

Therefore, the solution set when $x$ is an integer is $\{..., -2, -1, 0, 1, 2, 3, 4\}$.

Given:

The inequality $-12x > 30$.

To Find:

The solution set for the given inequality when:

(i) $x$ is a natural number.

(ii) $x$ is an integer.

Solution:

We are given the inequality:

$-12x > 30$

To solve for $x$, we divide both sides of the inequality by -12. Important: When dividing or multiplying both sides of an inequality by a negative number, we must reverse the direction of the inequality sign.

$\frac{-12x}{-12} < \frac{30}{-12}$

$x < -\frac{\cancel{30}^5}{\cancel{12}_2}$

$x < -\frac{5}{2}$

Converting the fraction to a decimal:

$x < -2.5$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is a natural number:

Natural numbers are the set $\{1, 2, 3, 4, ...\}$. These are all positive integers.

We need to find the natural numbers $x$ that satisfy $x < -2.5$.

Since all natural numbers are positive, there are no natural numbers less than -2.5.

Therefore, the solution set when $x$ is a natural number is the empty set, denoted as $\phi$ or { }.

(ii) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < -2.5$.

The integers strictly less than -2.5 are $-3, -4, -5, ...$ and so on, extending towards negative infinity.

Therefore, the solution set when $x$ is an integer is $\{..., -5, -4, -3\}$.

Given:

The inequality $5x - 3 < 7$.

To Find:

The solution set for the given inequality when:

(i) $x$ is an integer.

(ii) $x$ is a real number.

Solution:

We are given the inequality:

$5x - 3 < 7$

Add 3 to both sides of the inequality:

$5x - 3 + 3 < 7 + 3$

$5x < 10$

Divide both sides by 5. Since 5 is a positive number, the inequality sign remains unchanged.

$\frac{5x}{5} < \frac{10}{5}$

$x < 2$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < 2$.

The integers strictly less than 2 are all integers up to 1.

Therefore, the solution set when $x$ is an integer is $\{..., -2, -1, 0, 1\}$.

(ii) When $x$ is a real number:

Real numbers include all rational and irrational numbers.

We need to find the real numbers $x$ that satisfy $x < 2$.

This includes all real numbers less than 2.

In interval notation, the solution set is $(-\infty, 2)$.

Therefore, the solution set when $x$ is a real number is $\{x \in \mathbb{R} \mid x < 2\}$ or $(-\infty, 2)$.

Given:

The inequality $3x + 8 > 2$.

To Find:

The solution set for the given inequality when:

(i) $x$ is an integer.

(ii) $x$ is a real number.

Solution:

We are given the inequality:

$3x + 8 > 2$

Subtract 8 from both sides of the inequality:

$3x + 8 - 8 > 2 - 8$

$3x > -6$

Divide both sides by 3. Since 3 is a positive number, the inequality sign remains unchanged.

$\frac{3x}{3} > \frac{-6}{3}$

$x > -2$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x > -2$.

The integers strictly greater than -2 are $-1, 0, 1, 2, 3, ...$ and so on, extending towards positive infinity.

Therefore, the solution set when $x$ is an integer is $\{-1, 0, 1, 2, ...\}$.

(ii) When $x$ is a real number:

Real numbers include all rational and irrational numbers.

We need to find the real numbers $x$ that satisfy $x > -2$.

This includes all real numbers greater than -2.

In interval notation, the solution set is $(-2, \infty)$.

Therefore, the solution set when $x$ is a real number is $\{x \in \mathbb{R} \mid x > -2\}$ or $(-2, \infty)$.

Given:

The inequality $4x + 3 < 5x + 7$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$4x + 3 < 5x + 7$

To solve for $x$, we gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $4x$ from both sides:

$4x - 4x + 3 < 5x - 4x + 7$

$3 < x + 7$

Subtract 7 from both sides:

$3 - 7 < x + 7 - 7$

$-4 < x$

This is equivalent to $x > -4$.

Since $x$ must be a real number, the solution set includes all real numbers greater than -4.

Therefore, the solution set for the inequality is $x > -4$.

In interval notation, the solution set is $(-4, \infty)$.

Given:

The inequality $3x - 7 > 5x - 1$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$3x - 7 > 5x - 1$

To solve for $x$, we gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $5x$ from both sides:

$3x - 5x - 7 > 5x - 5x - 1$

$-2x - 7 > -1$

Add 7 to both sides:

$-2x - 7 + 7 > -1 + 7$

$-2x > 6$

Divide both sides by -2. Important: Remember to reverse the inequality sign when dividing by a negative number.

$\frac{-2x}{-2} < \frac{6}{-2}$

$x < -3$

Since $x$ must be a real number, the solution set includes all real numbers less than -3.

Therefore, the solution set for the inequality is $x < -3$.

In interval notation, the solution set is $(-\infty, -3)$.

Given:

The inequality $3(x - 1) \leq 2(x - 3)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$3(x - 1) \leq 2(x - 3)$

First, distribute the constants on both sides of the inequality:

$3 \times x - 3 \times 1 \leq 2 \times x - 2 \times 3$

$3x - 3 \leq 2x - 6$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $2x$ from both sides:

$3x - 2x - 3 \leq 2x - 2x - 6$

$x - 3 \leq -6$

Add 3 to both sides:

$x - 3 + 3 \leq -6 + 3$

$x \leq -3$

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to -3.

Therefore, the solution set for the inequality is $x \leq -3$.

In interval notation, the solution set is $(-\infty, -3]$.

Given:

The inequality $3(2 - x) \geq 2(1 - x)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$3(2 - x) \geq 2(1 - x)$

First, distribute the constants on both sides of the inequality:

$3 \times 2 - 3 \times x \geq 2 \times 1 - 2 \times x$

$6 - 3x \geq 2 - 2x$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Add $3x$ to both sides:

$6 - 3x + 3x \geq 2 - 2x + 3x$

$6 \geq 2 + x$

Subtract 2 from both sides:

$6 - 2 \geq 2 - 2 + x$

$4 \geq x$

This is equivalent to $x \leq 4$.

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 4.

Therefore, the solution set for the inequality is $x \leq 4$.

In interval notation, the solution set is $(-\infty, 4]$.

Given:

The inequality $x + \frac{x}{2} + \frac{x}{3} < 11$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$x + \frac{x}{2} + \frac{x}{3} < 11$

To solve this inequality, we first combine the terms on the left side. The least common multiple (LCM) of the denominators (1, 2, and 3) is 6.

Rewrite the left side with a common denominator of 6:

$\frac{6x}{6} + \frac{3x}{6} + \frac{2x}{6} < 11$

Combine the fractions:

$\frac{6x + 3x + 2x}{6} < 11$

$\frac{11x}{6} < 11$

Now, we want to isolate $x$. Multiply both sides of the inequality by 6. Since 6 is a positive number, the inequality sign remains the same.

$6 \times \left(\frac{11x}{6}\right) < 6 \times 11$

$11x < 66$

Finally, divide both sides by 11. Since 11 is a positive number, the inequality sign remains the same.

$\frac{11x}{11} < \frac{66}{11}$

$x < 6$

Since $x$ must be a real number, the solution set includes all real numbers less than 6.

Therefore, the solution set for the inequality is $x < 6$.

In interval notation, the solution set is $(-\infty, 6)$.

Given:

The inequality $\frac{x}{3} > \frac{x}{2} + 1$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{x}{3} > \frac{x}{2} + 1$

To eliminate the fractions, we can find the Least Common Multiple (LCM) of the denominators, 3 and 2. The LCM is 6.

Multiply both sides of the inequality by 6. Since 6 is a positive number, the inequality sign remains unchanged.

$6 \times \left(\frac{x}{3}\right) > 6 \times \left(\frac{x}{2} + 1\right)$

$\frac{\cancel{6}^2}{1} \times \frac{x}{\cancel{3}_1} > \frac{\cancel{6}^3}{1} \times \frac{x}{\cancel{2}_1} + 6 \times 1$

$2x > 3x + 6$

Now, we rearrange the inequality to isolate $x$. Subtract $3x$ from both sides:

$2x - 3x > 3x - 3x + 6$

$-x > 6$

To solve for $x$, multiply both sides by -1. Important: When multiplying (or dividing) by a negative number, we must reverse the direction of the inequality sign.

$(-1) \times (-x) < (-1) \times 6$

$x < -6$

Since $x$ must be a real number, the solution set includes all real numbers less than -6.

Therefore, the solution set for the inequality is $x < -6$.

In interval notation, the solution set is $(-\infty, -6)$.

Given:

The inequality $\frac{3(x - 2)}{5} \leq \frac{5(2 - x)}{3}$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{3(x - 2)}{5} \leq \frac{5(2 - x)}{3}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators, 5 and 3. The LCM is 15.

Multiply both sides of the inequality by 15. Since 15 is a positive number, the inequality sign remains unchanged.

$15 \times \left(\frac{3(x - 2)}{5}\right) \leq 15 \times \left(\frac{5(2 - x)}{3}\right)$

$\frac{\cancel{15}^3}{1} \times \frac{3(x - 2)}{\cancel{5}_1} \leq \frac{\cancel{15}^5}{1} \times \frac{5(2 - x)}{\cancel{3}_1}$

$3 \times 3(x - 2) \leq 5 \times 5(2 - x)$

$9(x - 2) \leq 25(2 - x)$

Distribute the constants on both sides:

$9x - 18 \leq 50 - 25x$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Add $25x$ to both sides:

$9x + 25x - 18 \leq 50 - 25x + 25x$

$34x - 18 \leq 50$

Add 18 to both sides:

$34x - 18 + 18 \leq 50 + 18$

$34x \leq 68$

Divide both sides by 34. Since 34 is a positive number, the inequality sign remains unchanged.

$\frac{34x}{34} \leq \frac{68}{34}$

$x \leq 2$

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 2.

Therefore, the solution set for the inequality is $x \leq 2$.

In interval notation, the solution set is $(-\infty, 2]$.

Given:

The inequality $\frac{1}{2}\left( \frac{3x}{5} + 4 \right) \geq \frac{1}{3}(x - 6)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{1}{2}\left( \frac{3x}{5} + 4 \right) \geq \frac{1}{3}(x - 6)$

First, distribute the constants on both sides of the inequality:

$\frac{1}{2} \times \frac{3x}{5} + \frac{1}{2} \times 4 \geq \frac{1}{3} \times x - \frac{1}{3} \times 6$

$\frac{3x}{10} + \frac{4}{2} \geq \frac{x}{3} - \frac{6}{3}$

Simplify the constants:

$\frac{3x}{10} + 2 \geq \frac{x}{3} - 2$

To eliminate the fractions, find the Least Common Multiple (LCM) of the denominators 10 and 3. The LCM is 30.

Multiply both sides of the inequality by 30. Since 30 is a positive number, the inequality sign remains unchanged.

$30 \times \left(\frac{3x}{10} + 2\right) \geq 30 \times \left(\frac{x}{3} - 2\right)$

$30 \times \frac{3x}{10} + 30 \times 2 \geq 30 \times \frac{x}{3} - 30 \times 2$

$\frac{\cancel{30}^3}{1} \times \frac{3x}{\cancel{10}_1} + 60 \geq \frac{\cancel{30}^{10}}{1} \times \frac{x}{\cancel{3}_1} - 60$

$3 \times 3x + 60 \geq 10 \times x - 60$

$9x + 60 \geq 10x - 60$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $9x$ from both sides:

$9x - 9x + 60 \geq 10x - 9x - 60$

$60 \geq x - 60$

Add 60 to both sides:

$60 + 60 \geq x - 60 + 60$

$120 \geq x$

This is equivalent to $x \leq 120$.

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 120.

Therefore, the solution set for the inequality is $x \leq 120$.

In interval notation, the solution set is $(-\infty, 120]$.

Given:

The inequality $2(2x + 3) - 10 < 6(x - 2)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$2(2x + 3) - 10 < 6(x - 2)$

First, distribute the constants on both sides of the inequality:

$2 \times 2x + 2 \times 3 - 10 < 6 \times x - 6 \times 2$

$4x + 6 - 10 < 6x - 12$

Simplify both sides:

$4x - 4 < 6x - 12$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $4x$ from both sides:

$4x - 4x - 4 < 6x - 4x - 12$

$-4 < 2x - 12$

Add 12 to both sides:

$-4 + 12 < 2x - 12 + 12$

$8 < 2x$

Divide both sides by 2. Since 2 is a positive number, the inequality sign remains unchanged.

$\frac{8}{2} < \frac{2x}{2}$

$4 < x$

This is equivalent to $x > 4$.

Since $x$ must be a real number, the solution set includes all real numbers greater than 4.

Therefore, the solution set for the inequality is $x > 4$.

In interval notation, the solution set is $(4, \infty)$.

Given:

The inequality $37 – (3x + 5) \geq 9x – 8 (x – 3)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$37 – (3x + 5) \geq 9x – 8 (x – 3)$

First, simplify both sides of the inequality by removing the parentheses.

Left side: $37 - 3x - 5$

Combine constant terms on the left side:

$32 - 3x$

Right side: $9x - 8(x) - 8(-3)$

$9x - 8x + 24$

Combine $x$ terms on the right side:

$x + 24$

So the inequality becomes:

$32 - 3x \geq x + 24$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Add $3x$ to both sides:

$32 - 3x + 3x \geq x + 3x + 24$

$32 \geq 4x + 24$

Subtract 24 from both sides:

$32 - 24 \geq 4x + 24 - 24$

$8 \geq 4x$

Divide both sides by 4. Since 4 is a positive number, the inequality sign remains unchanged.

$\frac{8}{4} \geq \frac{4x}{4}$

$2 \geq x$

This is equivalent to $x \leq 2$.

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 2.

Therefore, the solution set for the inequality is $x \leq 2$.

In interval notation, the solution set is $(-\infty, 2]$.

Given:

The inequality $\frac{x}{4} < \frac{(5x - 2)}{3} - \frac{(7x - 3)}{5}$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{x}{4} < \frac{5x - 2}{3} - \frac{7x - 3}{5}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators 4, 3, and 5.

LCM(4, 3, 5) = $4 \times 3 \times 5 = 60$.

Multiply both sides of the inequality by 60. Since 60 is a positive number, the inequality sign remains unchanged.

$60 \times \left(\frac{x}{4}\right) < 60 \times \left(\frac{5x - 2}{3} - \frac{7x - 3}{5}\right)$

$60 \times \frac{x}{4} < 60 \times \frac{5x - 2}{3} - 60 \times \frac{7x - 3}{5}$

$\frac{\cancel{60}^{15}}{1} \times \frac{x}{\cancel{4}_1} < \frac{\cancel{60}^{20}}{1} \times \frac{(5x - 2)}{\cancel{3}_1} - \frac{\cancel{60}^{12}}{1} \times \frac{(7x - 3)}{\cancel{5}_1}$

$15x < 20(5x - 2) - 12(7x - 3)$

Distribute the constants on the right side:

$15x < (20 \times 5x - 20 \times 2) - (12 \times 7x - 12 \times 3)$

$15x < (100x - 40) - (84x - 36)$

$15x < 100x - 40 - 84x + 36$

Combine like terms on the right side:

$15x < (100x - 84x) + (-40 + 36)$

$15x < 16x - 4$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $15x$ from both sides:

$15x - 15x < 16x - 15x - 4$

$0 < x - 4$

Add 4 to both sides:

$0 + 4 < x - 4 + 4$

$4 < x$

This is equivalent to $x > 4$.

Since $x$ must be a real number, the solution set includes all real numbers greater than 4.

Therefore, the solution set for the inequality is $x > 4$.

In interval notation, the solution set is $(4, \infty)$.

Given:

The inequality $\frac{(2x - 1)}{3} \geq \frac{(3x - 2)}{4} - \frac{(2 - x)}{5}$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{2x - 1}{3} \geq \frac{3x - 2}{4} - \frac{2 - x}{5}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators 3, 4, and 5. The LCM is $3 \times 4 \times 5 = 60$.

Multiply both sides of the inequality by 60. Since 60 is a positive number, the inequality sign remains unchanged.

$60 \times \left(\frac{2x - 1}{3}\right) \geq 60 \times \left(\frac{3x - 2}{4} - \frac{2 - x}{5}\right)$

$60 \times \frac{2x - 1}{3} \geq 60 \times \frac{3x - 2}{4} - 60 \times \frac{2 - x}{5}$

$\frac{\cancel{60}^{20}}{1} \times \frac{(2x - 1)}{\cancel{3}_1} \geq \frac{\cancel{60}^{15}}{1} \times \frac{(3x - 2)}{\cancel{4}_1} - \frac{\cancel{60}^{12}}{1} \times \frac{(2 - x)}{\cancel{5}_1}$

$20(2x - 1) \geq 15(3x - 2) - 12(2 - x)$

Distribute the constants on both sides:

$20 \times 2x - 20 \times 1 \geq (15 \times 3x - 15 \times 2) - (12 \times 2 - 12 \times x)$

$40x - 20 \geq (45x - 30) - (24 - 12x)$

Remove the parentheses on the right side, paying attention to the signs:

$40x - 20 \geq 45x - 30 - 24 + 12x$

Combine like terms on the right side:

$40x - 20 \geq (45x + 12x) + (-30 - 24)$

$40x - 20 \geq 57x - 54$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $40x$ from both sides:

$40x - 40x - 20 \geq 57x - 40x - 54$

$-20 \geq 17x - 54$

Add 54 to both sides:

$-20 + 54 \geq 17x - 54 + 54$

$34 \geq 17x$

Divide both sides by 17. Since 17 is a positive number, the inequality sign remains unchanged.

$\frac{34}{17} \geq \frac{17x}{17}$

$2 \geq x$

This is equivalent to $x \leq 2$.

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 2.

Therefore, the solution set for the inequality is $x \leq 2$.

In interval notation, the solution set is $(-\infty, 2]$.

Given:

The inequality $3x – 2 < 2x + 1$.

To Find:

The solution set for the given inequality and to show the graph of the solution on a number line.

Solution:

We are given the inequality:

$3x - 2 < 2x + 1$

To solve for $x$, we first move the terms involving $x$ to one side and the constant terms to the other side.

Subtract $2x$ from both sides of the inequality:

$3x - 2x - 2 < 2x - 2x + 1$

Simplifying both sides, we get:

$x - 2 < 1$

Now, add $2$ to both sides of the inequality:

$x - 2 + 2 < 1 + 2$

Simplifying both sides, we get:

$x < 3$

Therefore, the solution to the inequality is all real numbers less than 3.

In interval notation, this is represented as $(-\infty, 3)$.

Graph on Number Line:

The solution set $x < 3$ can be represented on a number line. We draw an open circle at the point 3 to indicate that 3 is not included in the solution. Then, we shade the region to the left of 3, as all numbers in this region are less than 3.

Given:

The inequality $5x – 3 \ge 3x – 5$.

To Find:

The solution set for the given inequality and to show the graph of the solution on a number line.

Solution:

We are given the inequality:

$5x - 3 \ge 3x - 5$

To solve for $x$, we move the terms involving $x$ to one side and the constant terms to the other side.

Subtract $3x$ from both sides of the inequality:

$5x - 3x - 3 \ge 3x - 3x - 5$

Simplifying both sides, we get:

$2x - 3 \ge -5$

Now, add $3$ to both sides of the inequality:

$2x - 3 + 3 \ge -5 + 3$

Simplifying both sides, we get:

$2x \ge -2$

Divide both sides by $2$. Since 2 is a positive number, the inequality sign does not change:

$\frac{2x}{2} \ge \frac{-2}{2}$

Simplifying both sides, we get:

$x \ge -1$

Therefore, the solution to the inequality is all real numbers greater than or equal to -1.

In interval notation, this is represented as $[-1, \infty)$.

Graph on Number Line:

The solution set $x \ge -1$ can be represented on a number line. We draw a closed circle at the point -1 to indicate that -1 is included in the solution. Then, we shade the region to the right of -1, as all numbers in this region are greater than or equal to -1.

Given:

The inequality $3 (1 – x) < 2 (x + 4)$.

To Find:

The solution set for the given inequality and to show the graph of the solution on a number line.

Solution:

We are given the inequality:

$3(1 - x) < 2(x + 4)$

First, distribute the constants on both sides of the inequality:

$3 - 3x < 2x + 8$

Now, we move the terms involving $x$ to one side and the constant terms to the other side.

Add $3x$ to both sides of the inequality:

$3 - 3x + 3x < 2x + 3x + 8$

Simplifying both sides, we get:

$3 < 5x + 8$

Subtract $8$ from both sides of the inequality:

$3 - 8 < 5x + 8 - 8$

Simplifying both sides, we get:

$-5 < 5x$

Divide both sides by $5$. Since 5 is a positive number, the inequality sign does not change:

$\frac{-5}{5} < \frac{5x}{5}$

Simplifying both sides, we get:

$-1 < x$

This is equivalent to $x > -1$.

Therefore, the solution to the inequality is all real numbers greater than -1.

In interval notation, this is represented as $(-1, \infty)$.

Graph on Number Line:

The solution set $x > -1$ can be represented on a number line. We draw an open circle at the point -1 to indicate that -1 is not included in the solution. Then, we shade the region to the right of -1, as all numbers in this region are greater than -1.

Given:

The inequality $\frac{x}{2} \ge \frac{5x - 2}{3} - \frac{7x - 3}{5}$.

To Find:

The solution set for the given inequality and to show the graph of the solution on a number line.

Solution:

We are given the inequality:

$\frac{x}{2} \ge \frac{5x - 2}{3} - \frac{7x - 3}{5}$

To eliminate the denominators, we multiply both sides of the inequality by the Least Common Multiple (LCM) of the denominators 2, 3, and 5. The LCM of 2, 3, and 5 is 30.

Multiplying the entire inequality by 30 (which is a positive number, so the inequality sign remains unchanged):

$30 \times \left(\frac{x}{2}\right) \ge 30 \times \left(\frac{5x - 2}{3} - \frac{7x - 3}{5}\right)$

$30\left(\frac{x}{2}\right) \ge 30\left(\frac{5x - 2}{3}\right) - 30\left(\frac{7x - 3}{5}\right)$

$15x \ge 10(5x - 2) - 6(7x - 3)$

Distribute the constants on the right side:

$15x \ge 50x - 20 - 42x + 18$

Combine like terms on the right side:

$15x \ge (50x - 42x) + (-20 + 18)$

$15x \ge 8x - 2$

Move the terms involving $x$ to the left side by subtracting $8x$ from both sides:

$15x - 8x \ge -2$

$7x \ge -2$

Divide both sides by 7 (a positive number, so the inequality sign remains unchanged):

$\frac{7x}{7} \ge \frac{-2}{7}$

$x \ge -\frac{2}{7}$

Therefore, the solution to the inequality is all real numbers greater than or equal to $-\frac{2}{7}$.

In interval notation, this is represented as $[-\frac{2}{7}, \infty)$.

Graph on Number Line:

The solution set $x \ge -\frac{2}{7}$ can be represented on a number line. We draw a closed circle at the point $-\frac{2}{7}$ (approximately -0.28) to indicate that this point is included in the solution. Then, we shade the region to the right of $-\frac{2}{7}$, as all numbers in this region are greater than or equal to $-\frac{2}{7}$.

Solution:

Let $x$ be the marks Ravi obtains in the third unit test.

The marks obtained in the first two unit tests are 70 and 75.

The average marks of the three tests is calculated by summing the marks and dividing by the number of tests (which is 3).

Average marks $= \frac{\text{Sum of marks}}{\text{Number of tests}}$

Average marks $= \frac{70 + 75 + x}{3}$

We are given that the average of the three tests must be at least 60 marks. "At least 60" means the average must be greater than or equal to 60.

So, we can write the inequality:

$\frac{70 + 75 + x}{3} \ge 60$

Now, we solve this inequality for $x$:

First, simplify the numerator on the left side:

$70 + 75 = 145$

So, the inequality becomes:

$\frac{145 + x}{3} \ge 60$

Multiply both sides of the inequality by $3$ (which is a positive number, so the inequality sign does not change):

$3 \times \frac{145 + x}{3} \ge 60 \times 3$

$145 + x \ge 180$

Subtract $145$ from both sides of the inequality:

$145 + x - 145 \ge 180 - 145$

Simplifying both sides, we get:

$x \ge 35$

The solution $x \ge 35$ means that the marks obtained in the third test must be 35 or more to have an average of at least 60.

Since marks cannot be less than 0, and the minimum requirement is 35, the minimum marks Ravi should get in the third test is 35.

Conclusion:

The minimum marks Ravi should get in the third test to have an average of at least 60 marks is 35.

Solution:

Let the marks Sunita obtains in the fifth examination be $x$.

The marks obtained in the first four examinations are 87, 92, 94, and 95.

The total marks in the five examinations will be the sum of the marks from the first four exams and the marks from the fifth exam:

Total marks $= 87 + 92 + 94 + 95 + x$

Calculate the sum of the marks from the first four exams:

$87 + 92 + 94 + 95 = 368$

So, the total marks $= 368 + x$.

The average marks in the five examinations is calculated by dividing the total marks by the number of examinations (which is 5).

Average marks $= \frac{\text{Total marks}}{\text{Number of examinations}}$

Average marks $= \frac{368 + x}{5}$

To receive Grade ‘A’, Sunita must obtain an average of 90 marks or more. This means the average marks must be greater than or equal to 90.

We can write this as an inequality:

$\frac{368 + x}{5} \ge 90$

Now, we solve this inequality for $x$:

Multiply both sides of the inequality by $5$ (which is a positive number, so the inequality sign does not change):

$5 \times \frac{368 + x}{5} \ge 90 \times 5$

$368 + x \ge 450$

Subtract $368$ from both sides of the inequality:

$368 + x - 368 \ge 450 - 368$

Simplifying both sides, we get:

$x \ge 82$

The inequality $x \ge 82$ tells us that Sunita must obtain 82 marks or more in the fifth examination to achieve an average of 90 or more.

Since the marks are out of 100, a score of 82 or higher is possible.

The minimum marks required is the smallest value of $x$ that satisfies the inequality, which is 82.

Conclusion:

The minimum marks that Sunita must obtain in the fifth examination to get grade ‘A’ in the course is 82.

Solution:

Let the first consecutive odd positive integer be $x$. Since the integers are consecutive and odd, the next consecutive odd positive integer will be $x+2$.

We are given two conditions for these integers:

1. Both integers must be smaller than 10.

This implies $x < 10$ and $x+2 < 10$.

From $x+2 < 10$, we subtract 2 from both sides:

$x < 10 - 2$

$x < 8$

So, we must have $x < 8$.

Since $x$ must be a positive odd integer, the possible positive odd integers smaller than 8 are 1, 3, 5, and 7.

2. Their sum must be more than 11.

The sum of the two integers is $x + (x+2)$.

The condition is:

Combine like terms:

Subtract 2 from both sides:

Divide both sides by 2 (a positive number, so the inequality sign does not change):

We need to find the positive odd integers $x$ that satisfy both $x < 8$ (from condition 1) and $x > 4.5$ (from inequality (iv)).

The positive odd integers less than 8 are {1, 3, 5, 7}. We check which of these are greater than 4.5.

For $x=1$: $1 \ngtr 4.5$

For $x=3$: $3 \ngtr 4.5$

For $x=5$: $5 > 4.5$. This value is valid for the first integer.

For $x=7$: $7 > 4.5$. This value is valid for the first integer.

Therefore, the possible values for the first integer $x$ are 5 and 7.

Now, we find the pairs $(x, x+2)$ based on these values of $x$:

If $x = 5$, the pair is $(5, 5+2) = (5, 7)$.

If $x = 7$, the pair is $(7, 7+2) = (7, 9)$.

Let's verify these pairs:

Pair (5, 7): Both are consecutive odd positive integers. Both 5 and 7 are less than 10. Their sum is $5+7=12$, which is greater than 11. This pair is valid.

Pair (7, 9): Both are consecutive odd positive integers. Both 7 and 9 are less than 10. Their sum is $7+9=16$, which is greater than 11. This pair is valid.

Conclusion:

The pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11 are (5, 7) and (7, 9).

Solution:

Let the first consecutive even positive integer be $x$.

Since the integers are consecutive and even, the next consecutive even positive integer will be $x+2$.

We are given two conditions for these integers:

1. Both integers must be larger than 5.

This implies $x > 5$ and $x+2 > 5$.

From $x+2 > 5$, we subtract 2 from both sides:

$x > 5 - 2$

$x > 3$

The condition $x > 5$ is stronger than $x > 3$. So, we must have $x > 5$.

Since $x$ must be an even positive integer, the possible values for $x$ that are greater than 5 are 6, 8, 10, 12, and so on.

2. Their sum must be less than 23.

The sum of the two integers is $x + (x+2)$.

The condition is:

$x + (x+2) < 23$

Combine like terms:

$2x + 2 < 23$

Subtract 2 from both sides:

$2x + 2 - 2 < 23 - 2$

$2x < 21$

Divide both sides by 2 (a positive number, so the inequality sign does not change):

$\frac{2x}{2} < \frac{21}{2}$

$x < 10.5$

We need to find the even positive integers $x$ that satisfy both $x > 5$ and $x < 10.5$.

The even positive integers greater than 5 are {6, 8, 10, 12, ...}.

The even positive integers less than 10.5 are {..., 6, 8, 10}.

Combining these conditions, the even positive integers $x$ that are greater than 5 and less than 10.5 are 6, 8, and 10.

Now, we form the pairs $(x, x+2)$ based on these possible values of $x$:

If $x = 6$, the pair is $(6, 6+2) = (6, 8)$.

If $x = 8$, the pair is $(8, 8+2) = (8, 10)$.

If $x = 10$, the pair is $(10, 10+2) = (10, 12)$.

Let's verify these pairs:

Pair (6, 8): Both are consecutive even positive integers. Both 6 and 8 are larger than 5. Their sum is $6+8=14$, which is less than 23. This pair is valid.

Pair (8, 10): Both are consecutive even positive integers. Both 8 and 10 are larger than 5. Their sum is $8+10=18$, which is less than 23. This pair is valid.

Pair (10, 12): Both are consecutive even positive integers. Both 10 and 12 are larger than 5. Their sum is $10+12=22$, which is less than 23. This pair is valid.

The next possible value for $x$ would be 12, but $12 \nless 10.5$. So, we have found all possible pairs.

Conclusion:

The pairs of consecutive even positive integers both of which are larger than 5 such that their sum is less than 23 are (6, 8), (8, 10), and (10, 12).

Solution:

Let the length of the shortest side of the triangle be $x$ cm.

Based on the problem description:

The longest side is 3 times the shortest side, so its length is $3x$ cm.

The third side is 2 cm shorter than the longest side, so its length is $(3x - 2)$ cm.

For these lengths to form a valid triangle, they must all be positive, and the triangle inequality must hold (the sum of any two sides must be greater than the third side).

Lengths must be positive:

$x > 0$

$3x > 0 \implies x > 0$

$3x - 2 > 0 \implies 3x > 2 \implies x > \frac{2}{3}$

Triangle inequality:

$x + 3x > 3x - 2 \implies 4x > 3x - 2 \implies x > -2$. (Already covered by $x > 2/3$)

$x + (3x - 2) > 3x \implies 4x - 2 > 3x \implies x > 2$.

$3x + (3x - 2) > x \implies 6x - 2 > x \implies 5x > 2 \implies x > \frac{2}{5}$. (Already covered by $x > 2$)

So, for a valid triangle, $x$ must be greater than 2.

The perimeter of the triangle is the sum of the lengths of its three sides:

Perimeter $= x + 3x + (3x - 2)$

Perimeter $= (x + 3x + 3x) - 2$

Perimeter $= 7x - 2$ cm

We are given that the perimeter of the triangle is at least 61 cm. "At least 61 cm" means the perimeter must be greater than or equal to 61 cm.

We can write this as an inequality:

$7x - 2 \ge 61$

Now, we solve this inequality for $x$:

Add 2 to both sides of the inequality:

$7x - 2 + 2 \ge 61 + 2$

$7x \ge 63$

Divide both sides by 7 (which is a positive number, so the inequality sign does not change):

$\frac{7x}{7} \ge \frac{63}{7}$

$x \ge 9$

This inequality $x \ge 9$ tells us that the length of the shortest side must be greater than or equal to 9 cm.

We also established that for a valid triangle, $x > 2$. The condition $x \ge 9$ satisfies $x > 2$.

The question asks for the minimum length of the shortest side.

The minimum value that satisfies $x \ge 9$ is 9.

Conclusion:

The minimum length of the shortest side of the triangle is 9 cm.

Let's verify the side lengths for $x=9$ cm:

Shortest side = 9 cm

Longest side = $3 \times 9 = 27$ cm

Third side = $3 \times 9 - 2 = 27 - 2 = 25$ cm

These lengths (9, 27, 25) form a valid triangle (9+25 > 27, 9+27 > 25, 25+27 > 9).

Perimeter = $9 + 27 + 25 = 61$ cm, which is "at least 61 cm".

Solution:

Let the length of the shortest piece of board be $x$ cm.

Based on the problem statement, the lengths of the three pieces are:

Shortest piece: $x$ cm

Second piece: $3$ cm longer than the shortest, so $(x + 3)$ cm

Third piece: twice as long as the shortest, so $2x$ cm

Since these are lengths, they must be positive:

$x > 0$

$x + 3 > 0 \implies x > -3$

$2x > 0 \implies x > 0$

The condition $x > 0$ satisfies $x > -3$. So, we must have $x > 0$.

We are given two main conditions:

1. The total length of the three pieces cut from a 91 cm board cannot exceed the length of the board. The sum of the lengths must be less than or equal to 91 cm.

$x + (x + 3) + 2x \le 91$

2. The third piece is to be at least 5 cm longer than the second piece.

$2x \ge (x + 3) + 5$

Let's solve the first inequality:

$x + x + 3 + 2x \le 91$

Combine like terms:

$4x + 3 \le 91$

Subtract 3 from both sides:

$4x \le 91 - 3$

$4x \le 88$

Divide both sides by 4 (a positive number):

$\frac{4x}{4} \le \frac{88}{4}$

$x \le 22$

Now let's solve the second inequality:

$2x \ge x + 3 + 5$

Combine constants on the right side:

$2x \ge x + 8$

Subtract $x$ from both sides:

$2x - x \ge 8$

$x \ge 8$

We also have the condition that the length must be positive:

We need to find the values of $x$ that satisfy all three conditions: $x \le 22$, $x \ge 8$, and $x > 0$.

Combining $x \ge 8$ and $x > 0$, we get $x \ge 8$.

Combining $x \ge 8$ and $x \le 22$, we get $8 \le x \le 22$.

The possible lengths of the shortest board are in the interval $[8, 22]$.

Conclusion:

The possible lengths of the shortest board range from 8 cm to 22 cm, inclusive. That is, the length $x$ must satisfy $8 \le x \le 22$.

Given:

The linear inequality $3x + 2y > 6$.

To Find:

The graphical solution of the given inequality.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

First, we consider the corresponding equation to the inequality, which will be the boundary line of the solution region.

$3x + 2y = 6$

We find two points on this line to plot it. We can create a table for this:

Step 2: Determine the type of line.

The given inequality is $3x + 2y > 6$, which is a strict inequality (it does not include equality). Therefore, the points on the line $3x + 2y = 6$ are not part of the solution. We represent this by drawing a dashed line through the points (0, 3) and (2, 0).

Step 3: Choose a test point and determine the solution region.

We select a test point that is not on the line, typically the origin $(0, 0)$, to determine which half-plane is the solution region.

Substitute $(0, 0)$ into the original inequality $3x + 2y > 6$:

$3(0) + 2(0) > 6$

$0 > 6$

This statement is false. This means the solution region is the half-plane that does not contain the test point $(0, 0)$. Therefore, we shade the region on the side of the dashed line opposite to the origin (i.e., the region above the line).

Graphical Representation:

The graph shows the dashed line $3x + 2y = 6$ passing through points $(2, 0)$ and $(0, 3)$. The shaded region above this line represents the solution set for the inequality $3x + 2y > 6$.

Given:

The linear inequality $3x - 6 \ge 0$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Simplify the inequality and find the boundary line.

First, we simplify the given inequality:

$3x \ge 6$

$x \ge 2$

The corresponding equation is $x = 2$. This equation represents a vertical line parallel to the y-axis, passing through the point $(2, 0)$.

Step 2: Determine the type of line.

The inequality is $x \ge 2$, which is inclusive (it includes equality). Therefore, the points on the line $x = 2$ are part of the solution. We represent this by drawing a solid line at $x=2$.

Step 3: Choose a test point and determine the solution region.

We select a test point that is not on the line, such as the origin $(0, 0)$.

Substitute $x=0$ into the simplified inequality $x \ge 2$:

$0 \ge 2$

This statement is false. This means the solution region is the half-plane that does not contain the test point $(0, 0)$. The line $x=2$ divides the plane into a left half (containing the origin) and a right half. Since the origin side is not the solution, we shade the region to the right of the solid line $x=2$.

Graphical Representation:

The graph shows a solid vertical line at $x=2$. The shaded region to the right of this line, including the line itself, represents the solution set for the inequality $3x - 6 \ge 0$.

Given:

The linear inequality $y < 2$.

To Find:

The graphical solution of the given inequality.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation to the inequality is $y = 2$. This equation represents a horizontal line parallel to the x-axis, passing through the point $(0, 2)$.

Step 2: Determine the type of line.

The inequality is $y < 2$, which is a strict inequality. Therefore, the points on the line $y = 2$ are not part of the solution. We represent this by drawing a dashed line at $y=2$.

Step 3: Choose a test point and determine the solution region.

We select a test point that is not on the line, such as the origin $(0, 0)$.

Substitute $y=0$ into the inequality $y < 2$:

$0 < 2$

This statement is true. This means the solution region is the half-plane that does contain the test point $(0, 0)$. The line $y=2$ divides the plane into an upper half and a lower half (containing the origin). Since the origin side is the solution, we shade the region below the dashed line $y=2$.

Graphical Representation:

The graph shows a dashed horizontal line at $y=2$. The shaded region below this line represents the solution set for the inequality $y < 2$.

Given:

The linear inequality $x + y < 5$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is:

$x + y = 5$

We find two points on this line to plot it.

Step 2: Determine the type of line.

The inequality is $x + y < 5$, which is a strict inequality. This means the points on the line are not part of the solution. Therefore, we draw a dashed line through the points (0, 5) and (5, 0).

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point because it does not lie on the line.

Substitute $(0, 0)$ into the original inequality $x + y < 5$:

$0 + 0 < 5$

$0 < 5$

This statement is true. Thus, the solution region is the half-plane that contains the origin $(0, 0)$. We shade this region.

Graphical Representation:

The graph shows a dashed line passing through (5, 0) and (0, 5). The region below this line (containing the origin) is shaded to represent the solution set of $x + y < 5$.

Given:

The linear inequality $2x + y \ge 6$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is:

$2x + y = 6$

We find two points on this line to plot it.

Step 2: Determine the type of line.

The inequality is $2x + y \ge 6$, which is an inclusive inequality. This means the points on the line are part of the solution. Therefore, we draw a solid line through the points (0, 6) and (3, 0).

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $2x + y \ge 6$:

$2(0) + 0 \ge 6$

$0 \ge 6$

This statement is false. Thus, the solution region is the half-plane that does not contain the origin $(0, 0)$. We shade the region on the opposite side of the line from the origin.

Graphical Representation:

The graph shows a solid line passing through (3, 0) and (0, 6). The region above this line (not containing the origin) is shaded to represent the solution set of $2x + y \ge 6$.

Given:

The linear inequality $3x + 4y \le 12$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is:

$3x + 4y = 12$

We find two points on this line to plot it.

Step 2: Determine the type of line.

The inequality is $3x + 4y \le 12$, which is an inclusive inequality. This means the points on the line are part of the solution. Therefore, we draw a solid line through the points (0, 3) and (4, 0).

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $3x + 4y \le 12$:

$3(0) + 4(0) \le 12$

$0 \le 12$

This statement is true. Thus, the solution region is the half-plane that contains the origin $(0, 0)$. We shade this region.

Graphical Representation:

The graph shows a solid line passing through (4, 0) and (0, 3). The region below this line (containing the origin) is shaded to represent the solution set of $3x + 4y \le 12$.

Given:

The linear inequality $y + 8 \ge 2x$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is:

$y + 8 = 2x$ (or $2x - y = 8$)

We find two points on this line to plot it.

Step 2: Determine the type of line.

The inequality is $y + 8 \ge 2x$, which is an inclusive inequality. This means the points on the line are part of the solution. Therefore, we draw a solid line through the points (0, -8) and (4, 0).

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $y + 8 \ge 2x$:

$0 + 8 \ge 2(0)$

$8 \ge 0$

This statement is true. Thus, the solution region is the half-plane that contains the origin $(0, 0)$. We shade this region.

Graphical Representation:

The graph shows a solid line passing through (4, 0) and (0, -8). The region above this line (containing the origin) is shaded to represent the solution set of $y + 8 \ge 2x$.

Given:

The linear inequality $x - y \le 2$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is:

$x - y = 2$

We find two points on this line to plot it.

Step 2: Determine the type of line.

The inequality is $x - y \le 2$, which is an inclusive inequality. This means the points on the line are part of the solution. Therefore, we draw a solid line through the points (0, -2) and (2, 0).

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $x - y \le 2$:

$0 - 0 \le 2$

$0 \le 2$

This statement is true. Thus, the solution region is the half-plane that contains the origin $(0, 0)$. We shade this region.

Graphical Representation:

The graph shows a solid line passing through (2, 0) and (0, -2). The region above this line (containing the origin) is shaded to represent the solution set of $x - y \le 2$.

Given:

The linear inequality $2x - 3y > 6$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is:

$2x - 3y = 6$

We find two points on this line to plot it.

Step 2: Determine the type of line.

The inequality is $2x - 3y > 6$, which is a strict inequality. This means the points on the line are not part of the solution. Therefore, we draw a dashed line through the points (0, -2) and (3, 0).

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $2x - 3y > 6$:

$2(0) - 3(0) > 6$

$0 > 6$

This statement is false. Thus, the solution region is the half-plane that does not contain the origin $(0, 0)$. We shade the region on the opposite side of the line from the origin (the region below the line).

Graphical Representation:

The graph shows a dashed line passing through (3, 0) and (0, -2). The region below this line is shaded to represent the solution set of $2x - 3y > 6$.

Given:

The linear inequality $-3x + 2y \ge -6$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is:

$-3x + 2y = -6$

We find two points on this line to plot it.

Step 2: Determine the type of line.

The inequality is $-3x + 2y \ge -6$, which is an inclusive inequality. This means the points on the line are part of the solution. Therefore, we draw a solid line through the points (0, -3) and (2, 0).

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $-3x + 2y \ge -6$:

$-3(0) + 2(0) \ge -6$

$0 \ge -6$

This statement is true. Thus, the solution region is the half-plane that contains the origin $(0, 0)$. We shade this region (the region above the line).

Graphical Representation:

The graph shows a solid line passing through (2, 0) and (0, -3). The region above this line is shaded to represent the solution set of $-3x + 2y \ge -6$.

Given:

The linear inequality $3y - 5x < 30$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is:

$3y - 5x = 30$

We find two points on this line to plot it.

Step 2: Determine the type of line.

The inequality is $3y - 5x < 30$, which is a strict inequality. This means the points on the line are not part of the solution. Therefore, we draw a dashed line through the points (0, 10) and (-6, 0).

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $3y - 5x < 30$:

$3(0) - 5(0) < 30$

$0 < 30$

This statement is true. Thus, the solution region is the half-plane that contains the origin $(0, 0)$. We shade this region (the region below the line).

Graphical Representation:

The graph shows a dashed line passing through (-6, 0) and (0, 10). The region below this line is shaded to represent the solution set of $3y - 5x < 30$.

Given:

The linear inequality $y < -2$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is $y = -2$. This equation represents a horizontal line parallel to the x-axis, passing through the point $(0, -2)$.

Step 2: Determine the type of line.

The inequality is $y < -2$, which is a strict inequality. This means the points on the line are not part of the solution. Therefore, we draw a dashed line at $y = -2$.

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $y < -2$:

$0 < -2$

This statement is false. Thus, the solution region is the half-plane that does not contain the origin $(0, 0)$. We shade the region below the line $y = -2$.

Graphical Representation:

The graph shows a dashed horizontal line at $y = -2$. The region below this line is shaded to represent the solution set of $y < -2$.

Given:

The linear inequality $x > -3$.

To Find:

The graphical solution of the given inequality in a two-dimensional plane.

Solution:

To solve the inequality graphically, we follow these steps:

Step 1: Draw the boundary line.

The corresponding equation for the inequality is $x = -3$. This equation represents a vertical line parallel to the y-axis, passing through the point $(-3, 0)$.

Step 2: Determine the type of line.

The inequality is $x > -3$, which is a strict inequality. This means the points on the line are not part of the solution. Therefore, we draw a dashed line at $x = -3$.

Step 3: Choose a test point and determine the solution region.

We select the origin $(0, 0)$ as a test point.

Substitute $(0, 0)$ into the original inequality $x > -3$:

$0 > -3$

This statement is true. Thus, the solution region is the half-plane that contains the origin $(0, 0)$. We shade the region to the right of the line $x = -3$.

Graphical Representation:

The graph shows a dashed vertical line at $x = -3$. The region to the right of this line is shaded to represent the solution set of $x > -3$.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

We will graph the solution for each inequality on the same coordinate plane. The common shaded region will be the solution to the system.

1. For the inequality $x + y \ge 5$:

The corresponding equation is $x + y = 5$. We find two points to draw this line.

Since the inequality is $\ge$ (inclusive), we draw a solid line through (0, 5) and (5, 0).

To find the solution region, we test the origin $(0, 0)$: $0 + 0 \ge 5 \implies 0 \ge 5$, which is false. Therefore, the solution region is the half-plane that does not contain the origin.

2. For the inequality $x - y \le 3$:

The corresponding equation is $x - y = 3$. We find two points to draw this line.

Since the inequality is $\le$ (inclusive), we draw a solid line through (0, -3) and (3, 0).

To find the solution region, we test the origin $(0, 0)$: $0 - 0 \le 3 \implies 0 \le 3$, which is true. Therefore, the solution region is the half-plane that contains the origin.

Graphical Representation:

The solution to the system is the common region shaded by both inequalities. This is the region above the line $x+y=5$ and also above the line $x-y=3$. The intersection of the lines is at $(4,1)$. The final solution is the region that satisfies both conditions simultaneously.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

We will find the common region that satisfies all three inequalities.

1. For the inequality $5x + 4y \le 40$:

The boundary line is $5x + 4y = 40$. We find points (8, 0) and (0, 10) on this line. Since the inequality is $\le$, the line is solid. Testing the origin $(0, 0)$ gives $0 \le 40$, which is true. So, the solution is the region containing the origin.

2. For the inequality $x \ge 2$:

The boundary line is $x = 2$, which is a solid vertical line. Testing the origin $(0, 0)$ gives $0 \ge 2$, which is false. So, the solution is the region to the right of the line $x=2$.

3. For the inequality $y \ge 3$:

The boundary line is $y = 3$, which is a solid horizontal line. Testing the origin $(0, 0)$ gives $0 \ge 3$, which is false. So, the solution is the region above the line $y=3$.

Graphical Representation:

The solution to the system is the triangular region common to all three half-planes. The vertices of this triangular region are found by the intersection of the boundary lines:

• Intersection of $x=2$ and $y=3$: Point is $(2, 3)$.
• Intersection of $x=2$ and $5x+4y=40$: $5(2)+4y=40 \implies 10+4y=40 \implies 4y=30 \implies y=7.5$. Point is $(2, 7.5)$.
• Intersection of $y=3$ and $5x+4y=40$: $5x+4(3)=40 \implies 5x+12=40 \implies 5x=28 \implies x=5.6$. Point is $(5.6, 3)$.

The shaded region bounded by these vertices is the feasible solution.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

1. For the inequality $x \ge 0$ and $y \ge 0$:

These two inequalities together restrict the solution to the first quadrant of the Cartesian plane, including the positive x-axis and y-axis.

2. For the inequality $8x + 3y \le 100$:

The boundary line is $8x + 3y = 100$. We find the intercepts to draw this line.

Since the inequality is $\le$, the line is solid. Testing the origin $(0, 0)$ gives $0 \le 100$, which is true. So, the solution is the region containing the origin.

Graphical Representation:

The solution to the system is the common region, which is the triangular area in the first quadrant bounded by the axes ($x=0, y=0$) and the line $8x + 3y = 100$. The vertices of this feasible region are the origin $(0, 0)$, the x-intercept $(12.5, 0)$, and the y-intercept $(0, 100/3)$. This entire triangular region, including its boundaries, is the solution.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

The inequalities $x > 0$ and $y > 0$ restrict the solution to the interior of the first quadrant. The boundary lines $x=0$ (y-axis) and $y=0$ (x-axis) are not included, so they will be dashed.

For $x + 2y \le 8$, the boundary line is the solid line $x + 2y = 8$, which passes through $(8, 0)$ and $(0, 4)$. The region is on the origin side.

For $2x + y \le 8$, the boundary line is the solid line $2x + y = 8$, which passes through $(4, 0)$ and $(0, 8)$. The region is on the origin side.

Graphical Representation:

The solution is the common region in the first quadrant bounded by the lines $x+2y=8$ and $2x+y=8$. To find the vertex of this region, we solve the two equations:

From $2x+y=8$, we get $y = 8 - 2x$. Substituting into the first equation:

$x + 2(8-2x) = 8$

$x + 16 - 4x = 8$

$-3x = -8 \implies x = 8/3$

$y = 8 - 2(8/3) = 8 - 16/3 = 8/3$

The intersection point is $(8/3, 8/3)$.

The final solution is the region within the polygon formed by vertices $(0,0)$, $(4,0)$, $(8/3, 8/3)$, and $(0,4)$. The boundary segments on the lines $x+2y=8$ and $2x+y=8$ are included, but the boundary segments on the axes are excluded due to the strict inequalities.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

We will graph each inequality on the same Cartesian plane and identify the common region.

1. For the inequality $x \geq 3$:

The boundary line is $x = 3$. This is a vertical line parallel to the y-axis.

Since the inequality is $\geq$ (inclusive), the line is solid.

The region satisfying $x \ge 3$ consists of all points on the line $x=3$ and all points to the right of it.

2. For the inequality $y \geq 2$:

The boundary line is $y = 2$. This is a horizontal line parallel to the x-axis.

Since the inequality is $\geq$ (inclusive), the line is solid.

The region satisfying $y \ge 2$ consists of all points on the line $y=2$ and all points above it.

Graphical Representation:

The solution to the system is the intersection of the two regions. This is the area that is simultaneously to the right of or on the line $x=3$ and above or on the line $y=2$. The corner of this unbounded region is the intersection of the two lines, which is the point $(3, 2)$.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

We will find the common region that satisfies all three inequalities.

1. For the inequality $3x + 2y \le 12$:

The boundary line is $3x + 2y = 12$. The points (4, 0) and (0, 6) are on this line. Since the inequality is $\le$, the line is solid. Testing the origin $(0, 0)$ gives $0 \le 12$, which is true. So, the solution is the region containing the origin (below the line).

2. For the inequality $x \ge 1$:

The boundary line is $x = 1$, which is a solid vertical line. The solution is the region to the right of this line.

3. For the inequality $y \ge 2$:

The boundary line is $y = 2$, which is a solid horizontal line. The solution is the region above this line.

Graphical Representation:

The solution to the system is the triangular region common to all three half-planes. The vertices of this triangular region are found by the intersection of the boundary lines:

• Intersection of $x=1$ and $y=2$: Point is $(1, 2)$.
• Intersection of $x=1$ and $3x+2y=12$: $3(1)+2y=12 \implies 3+2y=12 \implies 2y=9 \implies y=4.5$. Point is $(1, 4.5)$.
• Intersection of $y=2$ and $3x+2y=12$: $3x+2(2)=12 \implies 3x+4=12 \implies 3x=8 \implies x=8/3$. Point is $(\frac{8}{3}, 2)$.

The shaded region bounded by these vertices is the feasible solution.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

1. For the inequality $2x + y \ge 6$:

The boundary line is $2x + y = 6$. The points (3, 0) and (0, 6) are on this line. Since the inequality is $\ge$, the line is solid. Testing the origin $(0, 0)$ gives $0 \ge 6$, which is false. So, the solution is the region not containing the origin (above the line).

2. For the inequality $3x + 4y \le 12$:

The boundary line is $3x + 4y = 12$. The points (4, 0) and (0, 3) are on this line. Since the inequality is $\le$, the line is solid. Testing the origin $(0, 0)$ gives $0 \le 12$, which is true. So, the solution is the region containing the origin (below the line).

Graphical Representation:

The solution is the common region shaded by both inequalities. The point of intersection of the two lines is found by solving the system of equations:

$y = 6 - 2x$

$3x + 4(6 - 2x) = 12 \implies 3x + 24 - 8x = 12 \implies -5x = -12 \ $$ \implies x = \frac{12}{5}$

$y = 6 - 2(\frac{12}{5}) = \frac{30 - 24}{5} = \frac{6}{5}$

The intersection point is $(\frac{12}{5}, \frac{6}{5})$. The shaded feasible region is the area between the two lines.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

1. For the inequality $x + y \ge 4$:

The boundary line is $x + y = 4$. The points (4, 0) and (0, 4) are on this line. Since the inequality is $\ge$, the line is solid. Testing the origin $(0, 0)$ gives $0 \ge 4$, which is false. So, the solution is the region above the line.

2. For the inequality $2x - y < 0$:

The boundary line is $2x - y = 0$, or $y = 2x$. This line passes through the origin $(0,0)$ and the point $(1,2)$. Since the inequality is $<$, the line is dashed. We test a point not on the line, e.g., $(0, 2)$. $2(0) - 2 < 0 \implies -2 < 0$, which is true. So, the solution is the half-plane containing $(0, 2)$ (above the line).

Graphical Representation:

The solution is the common shaded region. The intersection of the boundary lines is at:

$x + (2x) = 4 \implies 3x = 4 \implies x = \frac{4}{3}$

$y = 2(\frac{4}{3}) = \frac{8}{3}$

The intersection point is $(\frac{4}{3}, \frac{8}{3})$. The feasible region is the area above the solid line $x+y=4$ and also above the dashed line $y=2x$.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

1. For the inequality $2x - y > 1$:

The boundary line is $2x - y = 1$. The points $(0, -1)$ and $(1, 1)$ are on this line. Since the inequality is $>$, the line is dashed. Testing the origin $(0, 0)$ gives $0 > 1$, which is false. So, the solution is the region below the line.

2. For the inequality $x - 2y < -1$:

The boundary line is $x - 2y = -1$. The points $(-1, 0)$ and $(1, 1)$ are on this line. Since the inequality is $<$, the line is dashed. Testing the origin $(0, 0)$ gives $0 < -1$, which is false. So, the solution is the region above the line.

Graphical Representation:

The solution is the common region that satisfies both conditions. The intersection of the boundary lines is found by solving the system of equations:

$y = 2x - 1$

$x - 2(2x - 1) = -1 \implies x - 4x + 2 = -1 \implies -3x = -3 \ $$ \implies x = 1$

$y = 2(1) - 1 = 1$

The intersection point is $(1, 1)$. Since both lines are dashed, this point is not included in the solution. The shaded feasible region is the area between the two dashed lines.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

We will graph both inequalities on the same coordinate plane.

1. For the inequality $x + y \le 6$:

The boundary line is $x + y = 6$. The points (6, 0) and (0, 6) are on this line. Since the inequality is $\le$, the line is solid. Testing the origin $(0, 0)$ gives $0 \le 6$, which is true. So, the solution is the region below or on the line $x+y=6$.

2. For the inequality $x + y \ge 4$:

The boundary line is $x + y = 4$. The points (4, 0) and (0, 4) are on this line. Since the inequality is $\ge$, the line is solid. Testing the origin $(0, 0)$ gives $0 \ge 4$, which is false. So, the solution is the region above or on the line $x+y=4$.

Graphical Representation:

The lines $x+y=6$ and $x+y=4$ are parallel to each other. The solution to the system is the region that lies between these two parallel lines, inclusive of the lines themselves. The shaded region represents all the points $(x,y)$ that satisfy both inequalities.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

1. For the inequality $2x + y \ge 8$:

The boundary line is $2x + y = 8$. The points (4, 0) and (0, 8) are on this line. It's a solid line. Testing the origin $(0, 0)$ gives $0 \ge 8$, which is false. So, the solution is the region above the line.

2. For the inequality $x + 2y \ge 10$:

The boundary line is $x + 2y = 10$. The points (10, 0) and (0, 5) are on this line. It's a solid line. Testing the origin $(0, 0)$ gives $0 \ge 10$, which is false. So, the solution is the region above the line.

Graphical Representation:

The solution is the common region above both lines. The intersection of the boundary lines is at:

$y = 8 - 2x$

$x + 2(8 - 2x) = 10 \implies x + 16 - 4x = 10 \implies -3x = -6 \ $$ \implies x = 2$

$y = 8 - 2(2) = 4$

The intersection point is $(2, 4)$. The feasible region is the area above both solid lines, starting from the corner point $(2, 4)$.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

1. For the inequality $x + y \le 9$:

The boundary is a solid line $x + y = 9$. The solution region is below this line (origin side).

2. For the inequality $y > x$:

The boundary is a dashed line $y = x$. The solution region is above this line.

3. For the inequality $x \ge 0$:

The boundary is the solid line $x = 0$ (the y-axis). The solution is the region to the right of the y-axis.

Graphical Representation:

The solution is the triangular region in the first and second quadrants that is below the solid line $x+y=9$, above the dashed line $y=x$, and to the right of the solid line $x=0$. The vertices of this region are at the intersections of the boundary lines: $(0,0)$, $(0,9)$, and $(4.5, 4.5)$. Due to the strict inequality $y > x$, the points on the line $y=x$ (including vertices $(0,0)$ and $(4.5, 4.5)$) are not included in the solution set.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

1. For the inequality $5x + 4y \le 20$:

The boundary is a solid line $5x + 4y = 20$. The solution region is below this line (origin side).

2. For the inequality $x \ge 1$:

The boundary is a solid vertical line $x = 1$. The solution is the region to the right.

3. For the inequality $y \ge 2$:

The boundary is a solid horizontal line $y = 2$. The solution is the region above.

Graphical Representation:

The solution is the triangular region bounded by the three solid lines. The vertices are the intersection points:

• Intersection of $x=1$ and $y=2$: $(1, 2)$.
• Intersection of $x=1$ and $5x+4y=20$: $5(1)+4y=20 \implies 4y=15 \implies y=3.75$. Point: $(1, 3.75)$.
• Intersection of $y=2$ and $5x+4y=20$: $5x+4(2)=20 \implies 5x=12 \implies x=2.4$. Point: $(2.4, 2)$.

The shaded region bounded by these vertices is the feasible solution.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

The inequalities $x \ge 0$ and $y \ge 0$ restrict the solution to the first quadrant.

1. For the inequality $3x + 4y \le 60$:

The boundary is a solid line $3x + 4y = 60$. The solution region is below this line.

2. For the inequality $x + 3y \le 30$:

The boundary is a solid line $x + 3y = 30$. The solution region is below this line.

Graphical Representation:

The solution is the common region in the first quadrant bounded by the lines. The vertices of this feasible region are:

• Intersection of axes: $(0, 0)$.
• Y-intercept of $x+3y=30$: $(0, 10)$.
• X-intercept of $3x+4y=60$: $(20, 0)$.
• Intersection of $3x+4y=60$ and $x+3y=30$: Solving gives $x=12, y=6$. Point: $(12, 6)$.

The shaded feasible region is the quadrilateral formed by the vertices $(0,0)$, $(20,0)$, $(12,6)$, and $(0,10)$.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

We graph the boundary line for each inequality and determine the feasible region.

1. For $2x + y \ge 4$: Boundary is the solid line $2x + y = 4$. The region is above this line (origin side is false).

2. For $x + y \le 3$: Boundary is the solid line $x + y = 3$. The region is below this line (origin side is true).

3. For $2x - 3y \le 6$: Boundary is the solid line $2x - 3y = 6$. The region is above this line (origin side is true).

Graphical Representation:

The solution is the triangular region formed by the intersection of these three half-planes. The vertices of this feasible region are the intersection points of the boundary lines:

• Intersection of $2x + y = 4$ and $x + y = 3$: Solving gives $x=1, y=2$. Point: $(1, 2)$.
• Intersection of $x + y = 3$ and $2x - 3y = 6$: Solving gives $x=3, y=0$. Point: $(3, 0)$.
• Intersection of $2x + y = 4$ and $2x - 3y = 6$: Solving gives $x=\frac{9}{4}, y=-\frac{1}{2}$. Point: $(\frac{9}{4}, -\frac{1}{2})$.

The shaded triangular region with these vertices is the solution set.

Given:

The following system of linear inequalities:

To Find:

The graphical solution for the given system of inequalities.

Solution:

We find the common region that satisfies all four inequalities. All boundary lines are solid.

1. For $x - 2y \le 3$: The region is above the line $x-2y=3$.

2. For $3x + 4y \ge 12$: The region is above the line $3x+4y=12$.

3. For $x \ge 0$: The region is to the right of the y-axis.

4. For $y \ge 1$: The region is above the line $y=1$.

Graphical Representation:

The solution is the common region. We find the vertices of this feasible region by finding the intersection of the boundary lines:

• Intersection of $y=1$ and $3x+4y=12$: $3x+4(1)=12 \implies 3x=8 \implies x=8/3$. Point: $(\frac{8}{3}, 1)$.
• Intersection of $y=1$ and $x-2y=3$: $x-2(1)=3 \implies x=5$. Point: $(5, 1)$.
• Intersection of $x=0$ and $3x+4y=12$: $3(0)+4y=12 \implies y=3$. Point: $(0, 3)$.

The intersection of $x-2y=3$ and $3x+4y=12$ is $(\frac{18}{5}, \frac{3}{10})$, which is not in the feasible region as $y < 1$. The feasible region is unbounded. The shaded region is defined by the vertices $(0,3)$, $(\frac{8}{3}, 1)$, and $(5,1)$, and extends upwards and to the right.

Given:

The following system of linear inequalities:

(The conditions $x \ge 0, y \ge 0$ are also given, but are redundant given $x \ge 3$ and $y \ge 2x$.)

To Find:

The graphical solution for the given system of inequalities.

Solution:

All boundary lines are solid. We find the common region.

1. For $4x + 3y \le 60$: The region is below the line $4x+3y=60$.

2. For $y \ge 2x$: The region is above the line $y=2x$.

3. For $x \ge 3$: The region is to the right of the line $x=3$.

Graphical Representation:

The solution is the triangular region bounded by the three solid lines. The vertices are the intersection points:

• Intersection of $x=3$ and $y=2x$: $y=2(3)=6$. Point: $(3, 6)$.
• Intersection of $x=3$ and $4x+3y=60$: $4(3)+3y=60 \implies 12+3y=60 \implies 3y=48 \implies y=16$. Point: $(3, 16)$.
• Intersection of $y=2x$ and $4x+3y=60$: $4x+3(2x)=60 \implies 10x=60 \implies x=6$. $y=2(6)=12$. Point: $(6, 12)$.

The shaded triangular region with these vertices is the solution set.

Given:

The system of linear inequalities:

$3x + 2y \le 150$, $x + 4y \le 80$, $x \le 15$, $x \ge 0$, $y \ge 0$.

To Find:

The graphical solution for the given system.

Solution:

The inequalities $x \ge 0, y \ge 0$ restrict the solution to the first quadrant. All boundary lines are solid.

1. For $3x + 2y \le 150$: Region is below the line.

2. For $x + 4y \le 80$: Region is below the line.

3. For $x \le 15$: Region is to the left of the line $x=15$.

Graphical Representation:

The solution is the common region in the first quadrant. The vertices of this feasible region are:

• Intersection of axes: $(0, 0)$.
• Intersection of $y=0$ and $x=15$: $(15, 0)$.
• Intersection of $x=15$ and $x+4y=80$: $15+4y=80 \implies 4y=65 \implies y=16.25$. Point: $(15, 16.25)$.
• Intersection of $x=0$ and $x+4y=80$: $y=20$. Point: $(0, 20)$.

Note that the line $3x+2y=150$ is redundant as it lies outside the region defined by the other constraints. For example, at $(15, 16.25)$, $3(15)+2(16.25)=45+32.5=77.5 \le 150$. At $(0,20)$, $3(0)+2(20)=40 \le 150$. The feasible region is a quadrilateral defined by the vertices $(0,0)$, $(15,0)$, $(15, 16.25)$, and $(0,20)$.

Given:

The system of inequalities:

$x + 2y \le 10$, $x + y \ge 1$, $x - y \le 0$, $x \ge 0$, $y \ge 0$.

To Find:

The graphical solution for the given system.

Solution:

We are in the first quadrant ($x \ge 0, y \ge 0$). All boundary lines are solid.

1. For $x + 2y \le 10$: Region is below the line.

2. For $x + y \ge 1$: Region is above the line.

3. For $x - y \le 0$ (or $y \ge x$): Region is above the line $y=x$.

Graphical Representation:

The solution is the common region. The vertices of this feasible region are:

• Intersection of $x+y=1$ and $x=0$: $(0, 1)$.
• Intersection of $x+2y=10$ and $x=0$: $(0, 5)$.
• Intersection of $y=x$ and $x+2y=10$: $x+2x=10 \implies 3x=10 \implies x=10/3$. Point: $(\frac{10}{3}, \frac{10}{3})$.
• Intersection of $y=x$ and $x+y=1$: $x+x=1 \implies 2x=1 \implies x=1/2$. Point: $(\frac{1}{2}, \frac{1}{2})$.

The shaded feasible region is the quadrilateral with vertices $(0,1)$, $(0,5)$, $(\frac{10}{3}, \frac{10}{3})$, and $(\frac{1}{2}, \frac{1}{2})$.

Given:

The inequality is $– 8 \le 5x – 3 < 7$.

Solution:

We need to find the values of $x$ that satisfy the compound inequality.

$– 8 \le 5x – 3 < 7$

$– 5 \le 5x < 10$

$– 1 \le x < 2$

Thus, the solution set consists of all real numbers $x$ such that $x$ is greater than or equal to $-1$ and less than $2$.

In interval notation, the solution set is $[-1, 2)$.

Alternate Solution:

The given inequality $– 8 \le 5x – 3 < 7$ can be treated as two separate inequalities which must hold simultaneously:

1) $– 8 \le 5x – 3$

2) $5x – 3 < 7$

Solving the first inequality:

$– 8 \le 5x – 3$

$– 5 \le 5x$

$– 1 \le x$ or $x \ge – 1$

Solving the second inequality:

$5x – 3 < 7$

$5x < 10$

$x < 2$

Combining the results from both inequalities, we need $x$ such that $x \ge – 1$ and $x < 2$.

Therefore, $– 1 \le x < 2$.

In interval notation, the solution set is $[-1, 2)$.

We are given the inequality:

$-5\leq\frac{5\;-\;3x}{2} \leq8$

To eliminate the denominator, multiply all parts of the inequality by $2$:

Simplifying this gives:

Now, subtract $5$ from all parts of the inequality (ii):

Simplifying this gives:

Finally, divide all parts of the inequality (iv) by $-3$. Remember to reverse the direction of the inequality signs when dividing by a negative number:

Simplifying this gives:

We can rewrite the inequality (vi) in the standard form, with the smaller value on the left:

$-\frac{11}{3} \leq x \leq 5$

This means that $x$ is greater than or equal to $-\frac{11}{3}$ and less than or equal to $5$. In interval notation, the solution is:

$\left[-\frac{11}{3}, 5\right]$

Given:

The system of linear inequalities:

To Find:

The solution set for the given system of inequalities and to represent the solution on the number line.

Solution:

We solve each inequality individually.

First, we solve the inequality (1):

$3x - 7 < 5 + x$

Subtract $x$ from both sides:

$3x - x - 7 < 5$

$2x - 7 < 5$

Add 7 to both sides:

$2x < 5 + 7$

$2x < 12$

Divide by 2:

Next, we solve the inequality (2):

$11 - 5x \le 1$

Subtract 11 from both sides:

$-5x \le 1 - 11$

$-5x \le -10$

Divide both sides by -5. When we divide an inequality by a negative number, the inequality sign is reversed.

$\frac{-5x}{-5} \ge \frac{-10}{-5}$

The solution of the system is the intersection of the solutions from (i) and (ii).

We need all values of $x$ that are less than 6 and greater than or equal to 2.

Combining these two conditions, we get:

$2 \le x < 6$

In interval notation, the solution set is $[2, 6)$.

Graph of the solution on the number line:

To represent the solution $2 \le x < 6$ on a number line:

1. We draw a closed circle at the point 2 to indicate that 2 is included in the solution set.

2. We draw an open circle at the point 6 to indicate that 6 is not included in the solution set.

3. We shade the portion of the number line between the closed circle at 2 and the open circle at 6.

We are given the range of temperature in Celsius:

The conversion formula between Celsius (C) and Fahrenheit (F) is given by:

Substitute the expression for C from (ii) into the inequality (i):

To isolate the term $(F - 32)$, multiply all parts of the inequality (iii) by $\frac{9}{5}$:

Simplify the multiplication:

Now, add $32$ to all parts of the inequality (v) to isolate F:

Simplify the addition:

Thus, the range of temperature in degree Fahrenheit is between $86^\circ$ F and $95^\circ$ F, exclusive of the endpoints.

The range is $86^\circ$ F $< F < 95^\circ$ F.

In interval notation, the range is $(86, 95)$.

Let $x$ be the number of litres of 30% acid solution that must be added.

Initial solution: 600 litres of 12% acid.

Amount of acid in the initial solution = $12\%$ of $600$ litres

Amount of acid = $0.12 \times 600 = 72$ litres.

Added solution: $x$ litres of 30% acid.

Amount of acid in the added solution = $30\%$ of $x$ litres

Amount of acid = $0.30 \times x = 0.3x$ litres.

Resulting mixture:

Total volume of the mixture = (Initial volume) + (Added volume)

Total amount of acid in the resulting mixture = (Acid in initial solution) + (Acid in added solution)

The acid content in the resulting mixture is given by:

Percentage of acid = $\frac{\text{Total amount of acid}}{\text{Total volume}} \times 100$

Percentage of acid = $\frac{72 + 0.3x}{600 + x} \times 100$

According to the problem, the acid content in the resulting mixture must be more than 15% but less than 18%. This gives us a system of two inequalities:

Inequality 1: Acid content > 15%

Inequality 2: Acid content < 18%

Since $x$ represents the amount of solution added, $x \geq 0$. The total volume $600+x$ is always positive. So we can multiply both sides of the inequalities by $600+x$ without changing the direction of the inequality sign.

Solving Inequality (A):

Subtract $0.15x$ from both sides:

Subtract $72$ from both sides:

Divide by $0.15$:

Solving Inequality (B):

Subtract $0.18x$ from both sides:

Subtract $72$ from both sides:

Divide by $0.12$:

For the acid content to be between 15% and 18%, the value of $x$ must satisfy both conditions (C) and (D).

Combining $x > 120$ and $x < 300$, we get:

$120 < x < 300$

Therefore, the manufacturer must add more than 120 litres but less than 300 litres of the 30% acid solution.

The range is $(120, 300)$ litres.

We are given the inequality:

$2 \leq 3x - 4 \leq 5$

To isolate the term with $x$, add $4$ to all parts of the inequality:

Simplifying this gives:

Now, divide all parts of the inequality (ii) by $3$ to solve for $x$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

The solution set consists of all real numbers $x$ that are greater than or equal to $2$ and less than or equal to $3$.

In interval notation, the solution is:

$\mathbf{[2, 3]}$

We are given the inequality:

$6 \leq -3(2x - 4) < 12$

First, distribute the $-3$ on the right side of the first part of the inequality:

So the inequality becomes:

To isolate the term with $x$, subtract $12$ from all parts of the inequality (i):

Simplifying this gives:

Now, divide all parts of the inequality (iii) by $-6$. Remember to reverse the direction of the inequality signs when dividing by a negative number:

Simplifying this gives:

We can rewrite the inequality (v) in the standard form, with the smaller value on the left:

$\mathbf{0 < x \leq 1}$

The solution set consists of all real numbers $x$ that are strictly greater than $0$ and less than or equal to $1$.

In interval notation, the solution is:

$\mathbf{(0, 1]}$

We are given the inequality:

$-3 \leq 4 - \frac{7x}{2} \leq 18$

To isolate the term with $x$, subtract $4$ from all parts of the inequality:

Simplifying this gives:

To eliminate the denominator, multiply all parts of the inequality (i) by $2$. Since $2$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

Now, divide all parts of the inequality (ii) by $-7$. Remember to reverse the direction of the inequality signs when dividing by a negative number:

Simplifying this gives:

We can rewrite the inequality (iii) in the standard form, with the smaller value on the left:

$\mathbf{-4 \leq x \leq 2}$

The solution set consists of all real numbers $x$ that are greater than or equal to $-4$ and less than or equal to $2$.

In interval notation, the solution is:

$\mathbf{[-4, 2]}$

We are given the inequality:

$-15 < \frac{3(x - 2)}{5} \leq 0$

To eliminate the denominator, multiply all parts of the inequality by $5$. Since $5$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

Now, distribute the $3$ on the right side of the left part of inequality (ii):

To isolate the term with $x$, add $6$ to all parts of the inequality (iii):

Simplifying this gives:

Finally, divide all parts of the inequality (iv) by $3$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

The solution set consists of all real numbers $x$ that are strictly greater than $-23$ and less than or equal to $2$.

In interval notation, the solution is:

$\mathbf{(-23, 2]}$

We are given the inequality:

$-12 < 4 - \frac{3x}{-5} \leq 2$

First, simplify the term $-\frac{3x}{-5} = -(-\frac{3x}{5}) = \frac{3x}{5}$. The inequality becomes:

To isolate the term with $x$, subtract $4$ from all parts of the inequality (i):

Simplifying this gives:

To eliminate the denominator, multiply all parts of the inequality (iii) by $5$. Since $5$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

Finally, divide all parts of the inequality (v) by $3$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

The solution set consists of all real numbers $x$ that are strictly greater than $-\frac{80}{3}$ and less than or equal to $-\frac{10}{3}$.

In interval notation, the solution is:

$\mathbf{\left(-\frac{80}{3}, -\frac{10}{3}\right]}$

We are given the inequality:

$7 \leq \frac{3x + 11}{2} \leq 11$

To eliminate the denominator, multiply all parts of the inequality by $2$. Since $2$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

To isolate the term with $x$, subtract $11$ from all parts of the inequality (ii):

Simplifying this gives:

Finally, divide all parts of the inequality (iv) by $3$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

The solution set consists of all real numbers $x$ that are greater than or equal to $1$ and less than or equal to $\frac{11}{3}$.

In interval notation, the solution is:

$\mathbf{\left[1, \frac{11}{3}\right]}$

Given:

The system of two linear inequalities:

To Find:

The solution set for the given system of inequalities and to represent it on a number line.

Solution:

First, we solve inequality (1):

$5x + 1 > -24$

Subtract 1 from both sides:

$5x > -24 - 1$

$5x > -25$

Divide by 5:

$x > -5$

Next, we solve inequality (2):

$5x - 1 < 24$

Add 1 to both sides:

$5x < 24 + 1$

$5x < 25$

Divide by 5:

$x < 5$

The solution to the system is the set of all $x$ that satisfy both conditions: $x > -5$ and $x < 5$.

This can be written as a single compound inequality:

$-5 < x < 5$

In interval notation, the solution is $(-5, 5)$.

Graphical Representation on Number Line:

To represent the solution $-5 < x < 5$ on the number line, we draw an open circle at -5 and an open circle at 5 to show that these endpoints are not included. Then, we shade the line segment between these two circles.

Given:

The system of two linear inequalities:

To Find:

The solution set for the given system and its representation on a number line.

Solution:

First, we solve inequality (1):

$2(x - 1) < x + 5$

$2x - 2 < x + 5$

$2x - x < 5 + 2$

$x < 7$

Next, we solve inequality (2):

$3(x + 2) > 2 - x$

$3x + 6 > 2 - x$

$3x + x > 2 - 6$

$4x > -4$

$x > -1$

The solution is the set of all $x$ satisfying both $x < 7$ and $x > -1$.

This can be written as:

$-1 < x < 7$

In interval notation, the solution is $(-1, 7)$.

Graphical Representation on Number Line:

We represent the solution $-1 < x < 7$ by drawing an open circle at -1 and an open circle at 7, and shading the line segment between them.

Given:

The system of two linear inequalities:

To Find:

The solution set for the given system and its representation on a number line.

Solution:

First, we solve inequality (1):

$3x - 7 > 2(x - 6)$

$3x - 7 > 2x - 12$

$3x - 2x > -12 + 7$

$x > -5$

Next, we solve inequality (2):

$6 - x > 11 - 2x$

$-x + 2x > 11 - 6$

$x > 5$

The solution is the set of all $x$ satisfying both $x > -5$ and $x > 5$. For a number to be greater than -5 and also greater than 5, it must simply be greater than 5. The intersection of these two sets is the set of numbers greater than 5.

Therefore, the solution is:

$x > 5$

In interval notation, the solution is $(5, \infty)$.

Graphical Representation on Number Line:

We represent the solution $x > 5$ by drawing an open circle at 5 and shading the line to the right of 5, with an arrow indicating it continues indefinitely.

Given:

The system of two linear inequalities:

To Find:

The solution set for the given system and its representation on a number line.

Solution:

First, we solve inequality (1):

$5(2x - 7) - 3(2x + 3) \le 0$

$10x - 35 - 6x - 9 \le 0$

$4x - 44 \le 0$

$4x \le 44$

$x \le 11$

Next, we solve inequality (2):

$2x + 19 \le 6x + 47$

$19 - 47 \le 6x - 2x$

$-28 \le 4x$

$-7 \le x$

The solution is the set of all $x$ satisfying both $x \le 11$ and $x \ge -7$.

This can be written as:

$-7 \le x \le 11$

In interval notation, the solution is $[-7, 11]$.

Graphical Representation on Number Line:

We represent the solution $-7 \le x \le 11$ by drawing a closed circle at -7 and a closed circle at 11, and shading the line segment between them.

We are given the range of temperature in Fahrenheit:

The conversion formula between Fahrenheit (F) and Celsius (C) is given by:

We need to find the range of temperature in Celsius. Substitute the expression for F from (ii) into the inequality (i):

To isolate the term with C, subtract $32$ from all parts of the inequality (iii):

Simplifying this gives:

Now, multiply all parts of the inequality (v) by $\frac{5}{9}$ to isolate C. Since $\frac{5}{9}$ is positive, the direction of the inequality signs remains unchanged:

Simplify the multiplication:

Thus, the range of temperature in degree Celsius is between $20^\circ$ C and $25^\circ$ C, exclusive of the endpoints.

The range is $\mathbf{20^\circ \text{C} < C < 25^\circ \text{C}}$.

In interval notation, the range is $\mathbf{(20, 25)}$.

Let $x$ be the number of litres of 2% boric acid solution that must be added.

Since we are adding a volume of solution, $x \geq 0$.

Initial solution: 640 litres of 8% boric acid.

Amount of boric acid in the initial solution = $8\%$ of $640$ litres

Amount of boric acid = $0.08 \times 640 = 51.2$ litres.

Added solution: $x$ litres of 2% boric acid.

Amount of boric acid in the added solution = $2\%$ of $x$ litres

Amount of boric acid = $0.02 \times x = 0.02x$ litres.

Resulting mixture:

Total volume of the mixture = (Initial volume) + (Added volume)

Total amount of boric acid in the resulting mixture = (Acid in initial solution) + (Acid in added solution)

The acid content in the resulting mixture is given by:

Percentage of acid = $\frac{\text{Total amount of acid}}{\text{Total volume}} \times 100$

Percentage of acid = $\frac{51.2 + 0.02x}{640 + x} \times 100$

According to the problem, the acid content in the resulting mixture must be more than 4% but less than 6%. This gives us a compound inequality:

Divide the inequality (iii) by 100:

Since $x \geq 0$, the total volume $640 + x$ is always positive. We can multiply all parts of the inequality (iv) by $(640 + x)$ without changing the direction of the inequality signs.

This compound inequality can be split into two separate inequalities:

Inequality 1: $0.04(640 + x) < 51.2 + 0.02x$

Distribute the 0.04:

Subtract $0.02x$ from both sides:

Subtract $25.6$ from both sides:

Divide by $0.02$ (positive number):

Inequality 2: $51.2 + 0.02x < 0.06(640 + x)$

Distribute the 0.06:

Subtract $0.02x$ from both sides:

Subtract $38.4$ from both sides:

Divide by $0.04$ (positive number):

To satisfy the conditions of the problem, $x$ must satisfy both inequality (A) and inequality (B).

From (A), $x < 1280$.

From (B), $x > 320$.

Combining these two conditions, we get:

$\mathbf{320 < x < 1280}$

Therefore, the manufacturer must add more than 320 litres but less than 1280 litres of the 2% boric acid solution.

The range of the amount of 2% solution to be added is $\mathbf{(320, 1280)}$ litres.

Let $x$ be the number of litres of water added to the 1125 litres of 45% acid solution.

Since $x$ is a volume of water added, $x \geq 0$.

Initial solution: 1125 litres of 45% acid.

Amount of acid in the initial solution = $45\%$ of $1125$ litres

$0.45 \times 1125 = 506.25$.

So, the initial amount of acid is $506.25$ litres.

Added solution: $x$ litres of water (0% acid).

Amount of acid in the added water = $0\%$ of $x$ litres = $0$ litres.

Resulting mixture:

Total volume of the mixture = (Initial volume) + (Volume of water added)

Total amount of acid in the resulting mixture = (Acid in initial solution) + (Acid in added water)

The acid content in the resulting mixture is given by:

Percentage of acid = $\frac{\text{Total amount of acid}}{\text{Total volume}} \times 100$

Percentage of acid = $\frac{506.25}{1125 + x} \times 100$

According to the problem, the acid content in the resulting mixture must be more than 25% but less than 30%. This gives us the compound inequality:

Divide the inequality (iv) by 100:

Since $x \geq 0$, the total volume $1125 + x$ is always positive. We can multiply all parts of the inequality (v) by $(1125 + x)$ without changing the direction of the inequality signs.

This compound inequality can be split into two separate inequalities:

Inequality 1: $0.25(1125 + x) < 506.25$

Distribute the 0.25:

Calculate $0.25 \times 1125 = \frac{1}{4} \times 1125 = 281.25$:

Subtract $281.25$ from both sides:

Divide by $0.25$ (positive number):

Inequality 2: $506.25 < 0.30(1125 + x)$

Distribute the 0.30:

Calculate $0.30 \times 1125 = 337.5$:

Subtract $337.5$ from both sides:

Divide by $0.30$ (positive number):

To satisfy the conditions of the problem, $x$ must satisfy both inequality (A) and inequality (B).

From (A), $x < 900$.

From (B), $x > 562.5$.

Combining these two conditions, we get:

$\mathbf{562.5 < x < 900}$

Therefore, the number of litres of water that must be added is more than 562.5 litres but less than 900 litres.

The range of water to be added is $\mathbf{(562.5, 900)}$ litres.

The formula for IQ is given by:

where MA is the mental age and CA is the chronological age.

We are given that the children are 12 years old, so their chronological age is $\text{CA} = 12$ years.

We are also given the range of IQ for this group of children:

Substitute the value of CA = 12 into the IQ formula (i):

Now, substitute the expression for IQ from (iii) into the inequality (ii):

Simplify the term with MA:

So the inequality (iv) becomes:

To isolate MA, first multiply all parts of the inequality (v) by $3$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

Now, divide all parts of the inequality (vii) by $25$. Since $25$ is positive, the direction of the inequality signs remains unchanged:

Simplify the fractions:

So the inequality (viii) becomes:

Therefore, the range of mental age (MA) for this group of 12-year-old children is between 9.6 years and 16.8 years, inclusive of the endpoints.

The range of their mental age is $\mathbf{9.6 \leq \text{MA} \leq 16.8}$ years.

In interval notation, the range is $\mathbf{[9.6, 16.8]}$.