Chapter 5 Linear Inequalities

Given:

The inequality is $30x < 200$.

Solution:

We need to solve the given inequality for $x$.

Divide both sides of the inequality by 30:

Simplifying the fraction on the right side:

This simplifies to:

The value of $\frac{20}{3}$ is approximately $6.66...$

Case (i): $x$ is a natural number.

Natural numbers are $\{1, 2, 3, 4, 5, 6, 7, 8, ...\}$.

From inequality (i), we are looking for natural numbers $x$ such that $x < \frac{20}{3}$.

Since $\frac{20}{3} \approx 6.66...$, the natural numbers that are less than $6.66...$ are $1, 2, 3, 4, 5, 6$.

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

Case (ii): $x$ is an integer.

Integers are $\{..., -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, ...\}$.

From inequality (i), we are looking for integers $x$ such that $x < \frac{20}{3}$.

Since $\frac{20}{3} \approx 6.66...$, the integers that are less than $6.66...$ are $6, 5, 4, 3, 2, 1, 0, -1, -2, ...$ and so on.

Therefore, the solution set when $x$ is an integer is $\{..., 4, 5, 6\}$.

Given:

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

Solution:

We need to solve the given inequality for $x$.

Subtract $3x$ from both sides:

Simplify both sides:

Add 3 to both sides:

Simplify both sides:

Divide both sides by 2:

This simplifies to:

Case (i): $x$ is an integer.

From inequality (i), we are looking for integers $x$ such that $x < 2$.

The integers less than 2 are $..., -3, -2, -1, 0, 1$.

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

Case (ii): $x$ is a real number.

From inequality (i), we are looking for real numbers $x$ such that $x < 2$.

The set of real numbers less than 2 can be represented in interval notation as $(-\infty, 2)$.

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

Given:

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

Solution:

We need to solve the given inequality for $x$.

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

Simplify both sides:

Subtract 7 from both sides of the inequality:

Simplify both sides:

Divide both sides of the inequality by 2:

Simplify both sides:

Inequality (i) can also be written as $x > -2$.

Since the domain of $x$ 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 is $(-2, \infty)$.

Therefore, the solution is $x > -2$, where $x$ is a real number.

Given:

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

Solution:

We need to solve the given inequality for $x$.

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, to eliminate the denominators:

Distribute the 6 on the right side:

Simplify the terms:

Apply the distributive property on the left side:

Add $4x$ to both sides of the inequality:

Simplify both sides:

Add 30 to both sides of the inequality:

Simplify both sides:

Divide both sides of the inequality by 5:

Simplify both sides:

Inequality (i) can also be written as $x \geq 8$.

Since the domain of $x$ is not specified, we assume $x$ is a real number.

The solution set consists of all real numbers greater than or equal to 8.

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

Therefore, the solution is $x \geq 8$, where $x$ is a real number.

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

To Solve: The given inequality for $x$.

To Show: The graph of the solutions on a number line.

Solution:

We have the inequality:

$7x + 3 < 5x + 9$

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

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

$2x + 3 < 9$

Subtracting $3$ from both sides of the inequality:

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

$2x < 6$

Dividing both sides by $2$ (since $2$ is a positive number, the inequality sign does not change):

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

$x < 3$

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

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

Graph of the solutions on number line:

To show the solution $x < 3$ on a number line, we represent all real numbers less than $3$.

1. Draw a number line.

2. Locate the number $3$ on the number line.

3. Since the inequality is strict ($<$, not $\leq$), the number $3$ itself is not included in the solution set. This is indicated by an open circle (or a hollow circle) at the point representing $3$ on the number line.

4. All numbers less than $3$ are part of the solution. This is indicated by drawing a thick line or shading the part of the number line to the left of $3$. An arrow should be drawn at the left end of the shaded line to show that the solution extends infinitely in the negative direction.

Given:

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

To Solve:

The given inequality for $x$.

To Show:

The graph of the solutions on a number line.

Solution:

We have the inequality:

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

First, simplify the right-hand side of the inequality by combining the terms:

$\frac{x + 1}{4} - 1 = \frac{x + 1}{4} - \frac{4}{4} = \frac{x + 1 - 4}{4} = \frac{x - 3}{4}$

So the inequality becomes:

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

To clear the fractions, we multiply both sides of the inequality by the Least Common Multiple (LCM) of the denominators $2$ and $4$, which is $4$. Since $4$ is a positive number, the inequality sign does not change.

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

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

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

$6x - 8 \geq x - 3$

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

Subtract $x$ from both sides of the inequality:

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

$5x - 8 \geq -3$

Add $8$ to both sides of the inequality:

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

$5x \geq 5$

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

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

$x \geq 1$

Thus, 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 solutions on number line:

To show the solution $x \geq 1$ on a number line, we represent all real numbers greater than or equal to $1$.

1. Draw a number line.

2. Locate the number $1$ on the number line.

3. Since the inequality is $\geq$, the number $1$ itself is included in the solution set. This is indicated by a closed circle (or a filled circle) at the point representing $1$ on the number line.

4. All numbers greater than $1$ are part of the solution. This is indicated by drawing a thick line or shading the part of the number line to the right of $1$. An arrow should be drawn at the right end of the shaded line to show that the solution extends infinitely in the positive direction.

Given:

Marks in the first terminal examination = $62$

Marks in the second terminal examination = $48$

Required average marks in three examinations $\geq 60$

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 total marks obtained in the three examinations (first terminal, second terminal, and annual) will be $62 + 48 + x$.

The average marks obtained in the three examinations is given by:

Average Marks $= \frac{\text{Sum of marks in three exams}}{\text{Number of exams}}$

Average Marks $= \frac{62 + 48 + x}{3}$

Average Marks $= \frac{110 + x}{3}$

According to the problem, the average marks must be at least $60$. This can be written as an inequality:

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

To solve for $x$, multiply both sides of the inequality by $3$:

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

$110 + x \geq 180$

Subtract $110$ from both sides of the inequality:

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

$x \geq 70$

This inequality tells us that the marks obtained in the annual examination must be greater than or equal to $70$.

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

The minimum marks required is 70.

Given:

Pairs of consecutive odd natural numbers.

Both numbers are larger than $10$.

Their sum is less than $40$.

To Find:

All such pairs of consecutive odd natural numbers.

Solution:

Let the first odd natural number 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 are larger than $10$.

So, we have two inequalities:

$x > 10$

and

$x + 2 > 10$

From the second inequality, $x + 2 > 10$, we subtract $2$ from both sides:

$x + 2 - 2 > 10 - 2$

$x > 8$

Combining the two conditions ($x > 10$ and $x > 8$), we find that $x$ must be greater than $10$.

Since $x$ is an odd natural number, the possible values for $x$ are $11, 13, 15, 17, 19, 21, \dots$

According to the second condition, the sum of the two numbers is less than $40$.

Sum $= x + (x + 2) = 2x + 2$

So, we have the inequality:

$2x + 2 < 40$

To solve this inequality, first subtract $2$ from both sides:

$2x + 2 - 2 < 40 - 2$

$2x < 38$

Now, divide both sides by $2$ (a positive number, so the inequality direction remains unchanged):

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

$x < 19$

We need to find odd natural numbers $x$ that satisfy both conditions:

$x > 10$ and $x < 19$

The odd natural numbers greater than $10$ and less than $19$ are $11, 13, 15, 17$.

Now, we find the corresponding consecutive odd natural number, $x + 2$, for each valid value of $x$ and list the pairs:

If $x = 11$, the pair is $(11, 11+2) = (11, 13)$. Both are > 10 ($11 > 10, 13 > 10$) and their sum $11 + 13 = 24 < 40$. This pair is valid.

If $x = 13$, the pair is $(13, 13+2) = (13, 15)$. Both are > 10 ($13 > 10, 15 > 10$) and their sum $13 + 15 = 28 < 40$. This pair is valid.

If $x = 15$, the pair is $(15, 15+2) = (15, 17)$. Both are > 10 ($15 > 10, 17 > 10$) and their sum $15 + 17 = 32 < 40$. This pair is valid.

If $x = 17$, the pair is $(17, 17+2) = (17, 19)$. Both are > 10 ($17 > 10, 19 > 10$) and their sum $17 + 19 = 36 < 40$. This pair is valid.

If $x = 19$, the next odd number greater than 17 is 19. But $x$ must be less than 19. So $x=19$ is not included.

Therefore, the pairs of consecutive odd natural numbers, both larger than $10$ and with a sum less than $40$, are $(11, 13), (13, 15), (15, 17), \text{ and } (17, 19)$.

Given:

The inequality $24x < 100$.

(i) $x$ is a natural number.

(ii) $x$ is an integer.

To Find:

The values of $x$ that satisfy the inequality for each case.

Solution:

We are given the inequality:

To solve for $x$, we need to isolate $x$. We can do this by dividing both sides of the inequality by 24.

Dividing both sides of (1) by 24 (which is a positive number, so the inequality sign does not change):

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

Simplifying the fraction $\frac{100}{24}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 4:

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

So, the inequality becomes:

We can also write $\frac{25}{6}$ as a mixed fraction or a decimal for better understanding. $\frac{25}{6} = 4 \frac{1}{6}$. As a decimal, $\frac{25}{6} = 4.166...$ (approximately).

So, the inequality is $x < 4 \frac{1}{6}$ or $x < 4.166...$

Solving for Case (i): x is a natural number.

Natural numbers are the counting numbers: $1, 2, 3, 4, 5, ...$.

We need to find the natural numbers $x$ such that $x < 4 \frac{1}{6}$.

The natural numbers less than $4 \frac{1}{6}$ are $1, 2, 3,$ and $4$.

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

Solving for Case (ii): x is an integer.

Integers include natural numbers, zero, and the negative of natural numbers: $..., -3, -2, -1, 0, 1, 2, 3, 4, 5, ...$.

We need to find the integers $x$ such that $x < 4 \frac{1}{6}$.

The integers less than $4 \frac{1}{6}$ are $4, 3, 2, 1, 0, -1, -2, ...$.

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

Answer:

(i) When $x$ is a natural number, the solution is $x \in \{1, 2, 3, 4\}$.

(ii) When $x$ is an integer, the solution is $x \in \{..., -2, -1, 0, 1, 2, 3, 4\}$.

Given: The inequality $-12x > 30$.

To Solve: The given inequality for $x$ under two different conditions for $x$.

Solution:

We have the inequality:

$-12x > 30$

To solve for $x$, divide both sides of the inequality by $-12$. When dividing by a negative number, the direction of the inequality sign must be reversed.

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

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

Simplify the fraction $\frac{30}{12}$ by dividing both the numerator and denominator by their greatest common divisor, which is $6$.

$\frac{30}{12} = \frac{30 \div 6}{12 \div 6} = \frac{5}{2}$

So the inequality simplifies to:

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

$x < -2.5$

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

(i) x is a natural number.

The set of natural numbers is $\{1, 2, 3, 4, \dots\}$.

We are looking for natural numbers $x$ such that $x < -2.5$.

Since all natural numbers are positive ($1, 2, 3, \dots$), none of them are less than $-2.5$.

Therefore, there is no natural number $x$ that satisfies the inequality.

The solution set when $x$ is a natural number is the empty set, denoted by $\emptyset$.

(ii) x is an integer.

The set of integers is $\{\dots, -3, -2, -1, 0, 1, 2, \dots\}$.

We are looking for integers $x$ such that $x < -2.5$.

The integers that are less than $-2.5$ are $-3, -4, -5, \dots$ and so on.

Therefore, the integers satisfying the inequality are $\dots, -5, -4, -3$.

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

Given: The inequality $5x - 3 < 7$.

To Solve: The given inequality for $x$ under two different conditions for $x$.

Solution:

We have 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 does not change):

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

$x < 2$

The solution to the inequality $5x - 3 < 7$ is $x < 2$. Now we consider the solution set based on the given conditions for $x$.

(i) x is an integer.

The set of integers is $\{\dots, -3, -2, -1, 0, 1, 2, \dots\}$.

We are looking for integers $x$ such that $x < 2$.

The integers that are strictly less than $2$ are $1, 0, -1, -2, \dots$ and so on.

Therefore, the integers satisfying the inequality are $\dots, -2, -1, 0, 1$.

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

(ii) x is a real number.

The set of real numbers includes all numbers on the number line, including integers, fractions, decimals, and irrational numbers.

We are looking for real numbers $x$ such that $x < 2$.

The real numbers that are strictly less than $2$ are all numbers to the left of $2$ on the number line.

This can be represented in interval notation as $(-\infty, 2)$.

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 Solve: The given inequality for $x$ under two different conditions for $x$.

Solution:

We have 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 does not change):

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

$x > -2$

The solution to the inequality $3x + 8 > 2$ is $x > -2$. Now we consider the solution set based on the given conditions for $x$.

(i) x is an integer.

The set of integers is $\{\dots, -3, -2, -1, 0, 1, 2, \dots\}$.

We are looking for integers $x$ such that $x > -2$.

The integers that are strictly greater than $-2$ are $-1, 0, 1, 2, 3, \dots$ and so on.

Therefore, the integers satisfying the inequality are $-1, 0, 1, 2, 3, \dots$.

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

(ii) x is a real number.

The set of real numbers includes all numbers on the number line.

We are looking for real numbers $x$ such that $x > -2$.

The real numbers that are strictly greater than $-2$ are all numbers to the right of $-2$ on the number line.

This can be represented in interval notation as $(-2, \infty)$.

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

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

$4x + 3 < 5x + 7$

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

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

$3 < x + 7$

Subtract $7$ from both sides of the inequality:

$3 - 7 < x + 7 - 7$

$-4 < x$

This can also be written as $x > -4$.

Since $x$ is a real number, the solution includes all real numbers strictly greater than $-4$.

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x > -4\}$ or $(-4, \infty)$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

$3x - 7 > 5x - 1$

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

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

$-7 > 2x - 1$

Add $1$ to both sides of the inequality:

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

$-6 > 2x$

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

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

$-3 > x$

This can also be written as $x < -3$.

Since $x$ is a real number, the solution includes all real numbers strictly less than $-3$.

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x < -3\}$ or $(-\infty, -3)$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

Distribute the numbers on both sides of the inequality:

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

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

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

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

$x - 3 \leq -6$

Add $3$ to both sides of the inequality:

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

$x \leq -3$

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

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x \leq -3\}$ or $(-\infty, -3]$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

Distribute the numbers on both sides of the inequality:

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

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

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

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

$6 \geq 2 + x$

Subtract $2$ from both sides of the inequality:

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

$4 \geq x$

This can also be written as $x \leq 4$.

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

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x \leq 4\}$ or $(-\infty, 4]$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

To combine the terms on the left side, find the Least Common Multiple (LCM) of the denominators ($1, 2, 3$), which is $6$. Rewrite each term with the denominator $6$:

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

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

Combine the numerators over the common denominator:

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

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

Multiply both sides of the inequality by $6$ (since $6$ is positive, the inequality sign does not change):

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

$11x < 66$

Divide both sides of the inequality by $11$ (since $11$ is positive, the inequality sign does not change):

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

$x < 6$

Since $x$ is a real number, the solution includes all real numbers strictly less than $6$.

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x < 6\}$ or $(-\infty, 6)$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

Subtract $\frac{x}{2}$ from both sides of the inequality:

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

Find a common denominator for the terms on the left side. The LCM of $3$ and $2$ is $6$.

Rewrite the fractions with the common denominator:

$\frac{2x}{6} - \frac{3x}{6} > 1$

Combine the terms on the left side:

$\frac{2x - 3x}{6} > 1$

$\frac{-x}{6} > 1$

Multiply both sides of the inequality by $6$ (since $6$ is positive, the inequality sign does not change):

$6 \times \frac{-x}{6} > 6 \times 1$

$-x > 6$

Multiply both sides of the inequality by $-1$. When multiplying by a negative number, the direction of the inequality sign must be reversed.

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

$x < -6$

Since $x$ is a real number, the solution includes all real numbers strictly less than $-6$.

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x < -6\}$ or $(-\infty, -6)$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

To eliminate the denominators, we multiply both sides of the inequality by the Least Common Multiple (LCM) of $5$ and $3$, which is $15$.

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

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

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

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

Distribute the numbers on both sides of the inequality:

$9 \times x - 9 \times 2 \leq 25 \times 2 - 25 \times x$

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

Add $25x$ to both sides of the inequality:

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

$34x - 18 \leq 50$

Add $18$ to both sides of the inequality:

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

$34x \leq 68$

Divide both sides by $34$ (since $34$ is a positive number, the inequality sign does not change):

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

$x \leq 2$

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

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x \leq 2\}$ or $(-\infty, 2]$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

First, simplify the expressions inside the parentheses on both sides.

Left side: $\frac{1}{2}\left( \frac{3x}{5} + \frac{4 \times 5}{5}\right) = \frac{1}{2}\left( \frac{3x + 20}{5}\right) = \frac{3x + 20}{10}$

Right side: $\frac{1}{3}(x-6) = \frac{x - 6}{3}$

The inequality becomes:

$\frac{3x + 20}{10} \geq \frac{x - 6}{3}$

To eliminate the denominators, multiply both sides of the inequality by the Least Common Multiple (LCM) of $10$ and $3$, which is $30$. Since $30$ is positive, the inequality sign does not change.

$30 \times \left(\frac{3x + 20}{10}\right) \geq 30 \times \left(\frac{x - 6}{3}\right)$

$\cancel{30}^3 \times \left(\frac{3x + 20}{\cancel{10}_1}\right) \geq \cancel{30}^{10} \times \left(\frac{x - 6}{\cancel{3}_1}\right)$

$3(3x + 20) \geq 10(x - 6)$

Distribute the numbers on both sides of the inequality:

$3 \times 3x + 3 \times 20 \geq 10 \times x - 10 \times 6$

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

Subtract $9x$ from both sides of the inequality:

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

$60 \geq x - 60$

Add $60$ to both sides of the inequality:

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

$120 \geq x$

This can also be written as $x \leq 120$.

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

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x \leq 120\}$ or $(-\infty, 120]$.

Given: The inequality $2 (2x + 3) – 10 < 6 (x – 2)$.

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

Distribute the numbers on both sides of the inequality:

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

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

Combine like terms on the left side:

$4x - 4 < 6x - 12$

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

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

$-4 < 2x - 12$

Add $12$ to both sides of the inequality:

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

$8 < 2x$

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

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

$4 < x$

This can also be written as $x > 4$.

Since $x$ is a real number, the solution includes all real numbers strictly greater than $4$.

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x > 4\}$ or $(4, \infty)$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

$37 - (3x + 5) \geq 9x - 8(x - 3)$

Simplify both sides of the inequality by removing parentheses:

$37 - 3x - 5 \geq 9x - (8 \times x - 8 \times 3)$

$37 - 3x - 5 \geq 9x - (8x - 24)$

$37 - 3x - 5 \geq 9x - 8x + 24$

Combine like terms on both sides:

$(37 - 5) - 3x \geq (9x - 8x) + 24$

$32 - 3x \geq x + 24$

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

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

$32 \geq 4x + 24$

Subtract $24$ from both sides of the inequality:

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

$8 \geq 4x$

Divide both sides by $4$ (since $4$ is a positive number, the inequality sign does not change):

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

$2 \geq x$

This can also be written as $x \leq 2$.

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

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x \leq 2\}$ or $(-\infty, 2]$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

To eliminate the denominators, we multiply both sides of the inequality by the Least Common Multiple (LCM) of $4$, $3$, and $5$.

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

Multiply both sides by $60$ (since $60$ is positive, the inequality sign does not change):

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

$\cancel{60}^{15} \times \frac{x}{\cancel{4}_1} < 60 \times \frac{5x - 2}{3} - 60 \times \frac{7x - 3}{5}$

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

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

Distribute the numbers 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:

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

$15x < 16x - 4$

Subtract $16x$ from both sides of the inequality:

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

$-x < -4$

Multiply both sides of the inequality by $-1$. When multiplying by a negative number, the direction of the inequality sign must be reversed.

$(-1) \times (-x) > (-1) \times (-4)$

$x > 4$

Since $x$ is a real number, the solution includes all real numbers strictly greater than $4$.

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x > 4\}$ or $(4, \infty)$.

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

To Solve: The given inequality for real $x$.

Solution:

We have the inequality:

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

To eliminate the denominators, we multiply both sides of the inequality by the Least Common Multiple (LCM) of $3$, $4$, and $5$.

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

Multiply both sides by $60$ (since $60$ is positive, the inequality sign does not change):

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

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

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

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

Distribute the numbers on both sides:

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

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

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

Combine like terms on the right side:

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

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

Subtract $40x$ from both sides of the inequality:

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

$-20 \geq 17x - 54$

Add $54$ to both sides of the inequality:

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

$34 \geq 17x$

Divide both sides by $17$ (since $17$ is a positive number, the inequality sign does not change):

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

$2 \geq x$

This can also be written as $x \leq 2$.

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

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

The solution set for real $x$ is $\{x \in \mathbb{R} \mid x \leq 2\}$ or $(-\infty, 2]$.

Given: The inequality $3x - 2 < 2x + 1$.

To Solve: The given inequality for real $x$.

To Show: The graph of the solution on a number line.

Solution:

We have the inequality:

$3x - 2 < 2x + 1$

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

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

$x - 2 < 1$

Add $2$ to both sides of the inequality:

$x - 2 + 2 < 1 + 2$

$x < 3$

Thus, the solution to the inequality $3x - 2 < 2x + 1$ is $x < 3$.

Since $x$ is a real number, the solution includes all real numbers strictly less than $3$.

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

Graph of the solution on number line:

To show the solution $x < 3$ on a number line, we represent all real numbers less than $3$.

1. Draw a number line.

2. Locate the number $3$ on the number line.

3. Since the inequality is strict ($<$, not $\leq$), the number $3$ itself is not included in the solution set. This is indicated by an open circle (or a hollow circle) at the point representing $3$ on the number line.

4. All numbers less than $3$ are part of the solution. This is indicated by drawing a thick line or shading the part of the number line to the left of $3$. An arrow should be drawn at the left end of the shaded line to show that the solution extends infinitely in the negative direction.

Answer:

The solution to the inequality $3x - 2 < 2x + 1$ for real $x$ is $x < 3$. The graph shows all real numbers to the left of $3$, with $3$ not included.

Given: The inequality $5x - 3 > 3x - 5$.

To Solve: The given inequality for real $x$.

To Show: The graph of the solution on a number line.

Solution:

We have the inequality:

To solve for $x$, we need to isolate $x$ on one side of the inequality.

First, subtract $3x$ from both sides of the inequality (1):

$5x - 3x - 3 > 3x - 3x - 5$

$2x - 3 > -5$

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

$2x - 3 + 3 > -5 + 3$

$2x > -2$

Finally, divide both sides by $2$. Since we are dividing by a positive number, the inequality sign remains the same:

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

$x > -1$

Thus, the solution to the inequality $5x - 3 > 3x - 5$ is:

Since $x$ is a real number, the solution includes all real numbers strictly greater than $-1$.

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

Graph of the solution on number line:

To show the solution $x > -1$ on a number line, we represent all real numbers greater than $-1$.

1. Draw a number line.

2. Locate the number $-1$ on the number line.

3. Since the inequality is strict ($>$, not $\geq$), the number $-1$ itself is not included in the solution set. This is indicated by an open circle (or a hollow circle) at the point representing $-1$ on the number line.

4. All numbers greater than $-1$ are part of the solution. This is indicated by drawing a thick line or shading the part of the number line to the right of $-1$. An arrow should be drawn at the right end of the shaded line to show that the solution extends infinitely in the positive direction.

Answer:

The solution to the inequality $5x - 3 > 3x - 5$ for real $x$ is $x > -1$. The graph shows all real numbers to the right of $-1$, with $-1$ not included.

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

To Solve: The given inequality for real $x$.

To Show: The graph of the solution on a number line.

Solution:

We have the inequality:

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

First, distribute the numbers on both sides of the inequality to remove the parentheses:

$3 \times 1 - 3 \times x < 2 \times x + 2 \times 4$

This simplifies to:

$3 - 3x < 2x + 8$

Now, we need to collect the terms with $x$ on one side and the constant terms on the other side. Let's add $3x$ to both sides of the inequality:

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

This simplifies to:

$3 < 5x + 8$

Next, subtract $8$ from both sides of the inequality:

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

This simplifies to:

$-5 < 5x$

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

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

This gives us:

This inequality $-1 < x$ can also be written as $x > -1$.

Thus, the solution to the inequality $3(1 - x) < 2(x + 4)$ is $x > -1$.

Since $x$ is a real number, the solution includes all real numbers strictly greater than $-1$.

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

Graph of the solution on number line:

To show the solution $x > -1$ on a number line, we represent all real numbers greater than $-1$.

1. Draw a number line.

2. Locate the number $-1$ on the number line.

3. Since the inequality is strict ($>$, not $\geq$), the number $-1$ itself is not included in the solution set. This is indicated by an open circle (or a hollow circle) at the point representing $-1$ on the number line.

4. All numbers greater than $-1$ are part of the solution. This is indicated by drawing a thick line or shading the part of the number line to the right of $-1$. An arrow should be drawn at the right end of the shaded line to show that the solution extends infinitely in the positive direction.

Answer:

The solution to the inequality $3 (1 – x) < 2 (x + 4)$ for real $x$ is $x > -1$. The graph shows all real numbers to the right of $-1$, with $-1$ not included.

Given:

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

To Solve:

The given inequality for real $x$.

To Show:

The graph of the solution on a number line.

Solution:

We have the inequality:

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

LCM($2, 3, 5$) $= 30$.

Multiply both sides of the inequality (1) by $30$. Since $30$ is a positive number, the inequality sign does not change:

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

Applying the multiplication on both sides:

$\cancel{30}^{15} \times \frac{x}{\cancel{2}_1} \geq \left(30 \times \frac{5x \;-\; 2}{3}\right) - \left(30 \times \frac{7x \;-\; 3}{5}\right)$

$15x \geq \left(\cancel{30}^{10} \times \frac{(5x \;-\; 2)}{\cancel{3}_1}\right) - \left(\cancel{30}^{6} \times \frac{(7x \;-\; 3)}{\cancel{5}_1}\right)$

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

Now, distribute the numbers outside the parentheses on the right side:

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

$15x \geq (50x - 20) - (42x - 18)$

Remove the parentheses on the right side. Remember that the minus sign before the second parenthesis changes the sign of each term inside:

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

Combine the like terms ($x$ terms and constant terms) on the right side:

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

$15x \geq 8x - 2$

Now, we need to collect the terms with $x$ on one side and the constant terms on the other side. Subtract $8x$ from both sides of the inequality:

$15x - 8x \geq 8x - 8x - 2$

This simplifies to:

$7x \geq -2$

Finally, divide both sides by $7$. Since $7$ is a positive number, the inequality sign remains the same:

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

This gives us:

Thus, the solution to the inequality $\frac{x}{2} \geq \frac{(5x \;-\; 2)}{3} - \frac{(7x \;-\; 3)}{5}$ is $x \geq -\frac{2}{7}$.

Since $x$ is a real number, the solution includes all real numbers greater than or equal to $-\frac{2}{7}$.

In interval notation, the solution set is $\left[-\frac{2}{7}, \infty\right)$.

Graph of the solution on number line:

To show the solution $x \geq -\frac{2}{7}$ on a number line, we represent all real numbers greater than or equal to $-\frac{2}{7}$.

1. Draw a number line.

2. Locate the number $-\frac{2}{7}$ (which is approximately $-0.286$) on the number line.

3. Since the inequality is $\geq$, the number $-\frac{2}{7}$ itself is included in the solution set. This is indicated by a closed circle (or a filled circle) at the point representing $-\frac{2}{7}$ on the number line.

4. All numbers greater than $-\frac{2}{7}$ are part of the solution. This is indicated by drawing a thick line or shading the part of the number line to the right of $-\frac{2}{7}$. An arrow should be drawn at the right end of the shaded line to show that the solution extends infinitely in the positive direction.

Answer:

The solution to the inequality $\frac{x}{2} \geq \frac{(5x \;-\; 2)}{3} - \frac{(7x \;-\; 3)}{5}$ for real $x$ is $x \geq -\frac{2}{7}$. The graph shows all real numbers to the right of and including $-\frac{2}{7}$.

Given:

Marks in the first unit test = $70$

Marks in the second unit test = $75$

Required average marks in three tests $\geq 60$

To Find:

The minimum marks Ravi should get in the third test.

Solution:

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

The total marks obtained in the three unit tests will be the sum of the marks in the first, second, and third tests.

Total marks $= 70 + 75 + x = 145 + x$

The average marks obtained in the three tests is given by:

Average Marks $= \frac{\text{Total marks in three tests}}{\text{Number of tests}}$

Average Marks $= \frac{145 + x}{3}$

According to the problem, the average marks must be at least $60$. This means the average must be greater than or equal to $60$.

So, we set up the inequality:

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

To solve for $x$, first multiply both sides of the inequality by $3$. Since $3$ is a positive number, the inequality sign does not change.

$3 \times \left(\frac{145 + x}{3}\right) \geq 3 \times 60$

$145 + x \geq 180$

Now, subtract $145$ from both sides of the inequality:

$145 + x - 145 \geq 180 - 145$

$x \geq 35$

This inequality $x \geq 35$ means that the marks Ravi gets in the third test must be $35$ or greater.

To have an average of at least $60$, the minimum marks he should get in the third test is the smallest value that satisfies this condition, which is $35$.

Therefore, the minimum marks Ravi should get in the third test is $35$.

The minimum marks required is 35.

Given:

Marks in the first four examinations: $87, 92, 94, 95$.

Number of examinations: $5$.

Maximum marks for each examination: $100$.

Required average marks for Grade 'A': $90$ or more.

To Find:

Minimum marks Sunita must obtain in the fifth examination.

Solution:

Let $x$ be the marks obtained by Sunita in the fifth examination.

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

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

Sum of marks in the first four exams is $368$.

Total marks in five exams $= 368 + x$

The average marks obtained in the five examinations is given by:

Average Marks $= \frac{\text{Total marks in five exams}}{\text{Number of exams}}$

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

According to the condition for receiving Grade 'A', the average marks must be $90$ or more.

So, we set up the inequality:

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

To solve for $x$, first multiply both sides of the inequality by $5$. Since $5$ is a positive number, the inequality sign does not change.

$5 \times \left(\frac{368 + x}{5}\right) \geq 5 \times 90$

$368 + x \geq 450$

Now, subtract $368$ from both sides of the inequality:

$368 + x - 368 \geq 450 - 368$

$x \geq 82$

This inequality $x \geq 82$ means that the marks Sunita gets in the fifth examination must be $82$ or greater.

To get Grade 'A', the minimum marks she must obtain in the fifth examination is the smallest value that satisfies this condition, which is $82$.

Also, since each examination is of 100 marks, the marks obtained cannot exceed 100. So, $x \leq 100$.

Combining the conditions, $x$ must satisfy $82 \leq x \leq 100$.

The minimum marks is $82$.

Therefore, the minimum marks Sunita must obtain in the fifth examination to get Grade 'A' is $82$.

The minimum marks required is 82.

Given:

Pairs of consecutive odd positive integers.

Both integers in the pair are smaller than $10$.

Their sum is more than $11$.

To Find:

All such pairs of consecutive odd positive integers.

Solution:

Let the first odd positive integer be $x$.

Since the integers are consecutive and odd, the next odd positive integer is $x + 2$.

According to the first condition, both integers are positive odd integers. The positive odd integers are $1, 3, 5, 7, 9, 11, \dots$

According to the second condition, both integers are smaller than $10$.

So, we have two inequalities:

$x < 10$

and

$x + 2 < 10$

Solving the second inequality:

$x + 2 < 10$

$x + 2 - 2 < 10 - 2$

$x < 8$

Combining $x < 10$ and $x < 8$, we get $x < 8$.

So, $x$ must be an odd positive integer less than $8$. The possible values for $x$ from this condition are $1, 3, 5, 7$.

According to the third condition, their sum is more than $11$.

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

So, we have the inequality:

$2x + 2 > 11$

Solving the inequality:

$2x + 2 - 2 > 11 - 2$

$2x > 9$

Divide both sides by $2$ (since $2$ is positive, the inequality sign does not change):

$\frac{2x}{2} > \frac{9}{2}$

$x > 4.5$

Now, we need to find the odd positive integers $x$ that satisfy both conditions: $x < 8$ and $x > 4.5$.

The odd positive integers are $1, 3, 5, 7, 9, \dots$

The odd positive integers less than $8$ are $1, 3, 5, 7$.

From this list, the ones that are also greater than $4.5$ are $5$ and $7$.

So, the possible values for the first odd positive integer $x$ are $5$ and $7$.

Now we find the corresponding pairs $(x, x+2)$:

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 with the original conditions:

For the pair $(5, 7)$: They are consecutive odd positive integers. Both are smaller than $10$ ($5 < 10$, $7 < 10$). Their sum is $5 + 7 = 12$, which is more than $11$ ($12 > 11$). This pair is valid.

For the pair $(7, 9)$: They are consecutive odd positive integers. Both are smaller than $10$ ($7 < 10$, $9 < 10$). Their sum is $7 + 9 = 16$, which is more than $11$ ($16 > 11$). This pair is valid.

The next odd positive integer greater than $7$ is $9$. If $x=9$, it does not satisfy $x < 8$.

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

The pairs are (5, 7) and (7, 9).

Given:

Pairs of consecutive even positive integers.

Both integers in the pair are larger than $5$.

Their sum is less than $23$.

To Find:

All such pairs of consecutive even positive integers.

Solution:

Let the first even positive integer be $x$.

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

According to the first condition, both integers are larger than $5$.

So, we have two inequalities:

$x > 5$

and

$x + 2 > 5$

Solving the second inequality for $x$:

$x + 2 - 2 > 5 - 2$

$x > 3$

Combining the two conditions ($x > 5$ and $x > 3$), we find that $x$ must be greater than $5$.

Since $x$ is an even positive integer, the possible values for $x$ from this condition are $6, 8, 10, 12, \dots$

According to the second condition, the sum of the two consecutive even positive integers is less than $23$.

The sum is $x + (x + 2) = 2x + 2$.

So, we have the inequality:

$2x + 2 < 23$

Solving this inequality for $x$:

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

$2x < 21$

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

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

$x < 10.5$

Now, we need to find the even positive integers $x$ that satisfy both conditions:

$x > 5$ and $x < 10.5$

The even positive integers are $2, 4, 6, 8, 10, 12, \dots$

We are looking for integers from this list that are greater than $5$ and less than $10.5$.

The even positive integers satisfying both conditions are $6, 8, 10$.

These are the possible values for the first integer, $x$. We now find the corresponding consecutive even positive integer, $x+2$, for each valid value of $x$ and list the pairs:

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 each pair:

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

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

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

The next even positive integer is $12$. If $x=12$, it does not satisfy $x < 10.5$.

Therefore, the pairs of consecutive even positive integers, both larger than $5$ and with a sum less than $23$, are $(6, 8), (8, 10), \text{ and } (10, 12)$.

The pairs are (6, 8), (8, 10), and (10, 12).

Given:

In a triangle:

The longest side is $3$ times the shortest side.

The third side is $2$ cm shorter than the longest side.

The perimeter of the triangle is at least $61$ cm.

To Find:

The minimum length of the shortest side.

Solution:

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

According to the problem, the longest side is $3$ times the shortest side.

Length of the longest side $= 3s$ cm.

The third side is $2$ cm shorter than the longest side.

Length of the third side $= (3s - 2)$ cm.

For these to be the sides of a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. We also know the sides must be positive lengths. Since $s$ is a length, $s > 0$.

The longest side is $3s$. For $3s-2$ to be a positive length, $3s - 2 > 0 \implies 3s > 2 \implies s > \frac{2}{3}$. Since we are looking for integer lengths, $s$ must be a positive integer. $s \geq 1$. Also $s>2/3$. $s=1$ is possible. If $s=1$, sides are 1, 3, 1. Perimeter 5. $1+1 > 3$ is false. So $s$ must be large enough to form a triangle.

The triangle inequality conditions are:

$s + (3s - 2) > 3s \implies 4s - 2 > 3s \implies s > 2$

$s + 3s > 3s - 2 \implies 4s > 3s - 2 \implies s > -2$ (This is true since $s$ must be positive)

$(3s - 2) + 3s > s \implies 6s - 2 > s \implies 5s > 2 \implies s > \frac{2}{5}$

Combining $s > 2$ and $s > 2/5$ and $s > 0$, we get $s > 2$.

Since $s$ is a length, $s$ must be a positive real number. The condition $s > 2$ also ensures that $3s > s$ and $3s-2 > s$ (since $s>2 \implies 2s>4 \implies 3s-2 = s + 2s - 2 > s + 4 - 2 = s+2 > s$) and $3s > 3s-2$. So the shortest side is indeed $s$, the longest is $3s$, and the third side $3s-2$ is between $s$ and $3s$ (since $s>2 \implies s < 3s-2 < 3s$).

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

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

Perimeter $= s + 3s + 3s - 2$

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

Perimeter $= 7s - 2$

According to the problem, the perimeter of the triangle is at least $61$ cm. This means the perimeter is greater than or equal to $61$ cm.

So, we set up the inequality:

$7s - 2 \geq 61$

To solve for $s$, first add $2$ to both sides of the inequality:

$7s - 2 + 2 \geq 61 + 2$

$7s \geq 63$

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

$\frac{7s}{7} \geq \frac{63}{7}$

$s \geq 9$

This inequality $s \geq 9$ tells us that the length of the shortest side must be $9$ cm or greater.

We also need to check if this satisfies the triangle inequality condition $s > 2$. Since $9 > 2$, the condition is satisfied.

The minimum length of the shortest side is the smallest value of $s$ that satisfies $s \geq 9$. This value is $9$.

Therefore, the minimum length of the shortest side is $9$ cm.

Let's verify the sides if the shortest side is 9 cm:

Shortest side = 9 cm

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

Third side = $27 - 2 = 25$ cm

Sides are 9 cm, 25 cm, and 27 cm.

Check triangle inequality: $9 + 25 = 34 > 27$ (True). $9 + 27 = 36 > 25$ (True). $25 + 27 = 52 > 9$ (True).

Check perimeter: $9 + 25 + 27 = 61$ cm. This satisfies the condition that the perimeter is at least 61 cm.

The minimum length of the shortest side is 9 cm.

Given:

Total length of the board = $91$ cm.

Three lengths are cut from the board.

The second length is $3$ cm longer than the shortest length.

The third length is twice as long as the shortest length.

The third piece is at least $5$ cm longer than the second piece.

To Find:

The possible lengths of the shortest board.

Solution:

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

Since $x$ represents a length, it must be a positive value, so $x > 0$.

According to the problem:

Length of the second piece = $x + 3$ cm.

Length of the third piece = $2x$ cm.

These three lengths must be positive:

$x > 0$

$x + 3 > 0 \implies x > -3$ (This is satisfied if $x > 0$)

$2x > 0 \implies x > 0$

So, the lengths are valid if $x > 0$.

According to the first condition, the sum of the three lengths cannot exceed the total length of the board ($91$ cm).

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

Simplify the inequality:

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

$4x + 3 \leq 91$

Subtract $3$ from both sides:

$4x + 3 - 3 \leq 91 - 3$

$4x \leq 88$

Divide by $4$ (a positive number):

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

$x \leq 22$

According to the second condition, the third piece is to be at least $5$ cm longer than the second piece.

This means the length of the third piece is greater than or equal to the length of the second piece plus $5$ cm.

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

Simplify the inequality:

$2x \geq x + 3 + 5$

$2x \geq x + 8$

Subtract $x$ from both sides:

$2x - x \geq x - x + 8$

$x \geq 8$

We also have the condition that the shortest length $x$ must be positive, $x > 0$. This is satisfied if $x \geq 8$.

We need to find the values of $x$ that satisfy both inequalities (i) and (ii):

$x \leq 22$

and

$x \geq 8$

Combining these two inequalities, we get $8 \leq x \leq 22$.

Since the lengths are cut from a physical board, the lengths should ideally be positive real numbers. However, in practical scenarios involving lengths of boards, they are often assumed to be positive measurements which can be any real number in the given range.

The possible lengths of the shortest board are any value $x$ such that $8 \leq x \leq 22$.

The possible lengths of the shortest board are in the range [8, 22] cm.

Given: The inequality $-8 \leq 5x - 3 < 7$.

To Solve: The given compound inequality for $x$.

Solution:

We have the compound inequality:

$-8 \leq 5x - 3 < 7$

To solve this compound inequality, we perform the same operations on all three parts of the inequality simultaneously.

First, add $3$ to all parts of the inequality:

$-8 + 3 \leq 5x - 3 + 3 < 7 + 3$

$-5 \leq 5x < 10$

Next, divide all parts of the inequality by $5$. Since $5$ is a positive number, the direction of the inequality signs does not change.

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

$-1 \leq x < 2$

This inequality $-1 \leq x < 2$ means that $x$ is greater than or equal to $-1$ and strictly less than $2$.

Assuming $x$ is a real number (as is common when solving inequalities unless otherwise specified), the solution set includes all real numbers between $-1$ (inclusive) and $2$ (exclusive).

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

The solution set for $x$ is $\{x \mid -1 \leq x < 2\}$ or $[-1, 2)$.

Given: The inequality $-5 \leq \frac{5 - 3x}{2} \leq 8$.

To Solve: The given compound inequality for $x$.

Solution:

We have the compound inequality:

$-5 \leq \frac{5 - 3x}{2} \leq 8$

To eliminate the denominator, multiply all three parts of the inequality by $2$. Since $2$ is a positive number, the direction of the inequality signs does not change.

$2 \times (-5) \leq 2 \times \left(\frac{5 - 3x}{2}\right) \leq 2 \times 8$

$-10 \leq 5 - 3x \leq 16$

Next, subtract $5$ from all parts of the inequality:

$-10 - 5 \leq 5 - 3x - 5 \leq 16 - 5$

$-15 \leq -3x \leq 11$

Now, divide all parts of the inequality by $-3$. When dividing by a negative number, the direction of the inequality signs must be reversed.

$\frac{-15}{-3} \geq \frac{-3x}{-3} \geq \frac{11}{-3}$

$5 \geq x \geq -\frac{11}{3}$

Rewrite the inequality with the smaller number on the left:

$-\frac{11}{3} \leq x \leq 5$

Note that $-\frac{11}{3}$ is approximately $-3.67$.

This inequality $-\frac{11}{3} \leq x \leq 5$ means that $x$ is greater than or equal to $-\frac{11}{3}$ and less than or equal to $5$.

Assuming $x$ is a real number, the solution set includes all real numbers between $-\frac{11}{3}$ and $5$, inclusive of both endpoints.

In interval notation, the solution set is $\left[-\frac{11}{3}, 5\right]$.

The solution set for $x$ is $\left\{x \mid -\frac{11}{3} \leq x \leq 5\right\}$ or $\left[-\frac{11}{3}, 5\right]$.

Given: The system of inequalities:

To Solve: The given system of inequalities for real $x$.

Solution:

We solve each inequality separately.

Consider the first inequality:

Subtract $x$ from both sides:

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

$2x - 7 < 5$

Add $7$ to both sides:

$2x - 7 + 7 < 5 + 7$

$2x < 12$

Divide both sides by $2$ (positive):

$\frac{2x}{2} < \frac{12}{2}$

Consider the second inequality:

Subtract $11$ from both sides:

$11 - 11 - 5x \leq 1 - 11$

$-5x \leq -10$

Divide both sides by $-5$. When dividing by a negative number, reverse the inequality sign:

$\frac{-5x}{-5} \geq \frac{-10}{-5}$

The solution to the system of inequalities is the set of all real numbers $x$ that satisfy both inequality (3) and inequality (4).

From (3), we have $x < 6$.

From (4), we have $x \geq 2$.

We need $x$ to be greater than or equal to $2$ AND less than $6$.

Combining these conditions, we get $2 \leq x < 6$.

The solution set for real $x$ is the interval $[2, 6)$.

In set-builder notation, this is $\{x \in \mathbb{R} \mid 2 \leq x < 6\}$.

The solution set is $[2, 6)$.

Given:

The temperature in Celsius (C) must be between 30°C and 35°C.

The conversion formula from Fahrenheit (F) to Celsius (C) is $C = \frac{5}{9}(F - 32)$.

To Find:

The range of temperature in degree Fahrenheit (F).

Solution:

The condition for the temperature in Celsius is given by the inequality:

$30 < C < 35$

We are given the conversion formula $C = \frac{5}{9}(F - 32)$. Substitute this expression for C into the inequality:

$30 < \frac{5}{9}(F - 32) < 35$

To isolate the term $(F - 32)$, we can multiply all three parts of the inequality by the reciprocal of $\frac{5}{9}$, which is $\frac{9}{5}$. Since $\frac{9}{5}$ is a positive number, the direction of the inequality signs does not change.

$\frac{9}{5} \times 30 < \frac{9}{5} \times \frac{5}{9}(F - 32) < \frac{9}{5} \times 35$

Calculate the products:

Left side: $\frac{9}{5} \times 30 = 9 \times \frac{30}{5} = 9 \times 6 = 54$

Right side: $\frac{9}{5} \times 35 = 9 \times \frac{35}{5} = 9 \times 7 = 63$

The inequality becomes:

$54 < F - 32 < 63$

To isolate $F$, add $32$ to all three parts of the inequality:

$54 + 32 < F - 32 + 32 < 63 + 32$

$86 < F < 95$

This inequality $86 < F < 95$ gives the range of the temperature in degrees Fahrenheit.

The temperature in Fahrenheit must be strictly between 86°F and 95°F.

In interval notation, the range is $(86, 95)$.

Therefore, the range of temperature in degree Fahrenheit is between 86°F and 95°F.

The range of temperature in Fahrenheit is (86, 95).

Given:

Initial amount of 12% acid solution = $600$ litres.

Concentration of the initial solution = $12\%$ acid.

Concentration of the added solution = $30\%$ acid.

Required concentration of the resulting mixture: more than $15\%$ but less than $18\%$ acid.

To Find:

The number of litres of the $30\%$ acid solution that must be added.

Solution:

Let $x$ be the amount (in litres) of the $30\%$ acid solution that is added.

Since $x$ represents an amount of solution, $x$ must be a positive value, so $x > 0$.

The amount of acid in the initial $12\%$ solution is $12\%$ of $600$ litres.

Amount of acid in initial solution $= 12\% \times 600 = 0.12 \times 600 = 72$ litres.

The amount of acid in the added $30\%$ solution is $30\%$ of $x$ litres.

Amount of acid in added solution $= 30\% \times x = 0.30x$ litres.

When $x$ litres of the $30\%$ solution are added to the $600$ litres of the $12\%$ solution, the total volume of the resulting mixture is $(600 + x)$ litres.

The total amount of acid in the resulting mixture is the sum of the acid from the initial solution and the added solution.

Total amount of acid in mixture $= 72 + 0.30x$ litres.

The concentration of acid in the resulting mixture is the total amount of acid divided by the total volume of the mixture, expressed as a percentage.

Concentration of acid in mixture $= \frac{\text{Total amount of acid}}{\text{Total volume of mixture}} \times 100\%$

Concentration of acid in mixture $= \frac{72 + 0.30x}{600 + x} \times 100\%$

According to the problem, the acid content in the resulting mixture must be more than $15\%$ but less than $18\%$. This gives us a compound inequality for the concentration:

$15 < \frac{72 + 0.30x}{600 + x} \times 100 < 18$

Divide all parts by $100$:

$\frac{15}{100} < \frac{72 + 0.30x}{600 + x} < \frac{18}{100}$

$0.15 < \frac{72 + 0.30x}{600 + x} < 0.18$

We need to solve this compound inequality for $x$. Since the volume of the mixture $(600 + x)$ is always positive (because $x > 0$), we can multiply all parts of the inequality by $(600 + x)$ without changing the direction of the inequality signs.

$0.15(600 + x) < 72 + 0.30x < 0.18(600 + x)$

This compound inequality can be split into two separate inequalities:

Inequality 1: $0.15(600 + x) < 72 + 0.30x$

Inequality 2: $72 + 0.30x < 0.18(600 + x)$

Solve Inequality 1:

$0.15 \times 600 + 0.15x < 72 + 0.30x$

$90 + 0.15x < 72 + 0.30x$

Subtract $0.15x$ from both sides:

$90 < 72 + 0.30x - 0.15x$

$90 < 72 + 0.15x$

Subtract $72$ from both sides:

$90 - 72 < 0.15x$

$18 < 0.15x$

Divide by $0.15$ (positive):

$\frac{18}{0.15} < x$

$\frac{18}{\frac{15}{100}} < x \implies 18 \times \frac{100}{15} < x$

$\cancel{18}^6 \times \frac{100}{\cancel{15}_5} < x \implies 6 \times \frac{100}{5} < x \implies 6 \times 20 < x \implies 120 < x$

So, $x > 120$.

Solve Inequality 2:

$72 + 0.30x < 0.18 \times 600 + 0.18x$

$72 + 0.30x < 108 + 0.18x$

Subtract $0.18x$ from both sides:

$72 + 0.30x - 0.18x < 108$

$72 + 0.12x < 108$

Subtract $72$ from both sides:

$0.12x < 108 - 72$

$0.12x < 36$

Divide by $0.12$ (positive):

$x < \frac{36}{0.12}$

$x < \frac{36}{\frac{12}{100}} \implies x < 36 \times \frac{100}{12}$

$x < \cancel{36}^3 \times \frac{100}{\cancel{12}_1} \implies x < 3 \times 100 \implies x < 300$

So, $x < 300$.

The value of $x$ must satisfy both conditions (A) and (B):

$x > 120$

and

$x < 300$

Combining these two inequalities, we get $120 < x < 300$.

Also, we initially established that $x > 0$, which is satisfied by $x > 120$.

The possible amounts (in litres) of the $30\%$ acid solution to be added are the real numbers between $120$ and $300$, exclusive of the endpoints.

In interval notation, the range is $(120, 300)$.

Therefore, the manufacturer must add more than $120$ litres but less than $300$ litres of the $30\%$ acid solution.

The amount of 30% acid solution to be added must be between 120 litres and 300 litres (exclusive).

The given inequality is:

$2 \leq 3x - 4 \leq 5$

To isolate the term containing $x$, we add 4 to all parts of the inequality:

$2 + 4 \leq 3x - 4 + 4 \leq 5 + 4$

$6 \leq 3x \leq 9$

Now, to isolate $x$, we divide all parts of the inequality by 3. Since 3 is a positive number, the direction of the inequality signs remains unchanged:

$\frac{6}{3} \leq \frac{3x}{3} \leq \frac{9}{3}$

$2 \leq x \leq 3$

Thus, the solution to the given inequality is all real numbers $x$ such that $x$ is greater than or equal to 2 and less than or equal to 3.

In interval notation, the solution set is $\left[2, 3\right]$.

The given inequality is:

$6 \leq -3(2x - 4) < 12$

First, distribute the $-3$ on the right side:

$6 \leq -6x + 12 < 12$

To isolate the term with $x$, subtract 12 from all parts of the inequality:

$6 - 12 \leq -6x + 12 - 12 < 12 - 12$

$-6 \leq -6x < 0$

Now, divide all parts of the inequality by $-6$. Remember to reverse the direction of the inequality signs when dividing by a negative number:

$\frac{-6}{-6} \geq \frac{-6x}{-6} > \frac{0}{-6}$

$1 \geq x > 0$

Rewriting the inequality in the standard form with the smaller number on the left:

$0 < x \leq 1$

Thus, the solution to the given inequality is all real numbers $x$ such that $x$ is strictly greater than 0 and less than or equal to 1.

In interval notation, the solution set is $\left(0, 1\right]$.

The given inequality is:

$-3 \leq 4 - \frac{7x}{2} \leq 18$

To isolate the term containing $x$, subtract 4 from all parts of the inequality:

$-3 - 4 \leq 4 - \frac{7x}{2} - 4 \leq 18 - 4$

$-7 \leq - \frac{7x}{2} \leq 14$

To remove the fraction and the negative sign from the $x$ term, multiply all parts of the inequality by -2. Remember to reverse the direction of the inequality signs when multiplying by a negative number:

$-7 \times (-2) \geq - \frac{7x}{2} \times (-2) \geq 14 \times (-2)$

$14 \geq 7x \geq -28$

To isolate $x$, divide all parts of the inequality by 7. Since 7 is a positive number, the direction of the inequality signs remains unchanged:

$\frac{14}{7} \geq \frac{7x}{7} \geq \frac{-28}{7}$

$2 \geq x \geq -4$

Rewriting the inequality in the standard form with the smaller number on the left:

$-4 \leq x \leq 2$

Thus, the solution to the given inequality is all real numbers $x$ such that $x$ is greater than or equal to -4 and less than or equal to 2.

In interval notation, the solution set is $\left[-4, 2\right]$.

The given inequality is:

$-15 < \frac{3(x - 2)}{5} \leq 0$

To eliminate the denominator, multiply all parts of the inequality by 5. Since 5 is positive, the inequality signs remain unchanged:

$-15 \times 5 < \frac{3(x - 2)}{5} \times 5 \leq 0 \times 5$

$-75 < 3(x - 2) \leq 0$

To isolate the term $(x - 2)$, divide all parts of the inequality by 3. Since 3 is positive, the inequality signs remain unchanged:

$\frac{-75}{3} < \frac{3(x - 2)}{3} \leq \frac{0}{3}$

$-25 < x - 2 \leq 0$

To isolate $x$, add 2 to all parts of the inequality:

$-25 + 2 < x - 2 + 2 \leq 0 + 2$

$-23 < x \leq 2$

Thus, the solution to the given inequality is all real numbers $x$ such that $x$ is strictly greater than -23 and less than or equal to 2.

In interval notation, the solution set is $\left(-23, 2\right]$.

The given inequality is:

$-12 < 4 - \frac{3x}{-5} \leq 2$

First, simplify the term $\frac{3x}{-5}$ which is equal to $-\frac{3x}{5}$. The inequality becomes:

$-12 < 4 + \frac{3x}{5} \leq 2$

To isolate the term containing $x$, subtract 4 from all parts of the inequality:

$-12 - 4 < 4 + \frac{3x}{5} - 4 \leq 2 - 4$

$-16 < \frac{3x}{5} \leq -2$

To eliminate the denominator, multiply all parts of the inequality by 5. Since 5 is positive, the inequality signs remain unchanged:

$-16 \times 5 < \frac{3x}{5} \times 5 \leq -2 \times 5$

$-80 < 3x \leq -10$

To isolate $x$, divide all parts of the inequality by 3. Since 3 is positive, the inequality signs remain unchanged:

$\frac{-80}{3} < \frac{3x}{3} \leq \frac{-10}{3}$

$-\frac{80}{3} < x \leq -\frac{10}{3}$

Thus, the solution to the given inequality is all real numbers $x$ such that $x$ is strictly greater than $-\frac{80}{3}$ and less than or equal to $-\frac{10}{3}$.

In interval notation, the solution set is $\left(-\frac{80}{3}, -\frac{10}{3}\right]$.

The given inequality is:

$7 \leq \frac{3x + 11}{2} \leq 11$

To eliminate the denominator, multiply all parts of the inequality by 2. Since 2 is positive, the inequality signs remain unchanged:

$7 \times 2 \leq \frac{3x + 11}{2} \times 2 \leq 11 \times 2$

$14 \leq 3x + 11 \leq 22$

To isolate the term with $x$, subtract 11 from all parts of the inequality:

$14 - 11 \leq 3x + 11 - 11 \leq 22 - 11$

$3 \leq 3x \leq 11$

To isolate $x$, divide all parts of the inequality by 3. Since 3 is positive, the inequality signs remain unchanged:

$\frac{3}{3} \leq \frac{3x}{3} \leq \frac{11}{3}$

$1 \leq x \leq \frac{11}{3}$

Thus, the solution to the given inequality is all real numbers $x$ such that $x$ is greater than or equal to 1 and less than or equal to $\frac{11}{3}$.

In interval notation, the solution set is $\left[1, \frac{11}{3}\right]$.

Given:

Two inequalities are given:

To Solve:

Find the values of $x$ that satisfy both inequalities simultaneously.

Solution:

We need to solve each inequality separately and then find the common solution.

Let's solve the first inequality (1):

$5x + 1 > -24$

Subtract $1$ from both sides of the inequality:

$5x + 1 - 1 > -24 - 1$

$5x > -25$

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

$\frac{5x}{5} > \frac{-25}{5}$

$x > -5$

Now, let's solve the second inequality (2):

$5x - 1 < 24$

Add $1$ to both sides of the inequality:

$5x - 1 + 1 < 24 + 1$

$5x < 25$

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

$\frac{5x}{5} < \frac{25}{5}$

$x < 5$

The solution to the system of inequalities is the set of all real numbers $x$ that satisfy both $x > -5$ and $x < 5$.

Combining these two conditions, we get:

Thus, the solution set consists of all real numbers between $-5$ and $5$, excluding $-5$ and $5$.

In interval notation, the solution set is $\left(-5, 5\right)$.

Graph of the solution on number line:

To show the solution $-5 < x < 5$ on a number line, we represent all real numbers between $-5$ and $5$.

1. Draw a number line.

2. Locate the numbers $-5$ and $5$ on the number line.

3. Since the inequalities are strict ($>$ and $<$), the numbers $-5$ and $5$ themselves are not included in the solution set. This is indicated by open circles (or hollow circles) at the points representing $-5$ and $5$ on the number line.

4. All numbers between $-5$ and $5$ are part of the solution. This is indicated by drawing a thick line or shading the region of the number line between $-5$ and $5$.

Answer:

The solution to the given system of inequalities for real $x$ is $-5 < x < 5$. The graph shows all real numbers strictly between $-5$ and $5$.

Given:

Two inequalities are given:

To Solve:

Find the values of $x$ that satisfy both inequalities simultaneously.

Solution:

We need to solve each inequality separately and then find the common solution.

Let's solve the first inequality (1):

$2(x - 1) < x + 5$

First, distribute the $2$ on the left side to remove the parentheses:

$2x - 2 < x + 5$

Now, we need to collect the terms with $x$ on one side and the constant terms on the other. Subtract $x$ from both sides of the inequality:

$2x - x - 2 < x - x + 5$

This simplifies to:

$x - 2 < 5$

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

$x - 2 + 2 < 5 + 2$

This gives us:

Now, let's solve the second inequality (2):

$3(x + 2) > 2 - x$

First, distribute the $3$ on the left side to remove the parentheses:

$3x + 6 > 2 - x$

Now, we need to collect the terms with $x$ on one side and the constant terms on the other. Add $x$ to both sides of the inequality:

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

This simplifies to:

$4x + 6 > 2$

Next, subtract $6$ from both sides of the inequality:

$4x + 6 - 6 > 2 - 6$

This simplifies to:

$4x > -4$

Finally, divide both sides by $4$. Since $4$ is a positive number, the inequality sign remains the same:

$\frac{4x}{4} > \frac{-4}{4}$

This gives us:

The solution to the system of inequalities is the set of all real numbers $x$ that satisfy both $x < 7$ (from inequality 3) and $x > -1$ (from inequality 4) simultaneously.

Combining these two conditions, we get:

Thus, the solution set consists of all real numbers strictly between $-1$ and $7$.

In interval notation, the solution set is $\left(-1, 7\right)$.

Graph of the solution on number line:

To show the solution $-1 < x < 7$ on a number line, we represent all real numbers between $-1$ and $7$.

1. Draw a number line.

2. Locate the numbers $-1$ and $7$ on the number line.

3. Since the inequalities are strict ($>$ and $<$), the numbers $-1$ and $7$ themselves are not included in the solution set. This is indicated by open circles (or hollow circles) at the points representing $-1$ and $7$ on the number line.

4. All numbers greater than $-1$ and less than $7$ are part of the solution. This is indicated by drawing a thick line or shading the region of the number line between $-1$ and $7$.

Answer:

The solution to the given system of inequalities for real $x$ is $-1 < x < 7$. The graph shows all real numbers strictly between $-1$ and $7$.

Given:

Two inequalities are given:

To Solve:

Find the values of $x$ that satisfy both inequalities simultaneously.

Solution:

We need to solve each inequality separately and then find the common solution.

Let's solve the first inequality (1):

$3x - 7 > 2(x - 6)$

First, distribute the $2$ on the right side to remove the parentheses:

$3x - 7 > 2x - 12$

Now, collect the terms with $x$ on one side and constants on the other. Subtract $2x$ from both sides:

$3x - 2x - 7 > 2x - 2x - 12$

This simplifies to:

$x - 7 > -12$

Next, add $7$ to both sides of the inequality:

$x - 7 + 7 > -12 + 7$

This gives us:

Now, let's solve the second inequality (2):

$6 - x > 11 - 2x$

Add $2x$ to both sides of the inequality:

$6 - x + 2x > 11 - 2x + 2x$

This simplifies to:

$6 + x > 11$

Next, subtract $6$ from both sides of the inequality:

$6 + x - 6 > 11 - 6$

This gives us:

The solution to the system of inequalities is the set of all real numbers $x$ that satisfy both $x > -5$ (from inequality 3) and $x > 5$ (from inequality 4) simultaneously.

For a number $x$ to be greater than $-5$ AND greater than $5$, it must be greater than the larger of the two numbers, which is $5$.

So, the combined condition is:

Thus, the solution set consists of all real numbers strictly greater than $5$.

In interval notation, the solution set is $\left(5, \infty\right)$.

Graph of the solution on number line:

To show the solution $x > 5$ on a number line, we represent all real numbers greater than $5$.

1. Draw a number line.

2. Locate the number $5$ on the number line.

3. Since the inequality is strict ($>$, not $\geq$), the number $5$ itself is not included in the solution set. This is indicated by an open circle (or a hollow circle) at the point representing $5$ on the number line.

4. All numbers greater than $5$ are part of the solution. This is indicated by drawing a thick line or shading the part of the number line to the right of $5$. An arrow should be drawn at the right end of the shaded line to show that the solution extends infinitely in the positive direction.

Answer:

The solution to the given system of inequalities for real $x$ is $x > 5$. The graph shows all real numbers strictly greater than $5$.

Given:

Two inequalities are given:

To Solve:

Find the values of $x$ that satisfy both inequalities simultaneously.

Solution:

We need to solve each inequality separately and then find the common solution.

Let's solve the first inequality (1):

$5(2x - 7) - 3(2x + 3) \leq 0$

First, distribute the numbers outside the parentheses:

$5 \times 2x - 5 \times 7 - 3 \times 2x - 3 \times 3 \leq 0$

This simplifies to:

$10x - 35 - 6x - 9 \leq 0$

Combine the like terms ($x$ terms and constant terms) on the left side:

$(10x - 6x) + (-35 - 9) \leq 0$

This gives:

$4x - 44 \leq 0$

Now, add $44$ to both sides of the inequality:

$4x - 44 + 44 \leq 0 + 44$

This simplifies to:

$4x \leq 44$

Finally, divide both sides by $4$. Since $4$ is a positive number, the inequality sign remains the same:

$\frac{4x}{4} \leq \frac{44}{4}$

This gives us:

Now, let's solve the second inequality (2):

$2x + 19 \leq 6x + 47$

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

$2x - 2x + 19 \leq 6x - 2x + 47$

This simplifies to:

$19 \leq 4x + 47$

Next, subtract $47$ from both sides of the inequality:

$19 - 47 \leq 4x + 47 - 47$

This simplifies to:

$-28 \leq 4x$

Finally, divide both sides by $4$. Since $4$ is a positive number, the inequality sign remains the same:

$\frac{-28}{4} \leq \frac{4x}{4}$

This gives us:

$-7 \leq x$

This inequality can also be written as:

The solution to the system of inequalities is the set of all real numbers $x$ that satisfy both $x \leq 11$ (from inequality 3) and $x \geq -7$ (from inequality 4) simultaneously.

Combining these two conditions, we get that $x$ must be greater than or equal to $-7$ AND less than or equal to $11$.

So, the combined condition is:

Thus, the solution set consists of all real numbers between $-7$ and $11$, including $-7$ and $11$.

In interval notation, the solution set is $\left[-7, 11\right]$.

Graph of the solution on number line:

To show the solution $-7 \leq x \leq 11$ on a number line, we represent all real numbers between $-7$ and $11$, including the endpoints.

1. Draw a number line.

2. Locate the numbers $-7$ and $11$ on the number line.

3. Since the inequalities are $\leq$ and $\geq$, the numbers $-7$ and $11$ themselves are included in the solution set. This is indicated by closed circles (or solid dots) at the points representing $-7$ and $11$ on the number line.

4. All numbers greater than or equal to $-7$ and less than or equal to $11$ are part of the solution. This is indicated by drawing a thick line or shading the region of the number line between $-7$ and $11$.

Answer:

The solution to the given system of inequalities for real $x$ is $-7 \leq x \leq 11$. The graph shows all real numbers between $-7$ and $11$, including $-7$ and $11$.

Given:

The temperature range in Fahrenheit is between 68° F and 77° F.

The conversion formula from Celsius (C) to Fahrenheit (F) is $F = \frac{9}{5} C + 32$.

To Find:

The range in temperature in degrees Celsius (C).

Solution:

The given temperature range in Fahrenheit can be expressed as the inequality:

$68 < F < 77$

Substitute the given conversion formula $F = \frac{9}{5} C + 32$ into the inequality:

$68 < \frac{9}{5} C + 32 < 77$

To solve for C, first subtract 32 from all parts of the inequality:

$68 - 32 < \frac{9}{5} C + 32 - 32 < 77 - 32$

$36 < \frac{9}{5} C < 45$

Now, multiply all parts of the inequality by $\frac{5}{9}$ to isolate C. Since $\frac{5}{9}$ is positive, the inequality signs remain unchanged:

$36 \times \frac{5}{9} < \frac{9}{5} C \times \frac{5}{9} < 45 \times \frac{5}{9}$

$\frac{\cancel{36}^{4} \times 5}{\cancel{9}_{1}} < C < \frac{\cancel{45}^{5} \times 5}{\cancel{9}_{1}}$

$4 \times 5 < C < 5 \times 5$

$20 < C < 25$

Thus, the temperature range in degrees Celsius is between 20°C and 25°C.

Given:

Volume of 8% boric acid solution = 640 litres.

Concentration of the first solution = 8% boric acid.

Concentration of the second solution (to be added) = 2% boric acid.

Desired concentration of the resulting mixture is more than 4% and less than 6% boric acid.

To Find:

The quantity (in litres) of the 2% boric acid solution that needs to be added.

Solution:

Let $x$ be the number of litres of the 2% boric acid solution to be added. We must have $x > 0$ since a quantity of solution is being added.

The amount of boric acid in 640 litres of 8% solution is:

$8\% \text{ of } 640 = \frac{8}{100} \times 640 = 0.08 \times 640 = 51.2$ litres.

The amount of boric acid in $x$ litres of 2% solution is:

$2\% \text{ of } x = \frac{2}{100} \times x = 0.02x$ litres.

The total volume of the resulting mixture will be the sum of the volumes of the two solutions:

Total volume $= (640 + x)$ litres.

The total amount of boric acid in the resulting mixture will be the sum of the amounts of boric acid from the two solutions:

Total amount of boric acid $= (51.2 + 0.02x)$ litres.

The concentration of boric acid in the resulting mixture is given by:

Concentration $= \frac{\text{Total amount of boric acid}}{\text{Total volume of mixture}} \times 100\%$

Concentration $= \frac{51.2 + 0.02x}{640 + x} \times 100\%$

According to the problem, the resulting mixture is to be more than 4% but less than 6% boric acid. This can be expressed as the following inequality:

$4 < \frac{51.2 + 0.02x}{640 + x} \times 100 < 6$

Divide all parts of the inequality by 100:

$\frac{4}{100} < \frac{51.2 + 0.02x}{640 + x} < \frac{6}{100}$

$0.04 < \frac{51.2 + 0.02x}{640 + x} < 0.06$

Since $x > 0$, the total volume $(640 + x)$ is positive. We can multiply all parts of the inequality by $(640 + x)$ without changing the direction of the inequality signs. This gives us two separate inequalities:

Inequality 1:

$0.04(640 + x) < 51.2 + 0.02x$

$0.04 \times 640 + 0.04x < 51.2 + 0.02x$

$25.6 + 0.04x < 51.2 + 0.02x$

Subtract $0.02x$ from both sides:

$25.6 + 0.04x - 0.02x < 51.2 + 0.02x - 0.02x$

$25.6 + 0.02x < 51.2$

Subtract 25.6 from both sides:

$25.6 + 0.02x - 25.6 < 51.2 - 25.6$

$0.02x < 25.6$

Divide both sides by 0.02 (positive number):

$x < \frac{25.6}{0.02}$

$x < \frac{2560}{2}$

$x < 1280$

Inequality 2:

$51.2 + 0.02x < 0.06(640 + x)$

$51.2 + 0.02x < 0.06 \times 640 + 0.06x$

$51.2 + 0.02x < 38.4 + 0.06x$

Subtract $0.02x$ from both sides:

$51.2 + 0.02x - 0.02x < 38.4 + 0.06x - 0.02x$

$51.2 < 38.4 + 0.04x$

Subtract 38.4 from both sides:

$51.2 - 38.4 < 38.4 + 0.04x - 38.4$

$12.8 < 0.04x$

Divide both sides by 0.04 (positive number):

$\frac{12.8}{0.04} < x$

$\frac{1280}{4} < x$

$320 < x$

Combining the results from Inequality 1 ($x < 1280$) and Inequality 2 ($x > 320$), we get the range for $x$:

$320 < x < 1280$

Thus, the number of litres of the 2% boric acid solution to be added must be more than 320 litres and less than 1280 litres.

Given:

Initial volume of 45% acid solution = 1125 litres.

Added solution = water (0% acid).

Desired resulting mixture concentration: more than 25% and less than 30% acid.

To Find:

The quantity (in litres) of water ($x$) to be added.

Solution:

Let $x$ be the number of litres of water added. We assume $x > 0$, as water is being added.

The amount of acid in the initial 1125 litres of 45% solution is:

Amount of acid $= 45\% \text{ of } 1125 = \frac{45}{100} \times 1125 = 0.45 \times 1125 = 506.25$ litres.

When $x$ litres of water (which contains 0% acid) are added, the amount of acid in the mixture remains $506.25$ litres, as only water is added.

The total volume of the resulting mixture is the sum of the initial volume and the added water volume:

Total volume $= (1125 + x)$ litres.

The concentration of acid in the resulting mixture is calculated as the ratio of the total amount of acid to the total volume, multiplied by 100:

Concentration $= \frac{\text{Amount of acid}}{\text{Total volume}} \times 100\% = \frac{506.25}{1125 + x} \times 100 \%$.

According to the problem, the resulting mixture must contain more than 25% but less than 30% acid content. This gives us the following compound inequality:

$25 < \frac{506.25}{1125 + x} \times 100 < 30$

Divide all parts of the inequality by 100:

$\frac{25}{100} < \frac{506.25}{1125 + x} < \frac{30}{100}$

$0.25 < \frac{506.25}{1125 + x} < 0.30$

Since $x$ represents a volume of water added, $x > 0$. Therefore, the total volume $(1125 + x)$ is positive. We can multiply all parts of the inequality by $(1125 + x)$ without changing the direction of the inequality signs. This results in two separate inequalities:

1) $0.25(1125 + x) < 506.25$

2) $506.25 < 0.30(1125 + x)$

Solve Inequality 1:

$0.25 \times 1125 + 0.25x < 506.25$

$281.25 + 0.25x < 506.25$

Subtract 281.25 from both sides:

$0.25x < 506.25 - 281.25$

$0.25x < 225$

Divide both sides by 0.25 (which is positive):

$x < \frac{225}{0.25}$

$x < \frac{225 \times 100}{0.25 \times 100}$

$x < \frac{22500}{25}$

$x < 900$

Solve Inequality 2:

$506.25 < 0.30(1125 + x)$

$506.25 < 0.30 \times 1125 + 0.30x$

$506.25 < 337.5 + 0.30x$

Subtract 337.5 from both sides:

$506.25 - 337.5 < 0.30x$

$168.75 < 0.30x$

Divide both sides by 0.30 (which is positive):

$\frac{168.75}{0.30} < x$

$\frac{168.75 \times 100}{0.30 \times 100} < x$

$\frac{16875}{30} < x$

Performing the division: $\frac{16875}{30} = 562.5$.

$562.5 < x$

Combining the results from both inequalities, $x < 900$ and $x > 562.5$, we find the range for $x$:

$562.5 < x < 900$

Thus, the number of litres of water to be added must be more than 562.5 litres and less than 900 litres.

Given:

IQ formula: $IQ = \frac{MA}{CA} \times 100$, where MA is mental age and CA is chronological age.

IQ range for a group of 12-year-old children: $80 \leq IQ \leq 140$.

Chronological Age (CA) of the children = 12 years.

To Find:

The range of their Mental Age (MA).

Solution:

The given IQ range is:

$80 \leq IQ \leq 140$

Substitute the formula for IQ and the given CA = 12 into the inequality:

$80 \leq \frac{MA}{12} \times 100 \leq 140$

Simplify the middle part:

$80 \leq \frac{100 MA}{12} \leq 140$

$80 \leq \frac{25 MA}{3} \leq 140$

To isolate MA, first multiply all parts of the inequality by 3. Since 3 is positive, the inequality signs remain unchanged:

$80 \times 3 \leq \frac{25 MA}{3} \times 3 \leq 140 \times 3$

$240 \leq 25 MA \leq 420$

Now, divide all parts of the inequality by 25. Since 25 is positive, the inequality signs remain unchanged:

$\frac{240}{25} \leq \frac{25 MA}{25} \leq \frac{420}{25}$

Simplify the fractions:

$\frac{48}{5} \leq MA \leq \frac{84}{5}$

Convert the fractions to decimal form:

$\frac{48}{5} = 9.6$

$\frac{84}{5} = 16.8$

So, the inequality for MA is:

$9.6 \leq MA \leq 16.8$

Thus, the range of mental age for the group of 12-year-old children is from 9.6 years to 16.8 years, inclusive.