Chapter 3 Pair Of Linear Equations In Two Variables

Algebraic Representation:

First, let's define the variables.

Let $x$ be the number of rides Akhila had on the Giant Wheel.

Let $y$ be the number of times she played Hoopla.

Now, we translate the given conditions into algebraic equations.

Condition 1: The number of times she played Hoopla ($y$) is half the number of rides she had on the Giant Wheel ($x$).

$y = \frac{1}{2}x$

Rearranging this equation to the standard form $ax+by+c=0$:

Condition 2: Each ride costs $\textsf{₹}$ 3, and each game of Hoopla costs $\textsf{₹}$ 4. She spent a total of $\textsf{₹}$ 20.

Total cost = (Cost per ride $\times$ Number of rides) + (Cost per game $\times$ Number of games)

$20 = (3 \times x) + (4 \times y)$

The pair of linear equations that algebraically represents the situation is:

$x - 2y = 0$

$3x + 4y = 20$

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Graphical Representation:

To represent the situation graphically, we need to find at least two points for each equation to draw the corresponding lines.

For equation (i): $x - 2y = 0 \implies x = 2y$

x	0	2	4
y = x/2	0	1	2

For equation (ii): $3x + 4y = 20 \implies 4y = 20 - 3x \implies y = \frac{20-3x}{4}$

x	0	4	$20/3$
y	5	2	0

Now, we plot these points on a graph and draw the two lines.

The two lines intersect at the point $(4, 2)$. This point is the unique solution to the pair of equations, which means Akhila had 4 rides on the Giant Wheel and played Hoopla 2 times.

Algebraic Representation:

Let the cost of 1 pencil be $\textsf{₹ } x$.

Let the cost of 1 eraser be $\textsf{₹ } y$.

Romila's purchase: 2 pencils and 3 erasers for $\textsf{₹ } 9$.

Sonali's purchase: 4 pencils and 6 erasers for $\textsf{₹ } 18$.

The pair of linear equations representing the situation is:

$2x + 3y = 9$

$4x + 6y = 18$

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Graphical Representation:

To represent the equations graphically, we find points on each line.

For equation (i): $2x + 3y = 9 \implies y = \frac{9 - 2x}{3}$

x	0	4.5	3
y	3	0	1

For equation (ii): $4x + 6y = 18$. We can simplify this by dividing the entire equation by 2, which gives $2x + 3y = 9$. This is the same as equation (i).

Since both equations are equivalent, they will be represented by the exact same line on the graph.

The graph shows that the two lines are coincident. This means that every point on the line is a solution to both equations. The system has infinitely many solutions.

Given Equations:

The two equations representing the rails are:

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Geometric Representation:

To represent this situation geometrically, we need to plot the graphs of these two linear equations.

For equation (i): $x + 2y = 4 \implies y = \frac{4 - x}{2}$

x	0	4	2
y	2	0	1

For equation (ii): $2x + 4y = 12$. Simplifying by dividing by 2 gives $x + 2y = 6 \implies y = \frac{6 - x}{2}$.

x	0	6	2
y	3	0	2

Now, we plot the points for each equation and draw the corresponding lines on a graph.

The geometric representation shows two parallel lines. This indicates that the rails never meet, and the system of equations has no solution.

Algebraic Representation:

Let the present age of Aftab be $x$ years.

Let the present age of his daughter be $y$ years.

Condition 1: Seven years ago

Aftab's age was $(x - 7)$ years.

His daughter's age was $(y - 7)$ years.

According to the problem, Aftab was seven times as old as his daughter:

$x - 7 = 7(y - 7)$

$x - 7 = 7y - 49$

Condition 2: Three years from now

Aftab's age will be $(x + 3)$ years.

His daughter's age will be $(y + 3)$ years.

According to the problem, Aftab will be three times as old as his daughter:

$x + 3 = 3(y + 3)$

$x + 3 = 3y + 9$

The algebraic representation of the situation is the pair of linear equations (i) and (ii).

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Graphical Representation:

To represent the equations graphically, we find points for each line.

For equation (i): $x - 7y = -42 \implies x = 7y - 42$

y	5	6	7
x	-7	0	7

For equation (ii): $x - 3y = 6 \implies x = 3y + 6$

y	0	-1	-2
x	6	3	0

We now plot these points on a graph and draw the lines. The x-axis represents Aftab's age and the y-axis represents his daughter's age.

The two lines intersect at the point $(42, 12)$, which is the solution to the problem. This means Aftab's present age is 42 years and his daughter's present age is 12 years.

Algebraic Representation:

Let the cost of one bat be $\textsf{₹ } x$.

Let the cost of one ball be $\textsf{₹ } y$.

First purchase: 3 bats and 6 balls for $\textsf{₹ } 3900$.

Second purchase: 1 bat and 3 balls for $\textsf{₹ } 1300$.

The algebraic representation is the pair of equations (i) and (ii).

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Geometric Representation:

We find points to plot for each line.

For equation (i): $3x + 6y = 3900$. We can simplify this by dividing by 3: $x + 2y = 1300 \implies x = 1300 - 2y$.

y	0	300	500
x	1300	700	300

For equation (ii): $x + 3y = 1300 \implies x = 1300 - 3y$

y	0	100	300
x	1300	1000	400

We plot these points and draw the lines on a graph, with the x-axis as the cost of bats and the y-axis as the cost of balls.

The two lines intersect at the point $(1300, 0)$. This is the unique solution where the cost of a bat is $\textsf{₹ } 1300$ and the cost of a ball is $\textsf{₹ } 0$.

Algebraic Representation:

Let the cost of 1 kg of apples be $\textsf{₹ } x$.

Let the cost of 1 kg of grapes be $\textsf{₹ } y$.

First situation: 2 kg of apples and 1 kg of grapes for $\textsf{₹ } 160$.

Second situation: 4 kg of apples and 2 kg of grapes for $\textsf{₹ } 300$.

The algebraic representation is the pair of equations (i) and (ii).

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Geometric Representation:

We find points to plot for each line.

For equation (i): $2x + y = 160 \implies y = 160 - 2x$

x	50	60	70
y	60	40	20

For equation (ii): $4x + 2y = 300$. Simplifying by dividing by 2 gives $2x + y = 150 \implies y = 150 - 2x$.

x	50	60	70
y	50	30	10

We plot these points and draw the lines on a graph, with the x-axis as the cost of apples and the y-axis as the cost of grapes.

The geometric representation shows two parallel lines. Since the lines never intersect, there is no solution to this system of equations, indicating an inconsistency in the pricing information provided.

To Check:

Whether the pair of equations $x + 3y = 6$ and $2x - 3y = 12$ is consistent by graphical method. If it is consistent, we need to find the solution.

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Solution:

A pair of linear equations is consistent if it has at least one solution. Graphically, this means the lines representing the equations either intersect at a single point or are coincident.

Let's find points to plot for each line.

For Equation (1): $x + 3y = 6 \implies y = \frac{6-x}{3}$

x	0	6
y	2	0

The points are A(0, 2) and B(6, 0).

For Equation (2): $2x - 3y = 12 \implies y = \frac{2x-12}{3}$

x	0	3	6
y	-4	-2	0

The points are C(0, -4) and D(6, 0).

Now, we plot these points on a graph and draw the lines.

From the graph, we can see that the two lines intersect at a single point, which is B(6, 0). Since the lines intersect, the system has a unique solution.

Therefore, the pair of equations is consistent.

The solution is the coordinates of the point of intersection.

Solution: $x = 6$, $y = 0$.

Given Equations:

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Analysis:

Let's simplify equation (ii) by multiplying the entire equation by 5 to remove the fractions:

$5 \cdot (3x - \frac{24}{5}y + \frac{3}{5}) = 5 \cdot 0$

$15x - 24y + 3 = 0$

Now, we can see that all terms in this equation are divisible by 3. Let's divide by 3:

$\frac{15x}{3} - \frac{24y}{3} + \frac{3}{3} = \frac{0}{3}$

$5x - 8y + 1 = 0$

This simplified equation is identical to equation (i). This means both equations represent the same line.

Let's find some points for the line $5x - 8y + 1 = 0$ (or $y = \frac{5x+1}{8}$) to plot it.

x	3	-5
y	2	-3

Points are (3, 2) and (-5, -3).

Since both equations represent the same line, the lines are coincident. This means they overlap at every point.

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Conclusion:

Because the lines are coincident, every point on the line is a solution to the system. Therefore, the pair of equations has infinitely many solutions.

Algebraic Representation:

Let $x$ be the number of pants purchased.

Let $y$ be the number of skirts purchased.

Statement 1: "The number of skirts is two less than twice the number of pants purchased."

Statement 2: "The number of skirts is four less than four times the number of pants purchased."

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Graphical Solution:

We will find the solution by plotting the graphs of these two equations.

Points for Equation (i): $y = 2x - 2$

x	0	1	2
y	-2	0	2

Points for Equation (ii): $y = 4x - 4$

x	0	1	2
y	-4	0	4

Now, we plot these points on a graph where the x-axis represents the number of pants and the y-axis represents the number of skirts.

The two lines intersect at the point (1, 0). This is the unique solution to the system of equations.

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Conclusion:

The solution is $x=1$ and $y=0$. This means:

Number of pants Champa bought = 1

Number of skirts Champa bought = 0

Part (i): Mathematics Quiz

1. Algebraic Representation:

Let $x$ be the number of boys and $y$ be the number of girls.

From the problem, we get two equations:

Total students are 10: $x + y = 10 \implies y = 10 - x$

Number of girls is 4 more than boys: $y = x + 4$

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2. Graphical Solution:

We find points for each line to plot them on a graph.

For $x + y = 10$:

x	0	10	3
y	10	0	7

For $y = x + 4$:

x	0	1	3
y	4	5	7

The two lines intersect at the point (3, 7).

Conclusion: The number of boys is 3 and the number of girls is 7.

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Part (ii): Pencils and Pens

1. Algebraic Representation:

Let the cost of one pencil be $\textsf{₹ } x$ and the cost of one pen be $\textsf{₹ } y$.

From the problem, we get two equations:

$5x + 7y = 50$

$7x + 5y = 46$

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2. Graphical Solution:

We find points for each line.

For $5x + 7y = 50 \implies y = \frac{50 - 5x}{7}$:

x	10	3	-4
y	0	5	10

For $7x + 5y = 46 \implies y = \frac{46 - 7x}{5}$:

x	3	8	-2
y	5	-2	12

The two lines intersect at the point (3, 5).

Conclusion: The cost of one pencil is $\textsf{₹ } 3$ and the cost of one pen is $\textsf{₹ } 5$.

For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we have the following conditions:

• Intersecting lines: $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
• Parallel lines: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
• Coincident lines: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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(i) $5x - 4y + 8 = 0$ and $7x + 6y - 9 = 0$

$a_1 = 5, b_1 = -4, c_1 = 8$

$a_2 = 7, b_2 = 6, c_2 = -9$

$\frac{a_{1}}{a_{2}} = \frac{5}{7}$

$\frac{b_{1}}{b_{2}} = \frac{-4}{6} = -\frac{2}{3}$

Since $\frac{5}{7} \neq -\frac{2}{3}$, we have $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$.

The lines are intersecting at a point.

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(ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$

$a_1 = 9, b_1 = 3, c_1 = 12$

$a_2 = 18, b_2 = 6, c_2 = 24$

$\frac{a_{1}}{a_{2}} = \frac{9}{18} = \frac{1}{2}$

$\frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}$

$\frac{c_{1}}{c_{2}} = \frac{12}{24} = \frac{1}{2}$

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$, the lines are coincident.

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(iii) $6x - 3y + 10 = 0$ and $2x - y + 9 = 0$

$a_1 = 6, b_1 = -3, c_1 = 10$

$a_2 = 2, b_2 = -1, c_2 = 9$

$\frac{a_{1}}{a_{2}} = \frac{6}{2} = 3$

$\frac{b_{1}}{b_{2}} = \frac{-3}{-1} = 3$

$\frac{c_{1}}{c_{2}} = \frac{10}{9}$

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$, the lines are parallel.

A pair of linear equations is consistent if it has one or infinitely many solutions. It is inconsistent if it has no solution.

• Consistent (Intersecting or Coincident lines): $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ or $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
• Inconsistent (Parallel lines): $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

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(i) $3x + 2y = 5$ and $2x - 3y = 7$

Here $a_1=3, b_1=2, a_2=2, b_2=-3$.

$\frac{a_{1}}{a_{2}} = \frac{3}{2}$

$\frac{b_{1}}{b_{2}} = \frac{2}{-3}$

Since $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$, the equations are consistent.

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(ii) $2x - 3y = 8$ and $4x - 6y = 9$

Here $a_1=2, b_1=-3, c_1=-8$ and $a_2=4, b_2=-6, c_2=-9$.

$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_{1}}{b_{2}} = \frac{-3}{-6} = \frac{1}{2}$

$\frac{c_{1}}{c_{2}} = \frac{-8}{-9} = \frac{8}{9}$

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$, the equations are inconsistent.

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(iii) $\frac{3}{2}x + \frac{5}{3}y = 7$ and $9x - 10y = 14$

Here $a_1 = \frac{3}{2}, b_1 = \frac{5}{3}, a_2 = 9, b_2 = -10$.

$\frac{a_{1}}{a_{2}} = \frac{3/2}{9} = \frac{3}{18} = \frac{1}{6}$

$\frac{b_{1}}{b_{2}} = \frac{5/3}{-10} = \frac{5}{-30} = -\frac{1}{6}$

Since $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$, the equations are consistent.

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(iv) $5x - 3y = 11$ and $-10x + 6y = -22$

Here $a_1=5, b_1=-3, c_1=-11$ and $a_2=-10, b_2=6, c_2=22$.

$\frac{a_{1}}{a_{2}} = \frac{5}{-10} = -\frac{1}{2}$

$\frac{b_{1}}{b_{2}} = \frac{-3}{6} = -\frac{1}{2}$

$\frac{c_{1}}{c_{2}} = \frac{-11}{22} = -\frac{1}{2}$

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$, the equations are consistent.

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(v) $\frac{4}{3}x + 2y = 8$ and $2x + 3y = 12$

Here $a_1 = \frac{4}{3}, b_1 = 2, c_1 = -8$ and $a_2=2, b_2=3, c_2=-12$.

$\frac{a_{1}}{a_{2}} = \frac{4/3}{2} = \frac{4}{6} = \frac{2}{3}$

$\frac{b_{1}}{b_{2}} = \frac{2}{3}$

$\frac{c_{1}}{c_{2}} = \frac{-8}{-12} = \frac{2}{3}$

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$, the equations are consistent.

(i) $x + y = 5$, $2x + 2y = 10$

$\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair is consistent with infinitely many solutions (coincident lines).

Graphical Solution: Both equations represent the same line $y=5-x$.

Any point on this line is a solution, e.g., (1, 4), (2, 3), (3, 2).

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(ii) $x – y = 8$, $3x – 3y = 16$

$\frac{a_1}{a_2} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}$, $\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair is inconsistent (parallel lines) and has no solution.

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(iii) $2x + y – 6 = 0$, $4x – 2y – 4 = 0$

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$.

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair is consistent with a unique solution (intersecting lines).

Graphical Solution:

For $y = 6 - 2x$: Points are (0,6), (3,0), (2,2).

For $y = 2x - 2$: Points are (0,-2), (1,0), (2,2).

The lines intersect at (2, 2). The solution is $x=2, y=2$.

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(iv) $2x – 2y – 2 = 0$, $4x – 4y – 5 = 0$

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair is inconsistent (parallel lines) and has no solution.

Algebraic Representation:

Let $l$ be the length and $w$ be the width of the rectangular garden.

Condition 1: Half the perimeter is 36 m.

The perimeter is $P = 2(l+w)$. Half the perimeter is $\frac{P}{2} = l+w$.

Condition 2: Length is 4 m more than its width.

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Solution:

We can solve this system of equations. Substitute the expression for $l$ from equation (ii) into equation (i):

$(w + 4) + w = 36$

$2w + 4 = 36$

$2w = 32$

$w = 16$

Now, find the length $l$ using equation (ii):

$l = w + 4 = 16 + 4 = 20$

The dimensions of the garden are 20 m and 16 m.

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Final Answer:

The length of the garden is 20 m.

The width of the garden is 16 m.

The given equation is $2x + 3y - 8 = 0$. Here, $a_1 = 2, b_1 = 3, c_1 = -8$.

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(i) Intersecting lines

The condition is $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$.

We need to choose $a_2$ and $b_2$ such that $\frac{2}{a_2} \neq \frac{3}{b_2}$. Many solutions are possible. For example, we can swap the coefficients.

A possible equation is $3x + 2y - 7 = 0$.

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(ii) Parallel lines

The condition is $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$.

We can multiply $a_1$ and $b_1$ by the same constant (e.g., 2) to get $a_2$ and $b_2$, and choose a different constant for $c_2$.

$a_2 = 2 \times 2 = 4$

$b_2 = 3 \times 2 = 6$

We need $c_2 \neq -8 \times 2$, so we can choose any number other than -16. Let's pick 5.

A possible equation is $4x + 6y + 5 = 0$.

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(iii) Coincident lines

The condition is $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$.

We can simply multiply the entire given equation by a constant (e.g., 2).

$2(2x + 3y - 8) = 2(0)$

A possible equation is $4x + 6y - 16 = 0$.

Given Equations:

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Finding Points for Graphing:

For equation (i): $y = x + 1$

x	-1	0	2
y	0	1	3

This line intersects the x-axis at (-1, 0).

For equation (ii): $y = \frac{12 - 3x}{2}$

x	0	4	2
y	6	0	3

This line intersects the x-axis at (4, 0).

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Vertices of the Triangle:

The vertices of the triangle are the points where the lines intersect each other and where they intersect the x-axis.

1. Intersection of the two lines: From the graph and the tables, the intersection point is (2, 3).

2. Intersection of the first line with the x-axis (where y=0): $x-0+1=0 \implies x=-1$. The point is (-1, 0).

3. Intersection of the second line with the x-axis (where y=0): $3x+2(0)-12=0 \implies 3x=12 \implies x=4$. The point is (4, 0).

The coordinates of the vertices of the triangle are (2, 3), (-1, 0), and (4, 0).

The triangular region formed by these vertices should be shaded on the graph.

Given Equations:

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Solution using Substitution Method:

Step 1: From equation (2), it is easy to express $x$ in terms of $y$.

Step 2: Substitute the expression for $x$ from equation (3) into equation (1).

$7(3 - 2y) - 15y = 2$

Step 3: Solve the resulting equation for $y$.

$21 - 14y - 15y = 2$

$21 - 29y = 2$

$-29y = 2 - 21$

$-29y = -19$

$y = \frac{-19}{-29} = \frac{19}{29}$

Step 4: Substitute the value of $y$ back into equation (3) to find $x$.

$x = 3 - 2y = 3 - 2\left(\frac{19}{29}\right)$

$x = 3 - \frac{38}{29} = \frac{3 \times 29 - 38}{29} = \frac{87 - 38}{29} = \frac{49}{29}$

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Result:

The solution is $x = \frac{49}{29}$ and $y = \frac{19}{29}$.

Algebraic Representation (from Q.1 of Ex 3.1):

Let Aftab's present age be $x$ and his daughter's present age be $y$.

The pair of linear equations representing the problem is:

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Solution using Substitution Method:

Step 1: From equation (ii), express $x$ in terms of $y$.

Step 2: Substitute this expression for $x$ into equation (i).

$(6 + 3y) - 7y = -42$

Step 3: Solve the resulting equation for $y$.

$6 - 4y = -42$

$-4y = -42 - 6$

$-4y = -48$

$y = 12$

Step 4: Substitute the value of $y$ back into equation (iii) to find $x$.

$x = 6 + 3(12)$

$x = 6 + 36$

$x = 42$

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Result:

The solution is $x = 42$ and $y = 12$.

Aftab's present age is 42 years and his daughter's present age is 12 years.

Algebraic Representation (from Example 2):

Let the cost of one pencil be $\textsf{₹ } x$ and one eraser be $\textsf{₹ } y$.

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Solution using Substitution Method:

Step 1: From equation (1), express $x$ in terms of $y$.

$2x = 9 - 3y \implies x = \frac{9 - 3y}{2}$

Step 2: Substitute this expression for $x$ into equation (2).

$4\left(\frac{9 - 3y}{2}\right) + 6y = 18$

Step 3: Solve the resulting equation.

$2(9 - 3y) + 6y = 18$

$18 - 6y + 6y = 18$

$18 = 18$

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Conclusion:

The resulting statement, $18 = 18$, is true for all values of $y$. This means that the two equations are dependent (they represent the same line).

Therefore, the system has infinitely many solutions, and we cannot determine a unique cost for each pencil and eraser from the given information.

Given Equations (from Example 3):

We need to determine if the rails will cross, which means we need to check if the system of equations has a solution.

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Solution using Substitution Method:

Step 1: From equation (1), express $x$ in terms of $y$.

$x = 4 - 2y$

Step 2: Substitute this expression for $x$ into equation (2).

$2(4 - 2y) + 4y - 12 = 0$

Step 3: Solve the resulting equation.

$8 - 4y + 4y - 12 = 0$

$-4 = 0$

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Conclusion:

The resulting statement, $-4 = 0$, is a false statement. This indicates that there is no value of $x$ and $y$ that can satisfy both equations simultaneously.

Therefore, the system has no solution. Geometrically, this means the lines are parallel.

The rails will not cross each other.

(i) $x + y = 14$ and $x - y = 4$

The given equations are:

From equation (2), we can express $x$ in terms of $y$:

Substitute the value of $x$ from equation (3) into equation (1):

$(4 + y) + y = 14$

$4 + 2y = 14$

$2y = 14 - 4$

$2y = 10$

$y = 5$

Substitute the value of $y = 5$ back into equation (3):

$x = 4 + 5$

$x = 9$

Therefore, the solution is $x = 9$ and $y = 5$.

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(ii) $s - t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$

The given equations are:

From equation (1), we express $s$ in terms of $t$:

Substitute the value of $s$ from equation (3) into equation (2):

$\frac{3 + t}{3} + \frac{t}{2} = 6$

To eliminate fractions, multiply the entire equation by the LCM of 3 and 2, which is 6:

$6 \left( \frac{3 + t}{3} \right) + 6 \left( \frac{t}{2} \right) = 6(6)$

$2(3 + t) + 3(t) = 36$

$6 + 2t + 3t = 36$

$6 + 5t = 36$

$5t = 36 - 6$

$5t = 30$

$t = 6$

Substitute the value of $t = 6$ back into equation (3):

$s = 3 + 6$

$s = 9$

Therefore, the solution is $s = 9$ and $t = 6$.

---

(iii) $3x - y = 3$ and $9x - 3y = 9$

The given equations are:

From equation (1), we express $y$ in terms of $x$:

Substitute the value of $y$ from equation (3) into equation (2):

$9x - 3(3x - 3) = 9$

$9x - 9x + 9 = 9$

$9 = 9$

This statement is true for all values of $x$. This means the equations are dependent (represent the same line).

Therefore, the pair of linear equations has infinitely many solutions. Any pair $(x, y)$ satisfying $y = 3x - 3$ is a solution.

---

(iv) $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$

The given equations are:

$0.2x + 0.3y = 1.3$

$0.4x + 0.5y = 2.3$

Multiply both equations by 10 to remove decimals:

From equation (1), express $x$ in terms of $y$:

$2x = 13 - 3y$

Substitute the value of $x$ from equation (3) into equation (2):

$4 \left( \frac{13 - 3y}{2} \right) + 5y = 23$

$2(13 - 3y) + 5y = 23$

$26 - 6y + 5y = 23$

$26 - y = 23$

$-y = 23 - 26$

$-y = -3$

$y = 3$

Substitute the value of $y = 3$ back into equation (3):

$x = \frac{13 - 3(3)}{2}$

$x = \frac{13 - 9}{2}$

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

$x = 2$

Therefore, the solution is $x = 2$ and $y = 3$.

---

(v) $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x - \sqrt{8}y = 0$

The given equations are:

From equation (1), express $x$ in terms of $y$:

$\sqrt{2}x = -\sqrt{3}y$

Substitute the value of $x$ from equation (3) into equation (2):

$\sqrt{3} \left( -\frac{\sqrt{3}}{\sqrt{2}} y \right) - \sqrt{8}y = 0$

$-\frac{3}{\sqrt{2}} y - \sqrt{8}y = 0$

Noting that $\sqrt{8} = 2\sqrt{2}$:

$-\frac{3}{\sqrt{2}} y - 2\sqrt{2}y = 0$

Multiply by $\sqrt{2}$ to clear the denominator:

$-3y - (2\sqrt{2})(\sqrt{2})y = 0$

$-3y - 4y = 0$

$-7y = 0$

$y = 0$

Substitute the value of $y = 0$ back into equation (3):

$x = -\frac{\sqrt{3}}{\sqrt{2}} (0)$

$x = 0$

Therefore, the solution is $x = 0$ and $y = 0$.

---

(vi) $\frac{3x}{2} - \frac{5y}{3} = -2$ and $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

The given equations are:

$\frac{3x}{2} - \frac{5y}{3} = -2$

$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

Clear the fractions by multiplying by the LCM of the denominators.

For the first equation, LCM(2, 3) = 6:

$6 \left( \frac{3x}{2} \right) - 6 \left( \frac{5y}{3} \right) = 6(-2)$

For the second equation, LCM(3, 2, 6) = 6:

$6 \left( \frac{x}{3} \right) + 6 \left( \frac{y}{2} \right) = 6 \left( \frac{13}{6} \right)$

From equation (2), express $x$ in terms of $y$:

$2x = 13 - 3y$

Substitute the value of $x$ from equation (3) into equation (1):

$9 \left( \frac{13 - 3y}{2} \right) - 10y = -12$

Multiply the entire equation by 2 to clear the fraction:

$9(13 - 3y) - 2(10y) = 2(-12)$

$117 - 27y - 20y = -24$

$117 - 47y = -24$

$-47y = -24 - 117$

$-47y = -141$

$y = \frac{-141}{-47}$

$y = 3$

Substitute the value of $y = 3$ back into equation (3):

$x = \frac{13 - 3(3)}{2}$

$x = \frac{13 - 9}{2}$

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

$x = 2$

Therefore, the solution is $x = 2$ and $y = 3$.

Given Equations:

And the relation:

---

Solving the pair of linear equations:

We can use the substitution method or elimination method. Let's use the substitution method.

From equation (i), express $x$ in terms of $y$:

$2x = 11 - 3y$

Substitute this expression for $x$ into equation (ii):

$2 \left( \frac{11 - 3y}{2} \right) - 4y = -24$

$(11 - 3y) - 4y = -24$

$11 - 7y = -24$

$-7y = -24 - 11$

$-7y = -35$

$y = \frac{-35}{-7}$

$y = 5$

Now substitute the value of $y = 5$ back into equation (iv) to find $x$:

$x = \frac{11 - 3(5)}{2}$

$x = \frac{11 - 15}{2}$

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

$x = -2$

So, the solution to the pair of linear equations is $x = -2$ and $y = 5$.

---

Finding the value of 'm':

Now substitute the values $x = -2$ and $y = 5$ into equation (iii):

$y = mx + 3$

$5 = m(-2) + 3$

$5 = -2m + 3$

$5 - 3 = -2m$

$2 = -2m$

$m = \frac{2}{-2}$

$m = -1$

---

Result:

The solution to the system of equations is $x = -2$, $y = 5$.

The value of $m$ for which $y = mx + 3$ is $m = -1$.

(i) Two Numbers

Let the two numbers be $x$ and $y$. Assume $x > y$.

According to the problem:

Substitute equation (2) into equation (1):

$(3y) - y = 26$

$2y = 26$

$y = \frac{26}{2}$

$y = 13$

Substitute the value of $y = 13$ back into equation (2):

$x = 3(13)$

$x = 39$

The two numbers are 39 and 13.

---

(ii) Supplementary Angles

Let the larger angle be $x$ degrees and the smaller angle be $y$ degrees.

Since the angles are supplementary:

The larger angle exceeds the smaller by 18 degrees:

Substitute equation (2) into equation (1):

$(y + 18) + y = 180$

$2y + 18 = 180$

$2y = 180 - 18$

$2y = 162$

$y = \frac{162}{2}$

$y = 81$

Substitute the value of $y = 81$ back into equation (2):

$x = 81 + 18$

$x = 99$

The two angles are 99 degrees and 81 degrees.

---

(iii) Bats and Balls

Let the cost of one bat be $\textsf{₹ } x$ and the cost of one ball be $\textsf{₹ } y$.

According to the problem:

From equation (2), express $x$ in terms of $y$:

$3x = 1750 - 5y$

Substitute the value of $x$ from equation (3) into equation (1):

$7 \left( \frac{1750 - 5y}{3} \right) + 6y = 3800$

Multiply by 3 to clear the fraction:

$7(1750 - 5y) + 3(6y) = 3(3800)$

$12250 - 35y + 18y = 11400$

$12250 - 17y = 11400$

$-17y = 11400 - 12250$

$-17y = -850$

$y = \frac{-850}{-17}$

$y = 50$

Substitute the value of $y = 50$ back into equation (3):

$x = \frac{1750 - 5(50)}{3}$

$x = \frac{1750 - 250}{3}$

$x = \frac{1500}{3}$

$x = 500$

The cost of one bat is $\textsf{₹ } 500$ and the cost of one ball is $\textsf{₹ } 50$.

---

(iv) Taxi Charges

Let the fixed charge be $\textsf{₹ } x$ and the charge per km be $\textsf{₹ } y$.

According to the problem:

From equation (1), express $x$ in terms of $y$:

Substitute the value of $x$ from equation (3) into equation (2):

$(105 - 10y) + 15y = 155$

$105 + 5y = 155$

$5y = 155 - 105$

$5y = 50$

$y = 10$

Substitute the value of $y = 10$ back into equation (3):

$x = 105 - 10(10)$

$x = 105 - 100$

$x = 5$

The fixed charge is $\textsf{₹ } 5$ and the charge per km is $\textsf{₹ } 10$.

Now, find the charge for travelling 25 km:

Total charge = Fixed charge + (Charge per km $\times$ Distance)

Total charge = $x + 25y$

Total charge = $5 + 25(10)$

Total charge = $5 + 250$

Total charge = $255$

The charge for travelling 25 km is $\textsf{₹ } 255$.

---

(v) Fraction

Let the fraction be $\frac{x}{y}$.

According to the problem:

If 2 is added to both numerator and denominator, the fraction becomes $\frac{9}{11}$:

$\frac{x+2}{y+2} = \frac{9}{11}$

$11(x+2) = 9(y+2)$

$11x + 22 = 9y + 18$

If 3 is added to both numerator and denominator, the fraction becomes $\frac{5}{6}$:

$\frac{x+3}{y+3} = \frac{5}{6}$

$6(x+3) = 5(y+3)$

$6x + 18 = 5y + 15$

From equation (2), express $x$ in terms of $y$:

$6x = 5y - 3$

Substitute the value of $x$ from equation (3) into equation (1):

$11 \left( \frac{5y - 3}{6} \right) - 9y = -4$

Multiply by 6 to clear the fraction:

$11(5y - 3) - 6(9y) = 6(-4)$

$55y - 33 - 54y = -24$

$y - 33 = -24$

$y = -24 + 33$

$y = 9$

Substitute the value of $y = 9$ back into equation (3):

$x = \frac{5(9) - 3}{6}$

$x = \frac{45 - 3}{6}$

$x = \frac{42}{6}$

$x = 7$

The fraction is $\frac{x}{y}$. Therefore, the fraction is $\frac{7}{9}$.

---

(vi) Ages

Let Jacob's present age be $x$ years and his son's present age be $y$ years.

Five years hence:

Jacob's age = $x+5$

Son's age = $y+5$

According to the problem: $x+5 = 3(y+5)$

$x+5 = 3y + 15$

Five years ago:

Jacob's age = $x-5$

Son's age = $y-5$

According to the problem: $x-5 = 7(y-5)$

$x-5 = 7y - 35$

From equation (1), express $x$ in terms of $y$:

Substitute the value of $x$ from equation (3) into equation (2):

$(3y + 10) - 7y = -30$

$-4y + 10 = -30$

$-4y = -30 - 10$

$-4y = -40$

$y = 10$

Substitute the value of $y = 10$ back into equation (3):

$x = 3(10) + 10$

$x = 30 + 10$

$x = 40$

Jacob's present age is 40 years and his son's present age is 10 years.

Given:

Ratio of monthly incomes of two persons = 9 : 7.

Ratio of their monthly expenditures = 4 : 3.

Monthly savings for each person = $\textsf{₹ } 2000$.

---

To Find:

The monthly incomes of the two persons.

---

Solution:

Let the monthly incomes of the two persons be $\textsf{₹ } 9x$ and $\textsf{₹ } 7x$ respectively.

Let their monthly expenditures be $\textsf{₹ } 4y$ and $\textsf{₹ } 3y$ respectively.

We know that: Income - Expenditure = Savings

For the first person:

For the second person:

Now, we need to solve this pair of linear equations for $x$. We can use the elimination method.

Multiply equation (i) by 3 and equation (ii) by 4 to make the coefficients of $y$ equal:

$3 \times (9x - 4y = 2000) \implies 27x - 12y = 6000$

$4 \times (7x - 3y = 2000) \implies 28x - 12y = 8000$

Subtract equation (iii) from equation (iv):

$(28x - 12y) - (27x - 12y) = 8000 - 6000$

$28x - 12y - 27x + 12y = 2000$

$x = 2000$

Now, calculate the monthly incomes:

Income of the first person = $9x = 9 \times 2000 = 18000$.

Income of the second person = $7x = 7 \times 2000 = 14000$.

---

Result:

The monthly incomes of the two persons are $\textsf{₹ } 18000$ and $\textsf{₹ } 14000$ respectively.

Given Equations:

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Solution using Elimination Method:

To use the elimination method, we make the coefficients of one variable equal. Let's make the coefficient of $x$ equal.

Multiply equation (1) by 2:

$2(2x + 3y) = 2(8)$

Now, we subtract equation (2) from our new equation (3):

$(4x + 6y) - (4x + 6y) = 16 - 7$

$0 = 9$

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Conclusion:

The result $0 = 9$ is a false statement. This indicates a contradiction, which means there is no pair of values $(x, y)$ that can satisfy both equations simultaneously.

Therefore, the given pair of linear equations has no solution.

Algebraic Representation:

Let the two-digit number be $10x + y$, where $x$ is the digit in the tens place and $y$ is the digit in the units place.

The number obtained by reversing the digits is $10y + x$.

According to the first condition, the sum of the number and the reversed number is 66:

$(10x + y) + (10y + x) = 66$

$11x + 11y = 66$

Divide the equation by 11:

According to the second condition, the digits differ by 2. This gives two possibilities:

Possibility 1: $x - y = 2$

Possibility 2: $y - x = 2$

---

Solving Case 1 (Equations (1) and (2)):

We have the system:

Using the elimination method, add equations (1) and (2):

$(x + y) + (x - y) = 6 + 2$

$2x = 8$

$x = 4$

Substitute $x = 4$ into equation (1):

$4 + y = 6$

$y = 2$

In this case, the number is $10x + y = 10(4) + 2 = 42$.

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Solving Case 2 (Equations (1) and (3)):

We have the system:

Using the elimination method, add equations (1) and (3):

$(x + y) + (-x + y) = 6 + 2$

$2y = 8$

$y = 4$

Substitute $y = 4$ into equation (1):

$x + 4 = 6$

$x = 2$

In this case, the number is $10x + y = 10(2) + 4 = 24$.

---

Conclusion:

The numbers satisfying the given conditions are 42 and 24.

There are two such numbers.

(i) $x + y = 5$ and $2x - 3y = 4$

The given equations are:

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Elimination Method:

Multiply equation (1) by 3:

Add equation (2) and equation (3):

$(2x - 3y) + (3x + 3y) = 4 + 15$

$5x = 19$

$x = \frac{19}{5}$

Substitute $x = \frac{19}{5}$ into equation (1):

$\frac{19}{5} + y = 5$

$y = 5 - \frac{19}{5}$

$y = \frac{25 - 19}{5}$

$y = \frac{6}{5}$

Solution by elimination: $x = \frac{19}{5}, y = \frac{6}{5}$.

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Substitution Method:

From equation (1), express $y$ in terms of $x$:

Substitute this value of $y$ into equation (2):

$2x - 3(5 - x) = 4$

$2x - 15 + 3x = 4$

$5x = 4 + 15$

$5x = 19$

$x = \frac{19}{5}$

Substitute $x = \frac{19}{5}$ back into equation (4):

$y = 5 - \frac{19}{5}$

$y = \frac{25 - 19}{5}$

$y = \frac{6}{5}$

Solution by substitution: $x = \frac{19}{5}, y = \frac{6}{5}$.

The solution is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

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(ii) $3x + 4y = 10$ and $2x - 2y = 2$

The given equations are:

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Elimination Method:

Multiply equation (2) by 2:

Add equation (1) and equation (3):

$(3x + 4y) + (4x - 4y) = 10 + 4$

$7x = 14$

$x = 2$

Substitute $x = 2$ into equation (2):

$2(2) - 2y = 2$

$4 - 2y = 2$

$-2y = 2 - 4$

$-2y = -2$

$y = 1$

Solution by elimination: $x = 2, y = 1$.

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Substitution Method:

From equation (2), divide by 2: $x - y = 1$.

Express $x$ in terms of $y$:

Substitute this value of $x$ into equation (1):

$3(y + 1) + 4y = 10$

$3y + 3 + 4y = 10$

$7y + 3 = 10$

$7y = 7$

$y = 1$

Substitute $y = 1$ back into equation (4):

$x = 1 + 1$

$x = 2$

Solution by substitution: $x = 2, y = 1$.

The solution is $x = 2$ and $y = 1$.

---

(iii) $3x - 5y - 4 = 0$ and $9x = 2y + 7$

Rewrite the equations in standard form:

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Elimination Method:

Multiply equation (1) by 3:

Subtract equation (3) from equation (2):

$(9x - 2y) - (9x - 15y) = 7 - 12$

$9x - 2y - 9x + 15y = -5$

$13y = -5$

$y = -\frac{5}{13}$

Substitute $y = -\frac{5}{13}$ into equation (1):

$3x - 5 \left( -\frac{5}{13} \right) = 4$

$3x + \frac{25}{13} = 4$

$3x = 4 - \frac{25}{13}$

$3x = \frac{52 - 25}{13}$

$3x = \frac{27}{13}$

$x = \frac{27}{13 \times 3}$

$x = \frac{9}{13}$

Solution by elimination: $x = \frac{9}{13}, y = -\frac{5}{13}$.

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Substitution Method:

From equation (1), express $x$ in terms of $y$:

$3x = 5y + 4$

Substitute this value of $x$ into equation (2):

$9 \left( \frac{5y + 4}{3} \right) - 2y = 7$

$3(5y + 4) - 2y = 7$

$15y + 12 - 2y = 7$

$13y + 12 = 7$

$13y = 7 - 12$

$13y = -5$

$y = -\frac{5}{13}$

Substitute $y = -\frac{5}{13}$ back into equation (4):

$x = \frac{5(-\frac{5}{13}) + 4}{3}$

$x = \frac{-\frac{25}{13} + 4}{3}$

$x = \frac{\frac{-25 + 52}{13}}{3}$

$x = \frac{\frac{27}{13}}{3}$

$x = \frac{9}{13}$

Solution by substitution: $x = \frac{9}{13}, y = -\frac{5}{13}$.

The solution is $x = \frac{9}{13}$ and $y = -\frac{5}{13}$.

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(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$

Clear the fractions first.

Multiply the first equation by 6 (LCM of 2 and 3):

$6 \left( \frac{x}{2} \right) + 6 \left( \frac{2y}{3} \right) = 6(-1)$

Multiply the second equation by 3:

$3(x) - 3 \left( \frac{y}{3} \right) = 3(3)$

---

Elimination Method:

Subtract equation (2) from equation (1):

$(3x + 4y) - (3x - y) = -6 - 9$

$3x + 4y - 3x + y = -15$

$5y = -15$

$y = -3$

Substitute $y = -3$ into equation (2):

$3x - (-3) = 9$

$3x + 3 = 9$

$3x = 6$

$x = 2$

Solution by elimination: $x = 2, y = -3$.

---

Substitution Method:

From equation (2), express $y$ in terms of $x$:

Substitute this value of $y$ into equation (1):

$3x + 4(3x - 9) = -6$

$3x + 12x - 36 = -6$

$15x = -6 + 36$

$15x = 30$

$x = 2$

Substitute $x = 2$ back into equation (3):

$y = 3(2) - 9$

$y = 6 - 9$

$y = -3$

Solution by substitution: $x = 2, y = -3$.

The solution is $x = 2$ and $y = -3$.

(i) Fraction Problem

Let the numerator of the fraction be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.

According to the first condition:

$\frac{x+1}{y-1} = 1$

$x+1 = y-1$

According to the second condition:

$\frac{x}{y+1} = \frac{1}{2}$

$2x = y+1$

Now, solve equations (1) and (2) using the elimination method.

Subtract equation (1) from equation (2):

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

$2x - y - x + y = 3$

$x = 3$

Substitute $x = 3$ into equation (1):

$3 - y = -2$

$-y = -2 - 3$

$-y = -5$

$y = 5$

Therefore, the fraction is $\frac{3}{5}$.

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(ii) Age Problem

Let Nuri's present age be $n$ years and Sonu's present age be $s$ years.

Five years ago:

Nuri's age = $n-5$

Sonu's age = $s-5$

Condition: $n-5 = 3(s-5)$

$n-5 = 3s - 15$

Ten years later:

Nuri's age = $n+10$

Sonu's age = $s+10$

Condition: $n+10 = 2(s+10)$

$n+10 = 2s + 20$

Now, solve equations (1) and (2) using the elimination method.

Subtract equation (1) from equation (2):

$(n - 2s) - (n - 3s) = 10 - (-10)$

$n - 2s - n + 3s = 20$

$s = 20$

Substitute $s = 20$ into equation (2):

$n - 2(20) = 10$

$n - 40 = 10$

$n = 50$

Therefore, Nuri's present age is 50 years and Sonu's present age is 20 years.

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(iii) Two-Digit Number Problem

Let the digit in the tens place be $x$ and the digit in the units place be $y$.

The two-digit number is $10x + y$.

The number obtained by reversing the digits is $10y + x$.

According to the first condition (sum of digits is 9):

According to the second condition (nine times the number is twice the reversed number):

$9(10x + y) = 2(10y + x)$

$90x + 9y = 20y + 2x$

$90x - 2x = 20y - 9y$

$88x = 11y$

$8x = y$

Now, solve equations (1) and (2) using the elimination method.

Add equation (1) and equation (2):

$(x + y) + (8x - y) = 9 + 0$

$9x = 9$

$x = 1$

Substitute $x = 1$ into equation (1):

$1 + y = 9$

$y = 8$

The number is $10x + y = 10(1) + 8 = 18$.

Therefore, the number is 18.

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(iv) Bank Notes Problem

Let the number of $\textsf{₹ } 50$ notes be $f$ and the number of $\textsf{₹ } 100$ notes be $h$.

According to the first condition (total number of notes is 25):

According to the second condition (total value is $\textsf{₹ } 2000$):

$50f + 100h = 2000$

Divide the equation by 50:

Now, solve equations (1) and (2) using the elimination method.

Subtract equation (1) from equation (2):

$(f + 2h) - (f + h) = 40 - 25$

$f + 2h - f - h = 15$

$h = 15$

Substitute $h = 15$ into equation (1):

$f + 15 = 25$

$f = 10$

Therefore, Meena received 10 notes of $\textsf{₹ } 50$ and 15 notes of $\textsf{₹ } 100$.

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(v) Library Charges Problem

Let the fixed charge for the first three days be $\textsf{₹ } x$.

Let the additional charge for each day thereafter be $\textsf{₹ } y$.

For Saritha (book kept for 7 days):

Number of extra days = $7 - 3 = 4$

Total charge = Fixed charge + (Charge per extra day $\times$ Number of extra days)

For Susy (book kept for 5 days):

Number of extra days = $5 - 3 = 2$

Now, solve equations (1) and (2) using the elimination method.

Subtract equation (2) from equation (1):

$(x + 4y) - (x + 2y) = 27 - 21$

$x + 4y - x - 2y = 6$

$2y = 6$

$y = 3$

Substitute $y = 3$ into equation (2):

$x + 2(3) = 21$

$x + 6 = 21$

$x = 15$

Therefore, the fixed charge is $\textsf{₹ } 15$ and the charge for each extra day is $\textsf{₹ } 3$.

Algebraic Representation:

Let the fare from the bus stand in Bangalore to Malleswaram be $\textsf{₹ } x$.

Let the fare from the bus stand in Bangalore to Yeshwanthpur be $\textsf{₹ } y$.

According to the first condition:

According to the second condition:

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Solution using Cross-Multiplication Method:

Rewrite the equations in the standard form $ax + by + c = 0$:

Comparing these with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we get:

$a_1 = 2, b_1 = 3, c_1 = -46$

$a_2 = 3, b_2 = 5, c_2 = -74$

Using the cross-multiplication formula:

$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

Substitute the values:

$\frac{x}{(3)(-74) - (5)(-46)} = \frac{y}{(-46)(3) - (-74)(2)} = \frac{1}{(2)(5) - (3)(3)}$

$\frac{x}{-222 - (-230)} = \frac{y}{-138 - (-148)} = \frac{1}{10 - 9}$

$\frac{x}{-222 + 230} = \frac{y}{-138 + 148} = \frac{1}{1}$

$\frac{x}{8} = \frac{y}{10} = \frac{1}{1}$

Now, solve for $x$ and $y$:

$\frac{x}{8} = \frac{1}{1} \implies x = 8$

$\frac{y}{10} = \frac{1}{1} \implies y = 10$

---

Alternate Solution using Elimination Method:

Multiply equation (1) by 3 and equation (2) by 2:

Subtract equation (3) from equation (4):

$(6x + 10y) - (6x + 9y) = 148 - 138$

$y = 10$

Substitute $y = 10$ into equation (1):

$2x + 3(10) = 46$

$2x + 30 = 46$

$2x = 16$

$x = 8$

---

Result:

The fare from the bus stand to Malleswaram is $\textsf{₹ } 8$.

The fare from the bus stand to Yeshwanthpur is $\textsf{₹ } 10$.

Given Equations:

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Condition for Unique Solution:

A pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has a unique solution if and only if:

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

---

Identifying Coefficients:

Comparing equation (1) with $a_1x + b_1y + c_1 = 0$, we get:

$a_1 = 4$, $b_1 = p$, $c_1 = 8$

Comparing equation (2) with $a_2x + b_2y + c_2 = 0$, we get:

$a_2 = 2$, $b_2 = 2$, $c_2 = 2$

---

Applying the Condition:

Substitute the values of the coefficients into the condition for a unique solution:

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

$\frac{4}{2} \neq \frac{p}{2}$

$2 \neq \frac{p}{2}$

Multiply both sides by 2:

$2 \times 2 \neq p$

$4 \neq p$

---

Conclusion:

The pair of linear equations will have a unique solution for all real values of $p$ except $p = 4$.

Thus, the condition is $p \neq 4$.

Given Equations:

The pair of linear equations is:

---

Condition for Infinitely Many Solutions:

A pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has infinitely many solutions if and only if:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

---

Identifying Coefficients:

Comparing equation (1) with $a_1x + b_1y + c_1 = 0$, we get:

$a_1 = k$, $b_1 = 3$, $c_1 = -(k - 3) = 3 - k$

Comparing equation (2) with $a_2x + b_2y + c_2 = 0$, we get:

$a_2 = 12$, $b_2 = k$, $c_2 = -k$

---

Applying the Condition:

For infinitely many solutions, we must have:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

$\frac{k}{12} = \frac{3}{k} = \frac{3 - k}{-k}$

Let's consider the first two parts:

$\frac{k}{12} = \frac{3}{k}$

$k^2 = 12 \times 3$

$k^2 = 36$

$k = \pm 6$

Now, let's consider the last two parts (assuming $k \neq 0$):

$\frac{3}{k} = \frac{3 - k}{-k}$

Multiply both sides by $-k$:

$3(-1) = 3 - k$

$-3 = 3 - k$

$k = 3 + 3$

$k = 6$

Now, let's consider the first and third parts (assuming $k \neq 0$):

$\frac{k}{12} = \frac{3 - k}{-k}$

$k(-k) = 12(3 - k)$

$-k^2 = 36 - 12k$

$k^2 - 12k + 36 = 0$

$(k - 6)^2 = 0$

$k = 6$

For the system to have infinitely many solutions, the value of $k$ must satisfy all the conditions. The common value obtained from all parts is $k=6$.

---

Verification for k=6:

Substitute $k=6$ into the ratios:

$\frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{3 - 6}{-6} = \frac{-3}{-6} = \frac{1}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ when $k = 6$, the system has infinitely many solutions.

---

Conclusion:

The value of $k$ for which the given pair of linear equations has infinitely many solutions is $k = 6$.

We compare the given pairs of linear equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. The conditions for the nature of solutions are:

• Unique solution: $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
• No solution: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
• Infinitely many solutions: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

---

(i) $x - 3y - 3 = 0$ and $3x - 9y - 2 = 0$

Here, $a_1 = 1, b_1 = -3, c_1 = -3$

And $a_2 = 3, b_2 = -9, c_2 = -2$

Compute the ratios:

$\frac{a_1}{a_2} = \frac{1}{3}$

$\frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}$

$\frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair of linear equations has no solution.

---

(ii) $2x + y = 5$ and $3x + 2y = 8$

Rewrite the equations in standard form:

$2x + y - 5 = 0$

$3x + 2y - 8 = 0$

Here, $a_1 = 2, b_1 = 1, c_1 = -5$

And $a_2 = 3, b_2 = 2, c_2 = -8$

Compute the ratios:

$\frac{a_1}{a_2} = \frac{2}{3}$

$\frac{b_1}{b_2} = \frac{1}{2}$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair of linear equations has a unique solution.

Using the cross-multiplication method:

$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

$\frac{x}{(1)(-8) - (2)(-5)} = \frac{y}{(-5)(3) - (-8)(2)} = \frac{1}{(2)(2) - (3)(1)}$

$\frac{x}{-8 - (-10)} = \frac{y}{-15 - (-16)} = \frac{1}{4 - 3}$

$\frac{x}{-8 + 10} = \frac{y}{-15 + 16} = \frac{1}{1}$

$\frac{x}{2} = \frac{y}{1} = \frac{1}{1}$

Equating $\frac{x}{2}$ with $\frac{1}{1}$:

$x = 2$

Equating $\frac{y}{1}$ with $\frac{1}{1}$:

$y = 1$

The unique solution is $x = 2, y = 1$.

---

(iii) $3x - 5y = 20$ and $6x - 10y = 40$

Rewrite the equations in standard form:

$3x - 5y - 20 = 0$

$6x - 10y - 40 = 0$

Here, $a_1 = 3, b_1 = -5, c_1 = -20$

And $a_2 = 6, b_2 = -10, c_2 = -40$

Compute the ratios:

$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair of linear equations has infinitely many solutions.

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(iv) $x - 3y - 7 = 0$ and $3x - 3y - 15 = 0$

Here, $a_1 = 1, b_1 = -3, c_1 = -7$

And $a_2 = 3, b_2 = -3, c_2 = -15$

Compute the ratios:

$\frac{a_1}{a_2} = \frac{1}{3}$

$\frac{b_1}{b_2} = \frac{-3}{-3} = 1$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair of linear equations has a unique solution.

Using the cross-multiplication method:

$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

$\frac{x}{(-3)(-15) - (-3)(-7)} = \frac{y}{(-7)(3) - (-15)(1)} = \frac{1}{(1)(-3) - (3)(-3)}$

$\frac{x}{45 - 21} = \frac{y}{-21 - (-15)} = \frac{1}{-3 - (-9)}$

$\frac{x}{24} = \frac{y}{-21 + 15} = \frac{1}{-3 + 9}$

$\frac{x}{24} = \frac{y}{-6} = \frac{1}{6}$

Equating $\frac{x}{24}$ with $\frac{1}{6}$:

$x = \frac{24}{6}$

$x = 4$

Equating $\frac{y}{-6}$ with $\frac{1}{6}$:

$y = \frac{-6}{6}$

$y = -1$

The unique solution is $x = 4, y = -1$.

(i) Infinite Number of Solutions

The given pair of linear equations is:

$2x + 3y = 7 \implies 2x + 3y - 7 = 0$

$(a - b)x + (a + b)y = 3a + b - 2 \ $$ \implies (a - b)x + (a + b)y - (3a + b - 2) = 0$

For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have an infinite number of solutions, the condition is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Here, $a_1 = 2, b_1 = 3, c_1 = -7$.

And $a_2 = a - b, b_2 = a + b, c_2 = -(3a + b - 2)$.

Applying the condition:

$\frac{2}{a - b} = \frac{3}{a + b} = \frac{-7}{-(3a + b - 2)} = \frac{7}{3a + b - 2}$

From the first two parts:

$\frac{2}{a - b} = \frac{3}{a + b}$

$2(a + b) = 3(a - b)$

$2a + 2b = 3a - 3b$

$5b = a$

From the last two parts:

$\frac{3}{a + b} = \frac{7}{3a + b - 2}$

$3(3a + b - 2) = 7(a + b)$

$9a + 3b - 6 = 7a + 7b$

$9a - 7a + 3b - 7b = 6$

$2a - 4b = 6$

Substitute the value of $a$ from equation (1) into equation (2):

$(5b) - 2b = 3$

$3b = 3$

$b = 1$

Substitute the value of $b = 1$ back into equation (1):

$a = 5(1)$

$a = 5$

Thus, the pair of linear equations will have an infinite number of solutions when $a = 5$ and $b = 1$.

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(ii) No Solution

The given pair of linear equations is:

$3x + y = 1 \implies 3x + y - 1 = 0$

$(2k - 1)x + (k - 1)y = 2k + 1 \ $$ \implies (2k - 1)x + (k - 1)y - (2k + 1) = 0$

For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have no solution, the condition is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Here, $a_1 = 3, b_1 = 1, c_1 = -1$.

And $a_2 = 2k - 1, b_2 = k - 1, c_2 = -(2k + 1)$.

Applying the condition:

$\frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)}$

First, consider the equality part:

$\frac{3}{2k - 1} = \frac{1}{k - 1}$

$3(k - 1) = 1(2k - 1)$

$3k - 3 = 2k - 1$

$3k - 2k = -1 + 3$

$k = 2$

Now, consider the inequality part with $k = 2$:

$\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

$\frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)} = \frac{1}{2k + 1}$

Substitute $k = 2$:

$\frac{1}{2 - 1} \neq \frac{1}{2(2) + 1}$

$\frac{1}{1} \neq \frac{1}{4 + 1}$

$1 \neq \frac{1}{5}$

This inequality is true for $k = 2$.

Thus, the pair of linear equations will have no solution when $k = 2$.

The given pair of linear equations is:

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1. Substitution Method:

From equation (2), express $y$ in terms of $x$:

$2y = 4 - 3x$

Substitute the value of $y$ from equation (3) into equation (1):

$8x + 5 \left( \frac{4 - 3x}{2} \right) = 9$

Multiply the entire equation by 2 to clear the fraction:

$2(8x) + 5(4 - 3x) = 2(9)$

$16x + 20 - 15x = 18$

$x + 20 = 18$

$x = 18 - 20$

$x = -2$

Substitute the value of $x = -2$ back into equation (3):

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

$y = \frac{4 + 6}{2}$

$y = \frac{10}{2}$

$y = 5$

So, by substitution method, the solution is $x = -2, y = 5$.

---

2. Cross-Multiplication Method:

Rewrite the equations in the standard form $ax + by + c = 0$:

Comparing these with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we get:

$a_1 = 8, b_1 = 5, c_1 = -9$

$a_2 = 3, b_2 = 2, c_2 = -4$

Using the cross-multiplication formula:

$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

Substitute the values:

$\frac{x}{(5)(-4) - (2)(-9)} = \frac{y}{(-9)(3) - (-4)(8)} = \frac{1}{(8)(2) - (3)(5)}$

$\frac{x}{-20 - (-18)} = \frac{y}{-27 - (-32)} = \frac{1}{16 - 15}$

$\frac{x}{-20 + 18} = \frac{y}{-27 + 32} = \frac{1}{1}$

$\frac{x}{-2} = \frac{y}{5} = \frac{1}{1}$

Now, solve for $x$ and $y$:

$\frac{x}{-2} = \frac{1}{1} \implies x = -2$

$\frac{y}{5} = \frac{1}{1} \implies y = 5$

So, by cross-multiplication method, the solution is $x = -2, y = 5$.

---

Result:

The solution to the pair of linear equations is $x = -2$ and $y = 5$.

(i) Hostel Charges

Formulation:

Let the fixed monthly hostel charge be $\textsf{₹ } x$.

Let the cost of food per day be $\textsf{₹ } y$.

For student A (20 days food):

For student B (26 days food):

Solution (using Elimination Method):

Subtract equation (1) from equation (2):

$(x + 26y) - (x + 20y) = 1180 - 1000$

$6y = 180$

$y = \frac{180}{6}$

$y = 30$

Substitute the value of $y = 30$ into equation (1):

$x + 20(30) = 1000$

$x + 600 = 1000$

$x = 1000 - 600$

$x = 400$

Result:

The fixed charges are $\textsf{₹ } 400$ and the cost of food per day is $\textsf{₹ } 30$.

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(ii) Fraction Problem

Formulation:

Let the numerator be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.

According to the first condition:

$\frac{x-1}{y} = \frac{1}{3}$

$3(x-1) = y$

According to the second condition:

$\frac{x}{y+8} = \frac{1}{4}$

$4x = y + 8$

Solution (using Elimination Method):

Subtract equation (1) from equation (2):

$(4x - y) - (3x - y) = 8 - 3$

$4x - y - 3x + y = 5$

$x = 5$

Substitute the value of $x = 5$ into equation (1):

$3(5) - y = 3$

$15 - y = 3$

$-y = 3 - 15$

$-y = -12$

$y = 12$

Result:

The fraction is $\frac{x}{y}$, which is $\frac{5}{12}$.

---

(iii) Test Score Problem

Formulation:

Let the number of right answers be $r$.

Let the number of wrong answers be $w$.

According to the first condition:

According to the second condition:

$4r - 2w = 50$

Divide by 2:

Solution (using Elimination Method):

Subtract equation (2) from equation (1):

$(3r - w) - (2r - w) = 40 - 25$

$3r - w - 2r + w = 15$

$r = 15$

Substitute the value of $r = 15$ into equation (1):

$3(15) - w = 40$

$45 - w = 40$

$-w = 40 - 45$

$-w = -5$

$w = 5$

The total number of questions is the sum of right and wrong answers.

Total questions = $r + w = 15 + 5 = 20$

Result:

There were 20 questions in the test.

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(iv) Highway Cars Problem

Formulation:

Let the speed of the car starting from A be $u$ km/h.

Let the speed of the car starting from B be $v$ km/h.

Assume $u > v$ for them to meet when travelling in the same direction.

Case 1: Same direction

Relative speed = $u - v$ km/h.

Distance = Speed $\times$ Time

$100 = (u - v) \times 5$

Case 2: Towards each other

Relative speed = $u + v$ km/h.

Distance = Speed $\times$ Time

$100 = (u + v) \times 1$

Solution (using Elimination Method):

Add equation (1) and equation (2):

$(u - v) + (u + v) = 20 + 100$

$2u = 120$

$u = 60$

Substitute the value of $u = 60$ into equation (2):

$60 + v = 100$

$v = 100 - 60$

$v = 40$

Result:

The speeds of the two cars are 60 km/h and 40 km/h.

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(v) Rectangle Area Problem

Formulation:

Let the length of the rectangle be $l$ units.

Let the breadth of the rectangle be $b$ units.

The original area is $A = l \times b$.

Condition 1: Length reduced by 5, breadth increased by 3, area reduced by 9.

New length = $l - 5$

New breadth = $b + 3$

New area = $(l - 5)(b + 3)$

New area = Original area - 9

$(l - 5)(b + 3) = lb - 9$

$lb + 3l - 5b - 15 = lb - 9$

$3l - 5b = -9 + 15$

Condition 2: Length increased by 3, breadth increased by 2, area increased by 67.

New length = $l + 3$

New breadth = $b + 2$

New area = $(l + 3)(b + 2)$

New area = Original area + 67

$(l + 3)(b + 2) = lb + 67$

$lb + 2l + 3b + 6 = lb + 67$

$2l + 3b = 67 - 6$

Solution (using Elimination Method):

Multiply equation (1) by 2 and equation (2) by 3 to eliminate $l$.

Subtract equation (3) from equation (4):

$(6l + 9b) - (6l - 10b) = 183 - 12$

$6l + 9b - 6l + 10b = 171$

$19b = 171$

$b = \frac{171}{19}$

$b = 9$

Substitute the value of $b = 9$ into equation (2):

$2l + 3(9) = 61$

$2l + 27 = 61$

$2l = 61 - 27$

$2l = 34$

$l = 17$

Result:

The dimensions of the rectangle are length = 17 units and breadth = 9 units.

Given Equations:

These equations are not linear in $x$ and $y$. We can reduce them to a linear form.

---

Reduction to Linear Form:

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.

Substituting these into equations (i) and (ii), we get:

Now we have a pair of linear equations in $u$ and $v$.

---

Solving for $u$ and $v$ (using Elimination Method):

Multiply equation (iii) by 4 and equation (iv) by 3 to eliminate $v$:

$4 \times (2u + 3v = 13) \implies 8u + 12v = 52$

$3 \times (5u - 4v = -2) \implies 15u - 12v = -6$

Add equation (v) and equation (vi):

$(8u + 12v) + (15u - 12v) = 52 + (-6)$

$23u = 46$

$u = \frac{46}{23}$

$u = 2$

Substitute the value $u = 2$ into equation (iii):

$2(2) + 3v = 13$

$4 + 3v = 13$

$3v = 13 - 4$

$3v = 9$

$v = \frac{9}{3}$

$v = 3$

---

Finding $x$ and $y$:

Now substitute back the values of $u$ and $v$:

$u = \frac{1}{x} \implies 2 = \frac{1}{x} \implies x = \frac{1}{2}$

$v = \frac{1}{y} \implies 3 = \frac{1}{y} \implies y = \frac{1}{3}$

---

Result:

The solution to the given pair of equations is $x = \frac{1}{2}$ and $y = \frac{1}{3}$.

Given Equations:

These equations are not linear in $x$ and $y$. We need to reduce them to a linear form.

---

Reduction to Linear Form:

Let $u = \frac{1}{x - 1}$ and $v = \frac{1}{y - 2}$.

Substituting these into equations (i) and (ii), we get a pair of linear equations in $u$ and $v$:

---

Solving for $u$ and $v$ (using Elimination Method):

Multiply equation (iii) by 3 to make the coefficients of $v$ equal in magnitude:

$3 \times (5u + v = 2) \implies 15u + 3v = 6$

Now, add equation (iv) and equation (v):

$(6u - 3v) + (15u + 3v) = 1 + 6$

$21u = 7$

$u = \frac{7}{21}$

$u = \frac{1}{3}$

Substitute the value $u = \frac{1}{3}$ into equation (iii):

$5 \left( \frac{1}{3} \right) + v = 2$

$\frac{5}{3} + v = 2$

$v = 2 - \frac{5}{3}$

$v = \frac{6 - 5}{3}$

$v = \frac{1}{3}$

---

Finding $x$ and $y$:

Now substitute back the original definitions of $u$ and $v$:

$u = \frac{1}{x - 1} \implies \frac{1}{3} = \frac{1}{x - 1}$

Cross-multiplying gives:

$x - 1 = 3$

$x = 3 + 1$

$x = 4$

$v = \frac{1}{y - 2} \implies \frac{1}{3} = \frac{1}{y - 2}$

Cross-multiplying gives:

$y - 2 = 3$

$y = 3 + 2$

$y = 5$

---

Result:

The solution to the given pair of equations is $x = 4$ and $y = 5$.

Formulation:

Let the speed of the boat in still water be $x$ km/h.

Let the speed of the stream be $y$ km/h.

Then, the speed of the boat downstream = $(x + y)$ km/h.

And the speed of the boat upstream = $(x - y)$ km/h.

We know that Time = $\frac{\text{Distance}}{\text{Speed}}$.

Scenario 1: 30 km upstream and 44 km downstream in 10 hours.

Time taken for 30 km upstream = $\frac{30}{x - y}$ hours.

Time taken for 44 km downstream = $\frac{44}{x + y}$ hours.

Total time = 10 hours.

Scenario 2: 40 km upstream and 55 km downstream in 13 hours.

Time taken for 40 km upstream = $\frac{40}{x - y}$ hours.

Time taken for 55 km downstream = $\frac{55}{x + y}$ hours.

Total time = 13 hours.

---

Reduction to Linear Form:

Let $u = \frac{1}{x - y}$ and $v = \frac{1}{x + y}$.

Substituting these into equations (i) and (ii), we get:

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Solving for $u$ and $v$ (using Elimination Method):

Multiply equation (iii) by 4 and equation (iv) by 3 to make the coefficients of $u$ equal.

$4 \times (30u + 44v = 10) \implies 120u + 176v = 40$

$3 \times (40u + 55v = 13) \implies 120u + 165v = 39$

Subtract equation (vi) from equation (v):

$(120u + 176v) - (120u + 165v) = 40 - 39$

$11v = 1$

$v = \frac{1}{11}$

Substitute $v = \frac{1}{11}$ into equation (iii):

$30u + 44 \left( \frac{1}{11} \right) = 10$

$30u + 4 = 10$

$30u = 10 - 4$

$30u = 6$

$u = \frac{6}{30}$

$u = \frac{1}{5}$

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Finding $x$ and $y$:

Now substitute back the original definitions of $u$ and $v$:

$u = \frac{1}{x - y} \implies \frac{1}{5} = \frac{1}{x - y}$

$v = \frac{1}{x + y} \implies \frac{1}{11} = \frac{1}{x + y}$

Add equations (vii) and (viii):

$(x - y) + (x + y) = 5 + 11$

$2x = 16$

$x = 8$

Substitute $x = 8$ into equation (viii):

$8 + y = 11$

$y = 11 - 8$

$y = 3$

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Result:

The speed of the boat in still water is 8 km/h.

The speed of the stream is 3 km/h.

(i) $\frac{1}{2x} + \frac{1}{3y} = 2$ and $\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The given equations become:

$\frac{u}{2} + \frac{v}{3} = 2 \implies 3u + 2v = 12$

$\frac{u}{3} + \frac{v}{2} = \frac{13}{6} \implies 2u + 3v = 13$

The new pair of linear equations is:

Multiply equation (1) by 2 and equation (2) by 3:

Subtract equation (3) from equation (4):

$(6u + 9v) - (6u + 4v) = 39 - 24 \implies 5v = 15 \implies v = 3$.

Substitute $v = 3$ into equation (1):

$3u + 2(3) = 12 \implies 3u + 6 = 12 \implies 3u = 6 \implies u = 2$.

Now, substitute back the original variables:

$u = \frac{1}{x} \implies 2 = \frac{1}{x} \implies x = \frac{1}{2}$.

$v = \frac{1}{y} \implies 3 = \frac{1}{y} \implies y = \frac{1}{3}$.

Solution: $x = \frac{1}{2}, y = \frac{1}{3}$.

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(ii) $\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2$ and $\frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1$

Let $u = \frac{1}{\sqrt{x}}$ and $v = \frac{1}{\sqrt{y}}$. The equations become:

Multiply equation (1) by 3:

Add equation (2) and equation (3):

$(4u - 9v) + (6u + 9v) = -1 + 6 \implies 10u = 5 \implies u = \frac{1}{2}$.

Substitute $u = \frac{1}{2}$ into equation (1):

$2(\frac{1}{2}) + 3v = 2 \implies 1 + 3v = 2 \implies 3v = 1 \implies v = \frac{1}{3}$.

$u = \frac{1}{\sqrt{x}} \implies \frac{1}{2} = \frac{1}{\sqrt{x}} \implies \sqrt{x} = 2 \implies x = 4$.

$v = \frac{1}{\sqrt{y}} \implies \frac{1}{3} = \frac{1}{\sqrt{y}} \implies \sqrt{y} = 3 \implies y = 9$.

Solution: $x = 4, y = 9$.

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(iii) $\frac{4}{x} + 3y = 14$ and $\frac{3}{x} - 4y = 23$

Let $u = \frac{1}{x}$. The equations become:

Multiply equation (1) by 4 and equation (2) by 3:

Add equation (3) and equation (4):

$(16u + 12y) + (9u - 12y) = 56 + 69 \implies 25u = 125 \implies u = 5$.

Substitute $u = 5$ into equation (1):

$4(5) + 3y = 14 \implies 20 + 3y = 14 \implies 3y = -6 \implies y = -2$.

$u = \frac{1}{x} \implies 5 = \frac{1}{x} \implies x = \frac{1}{5}$.

Solution: $x = \frac{1}{5}, y = -2$.

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(iv) $\frac{5}{x - 1} + \frac{1}{y - 2} = 2$ and $\frac{6}{x - 1} - \frac{3}{y - 2} = 1$

Let $u = \frac{1}{x - 1}$ and $v = \frac{1}{y - 2}$. The equations become:

Multiply equation (1) by 3:

Add equation (2) and equation (3):

$(6u - 3v) + (15u + 3v) = 1 + 6 \implies 21u = 7 \implies u = \frac{1}{3}$.

Substitute $u = \frac{1}{3}$ into equation (1):

$5(\frac{1}{3}) + v = 2 \implies v = 2 - \frac{5}{3} = \frac{1}{3}$.

$u = \frac{1}{x - 1} \implies \frac{1}{3} = \frac{1}{x - 1} \implies x - 1 = 3 \implies x = 4$.

$v = \frac{1}{y - 2} \implies \frac{1}{3} = \frac{1}{y - 2} \implies y - 2 = 3 \implies y = 5$.

Solution: $x = 4, y = 5$.

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(v) $\frac{7x - 2y}{xy} = 5$ and $\frac{8x + 7y}{xy} = 15$

Split the fractions (assuming $x \neq 0, y \neq 0$):

$\frac{7x}{xy} - \frac{2y}{xy} = 5 \implies \frac{7}{y} - \frac{2}{x} = 5$

$\frac{8x}{xy} + \frac{7y}{xy} = 15 \implies \frac{8}{y} + \frac{7}{x} = 15$

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:

Multiply equation (1) by 7 and equation (2) by 2:

Add equation (3) and equation (4):

$(-14u + 49v) + (14u + 16v) = 35 + 30 \implies 65v = 65 \implies v = 1$.

Substitute $v = 1$ into equation (2):

$7u + 8(1) = 15 \implies 7u = 7 \implies u = 1$.

$u = \frac{1}{x} \implies x = 1$.

$v = \frac{1}{y} \implies y = 1$.

Solution: $x = 1, y = 1$.

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(vi) $6x + 3y = 6xy$ and $2x + 4y = 5xy$

Divide both equations by $xy$ (assuming $x, y \neq 0$):

$\frac{6}{y} + \frac{3}{x} = 6$

$\frac{2}{y} + \frac{4}{x} = 5$

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:

Subtract equation (1) from equation (2):

$(4u + 2v) - (u + 2v) = 5 - 2 \implies 3u = 3 \implies u = 1$.

Substitute $u = 1$ into equation (1):

$1 + 2v = 2 \implies 2v = 1 \implies v = \frac{1}{2}$.

$u=1 \implies x=1$.

$v=\frac{1}{2} \implies y=2$.

Solution: $x = 1, y = 2$.

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(vii) $\frac{10}{x + y} + \frac{2}{x - y} = 4$ and $\frac{15}{x + y} - \frac{5}{x - y} = -2$

Let $u = \frac{1}{x + y}$ and $v = \frac{1}{x - y}$. The equations become:

From (1), $v = 2 - 5u$. Substitute this into (2):

$15u - 5(2 - 5u) = -2 \implies 15u - 10 + 25u = -2 \ $$ \implies 40u = 8 \implies u = \frac{1}{5}$.

Substitute back to find $v$: $v = 2 - 5(\frac{1}{5}) = 2 - 1 = 1$.

$u=\frac{1}{5} \implies x+y=5$

$v=1 \implies x-y=1$

Add (3) and (4): $2x=6 \implies x=3$.

Substitute $x=3$ into (3): $3+y=5 \implies y=2$.

Solution: $x = 3, y = 2$.

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(viii) $\frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4}$ and $\frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = -\frac{1}{8}$

Let $u = \frac{1}{3x + y}$ and $v = \frac{1}{3x - y}$. The equations become:

Add (1) and (2): $2u = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \implies u = \frac{1}{4}$.

Substitute $u=\frac{1}{4}$ into (1): $\frac{1}{4} + v = \frac{3}{4} \implies v = \frac{2}{4} = \frac{1}{2}$.

$u=\frac{1}{4} \implies 3x+y=4$

$v=\frac{1}{2} \implies 3x-y=2$

Add (3) and (4): $6x = 6 \implies x=1$.

Substitute $x=1$ into (3): $3(1)+y=4 \implies y=1$.

Solution: $x = 1, y = 1$.

(i) Ritu's Rowing Speed

Formulation:

Let the speed of Ritu's rowing in still water be $x$ km/h.

Let the speed of the current be $y$ km/h.

Speed downstream = $(x + y)$ km/h.

Speed upstream = $(x - y)$ km/h.

We use the formula: Time = $\frac{\text{Distance}}{\text{Speed}}$.

Condition 1: Downstream 20 km in 2 hours.

$2 = \frac{20}{x + y} \implies x + y = \frac{20}{2}$

Condition 2: Upstream 4 km in 2 hours.

$2 = \frac{4}{x - y} \implies x - y = \frac{4}{2}$

Solution (using Elimination Method):

Add equation (1) and equation (2):

$(x + y) + (x - y) = 10 + 2$

$2x = 12$

$x = 6$

Substitute the value of $x = 6$ into equation (1):

$6 + y = 10$

$y = 4$

Result:

Ritu's speed of rowing in still water is 6 km/h and the speed of the current is 4 km/h.

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(ii) Embroidery Work

Formulation:

Let the time taken by 1 woman alone to finish the work be $w$ days.

Let the time taken by 1 man alone to finish the work be $m$ days.

Work done by 1 woman in 1 day = $\frac{1}{w}$.

Work done by 1 man in 1 day = $\frac{1}{m}$.

Condition 1: 2 women and 5 men finish the work in 4 days.

Work done by 2 women and 5 men in 1 day = $\frac{1}{4}$.

$2 \left(\frac{1}{w}\right) + 5 \left(\frac{1}{m}\right) = \frac{1}{4}$

Condition 2: 3 women and 6 men finish the work in 3 days.

Work done by 3 women and 6 men in 1 day = $\frac{1}{3}$.

$3 \left(\frac{1}{w}\right) + 6 \left(\frac{1}{m}\right) = \frac{1}{3}$

Let $u = \frac{1}{w}$ and $v = \frac{1}{m}$. The equations become:

Solution (using Elimination Method):

Multiply equation (1) by 9 and equation (2) by 8:

Subtract equation (4) from equation (3):

$(72u + 180v) - (72u + 144v) = 9 - 8$

$36v = 1 \implies v = \frac{1}{36}$

Substitute $v = \frac{1}{36}$ into equation (1):

$8u + 20 \left(\frac{1}{36}\right) = 1$

$8u + \frac{5}{9} = 1$

$8u = 1 - \frac{5}{9} = \frac{4}{9}$

$u = \frac{4}{9 \times 8} = \frac{1}{18}$

Since $u = \frac{1}{w}$, $w = \frac{1}{u} = 18$.

Since $v = \frac{1}{m}$, $m = \frac{1}{v} = 36$.

Result:

The time taken by 1 woman alone is 18 days.

The time taken by 1 man alone is 36 days.

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(iii) Roohi's Travel

Formulation:

Let the speed of the train be $t$ km/h.

Let the speed of the bus be $b$ km/h.

Total distance = 300 km.

We use the formula: Time = $\frac{\text{Distance}}{\text{Speed}}$.

Condition 1: 60 km by train and (300 - 60 = 240) km by bus takes 4 hours.

Condition 2: 100 km by train and (300 - 100 = 200) km by bus takes 4 hours and 10 minutes.

10 minutes = $\frac{10}{60} = \frac{1}{6}$ hours.

Total time = $4 + \frac{1}{6} = \frac{25}{6}$ hours.

Let $u = \frac{1}{t}$ and $v = \frac{1}{b}$. The equations become:

$60u + 240v = 4$

Divide by 4: $15u + 60v = 1$ ...(1)

$100u + 200v = \frac{25}{6}$

Multiply by 6: $600u + 1200v = 25$

Divide by 25: $24u + 48v = 1$ ...(2)

Solution (using Substitution Method):

From equation (1): $15u = 1 - 60v \implies u = \frac{1 - 60v}{15}$.

Substitute this into equation (2):

$24 \left( \frac{1 - 60v}{15} \right) + 48v = 1$

Simplify $\frac{24}{15} = \frac{8}{5}$:

$\frac{8(1 - 60v)}{5} + 48v = 1$

Multiply by 5:

$8(1 - 60v) + 240v = 5$

$8 - 480v + 240v = 5$

$8 - 240v = 5$

$3 = 240v$

$v = \frac{3}{240} = \frac{1}{80}$

Substitute $v = \frac{1}{80}$ into equation (1):

$15u + 60 \left( \frac{1}{80} \right) = 1$

$15u + \frac{6}{8} = 1$

$15u + \frac{3}{4} = 1$

$15u = 1 - \frac{3}{4} = \frac{1}{4}$

$u = \frac{1}{4 \times 15} = \frac{1}{60}$

Since $u = \frac{1}{t}$, $t = \frac{1}{u} = 60$.

Since $v = \frac{1}{b}$, $b = \frac{1}{v} = 80$.

Result:

The speed of the train is 60 km/h and the speed of the bus is 80 km/h.

Formulation:

Let the age of Ani be $x$ years.

Let the age of Biju be $y$ years.

The ages of Ani and Biju differ by 3 years. This gives two possible cases:

Case 1: Ani is older than Biju.

Case 2: Biju is older than Ani.

Ani's father Dharam's age = $2 \times$ Ani's age = $2x$ years.

Biju is twice as old as his sister Cathy, so Cathy's age = $\frac{\text{Biju's age}}{2} = \frac{y}{2}$ years.

The ages of Cathy and Dharam differ by 30 years. Since Dharam is a father and Cathy is a sister, it is reasonable to assume Dharam is older than Cathy.

Dharam's age - Cathy's age = 30

$2x - \frac{y}{2} = 30$

Multiply by 2 to clear the fraction:

We need to solve the system for both cases.

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Solution for Case 1:

Using equations (i) and (iii):

Subtract equation (i) from equation (iii):

$(4x - y) - (x - y) = 60 - 3$

$4x - y - x + y = 57$

$3x = 57$

$x = \frac{57}{3}$

$x = 19$

Substitute $x = 19$ into equation (i):

$19 - y = 3$

$y = 19 - 3$

$y = 16$

In this case, Ani's age is 19 years and Biju's age is 16 years.

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Solution for Case 2:

Using equations (ii) and (iii):

Add equation (ii) and equation (iii):

$(-x + y) + (4x - y) = 3 + 60$

$3x = 63$

$x = \frac{63}{3}$

$x = 21$

Substitute $x = 21$ into equation (ii):

$y - 21 = 3$

$y = 24$

In this case, Ani's age is 21 years and Biju's age is 24 years.

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Result:

There are two possible solutions:

1. Ani's age is 19 years and Biju's age is 16 years.

2. Ani's age is 21 years and Biju's age is 24 years.

Formulation:

Let the capital of the first friend be $\textsf{₹ } x$.

Let the capital of the second friend be $\textsf{₹ } y$.

First Statement: "Give me a hundred, friend! I shall then become twice as rich as you".

If the second friend gives $\textsf{₹ } 100$ to the first friend:

First friend's capital = $x + 100$

Second friend's capital = $y - 100$

According to the statement: $x + 100 = 2(y - 100)$

$x + 100 = 2y - 200$

$x - 2y = -200 - 100$

Second Statement: "If you give me ten, I shall be six times as rich as you".

If the first friend gives $\textsf{₹ } 10$ to the second friend:

First friend's capital = $x - 10$

Second friend's capital = $y + 10$

According to the statement: $y + 10 = 6(x - 10)$

$y + 10 = 6x - 60$

$-6x + y = -60 - 10$

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Solution (using Substitution Method):

From equation (2), express $y$ in terms of $x$:

Substitute this expression for $y$ into equation (1):

$x - 2(6x - 70) = -300$

$x - 12x + 140 = -300$

$-11x = -300 - 140$

$-11x = -440$

$x = \frac{-440}{-11}$

$x = 40$

Substitute the value of $x = 40$ back into equation (3):

$y = 6(40) - 70$

$y = 240 - 70$

$y = 170$

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Result:

The capital of the first friend is $\textsf{₹ } 40$.

The capital of the second friend is $\textsf{₹ } 170$.

Formulation:

Let the uniform speed of the train be $s$ km/h.

Let the scheduled time taken be $t$ hours.

Let the distance covered be $d$ km.

We know the relationship: Distance = Speed $\times$ Time

$d = s \times t$

Condition 1: Speed is $(s + 10)$ km/h, Time is $(t - 2)$ hours.

Distance $d = (s + 10)(t - 2)$

Using (*), $st = (s + 10)(t - 2)$

$st = st - 2s + 10t - 20$

$0 = -2s + 10t - 20$

$2s - 10t = -20$

Divide by 2:

Condition 2: Speed is $(s - 10)$ km/h, Time is $(t + 3)$ hours.

Distance $d = (s - 10)(t + 3)$

Using (*), $st = (s - 10)(t + 3)$

$st = st + 3s - 10t - 30$

$0 = 3s - 10t - 30$

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Solution (using Elimination Method):

We have the pair of linear equations:

Multiply equation (1) by 2:

Subtract equation (3) from equation (2):

$(3s - 10t) - (2s - 10t) = 30 - (-20)$

$3s - 10t - 2s + 10t = 30 + 20$

$s = 50$

Substitute the value $s = 50$ into equation (1):

$50 - 5t = -10$

$-5t = -10 - 50$

$-5t = -60$

$t = \frac{-60}{-5}$

$t = 12$

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Finding the Distance:

Now, calculate the distance using the formula $d = st$.

$d = 50 \times 12$

$d = 600$

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Result:

The distance covered by the train is 600 km.

Formulation:

Let the number of rows be $r$.

Let the number of students in each row be $s$.

The total number of students in the class = Number of rows $\times$ Number of students per row

Total number of students = $r \times s$.

Condition 1: If 3 students are extra in a row $(s + 3)$, there would be 1 row less $(r - 1)$.

Total number of students = $(r - 1)(s + 3)$.

Since the total number of students remains the same:

$rs = (r - 1)(s + 3)$

$rs = rs + 3r - s - 3$

$0 = 3r - s - 3$

Condition 2: If 3 students are less in a row $(s - 3)$, there would be 2 rows more $(r + 2)$.

Total number of students = $(r + 2)(s - 3)$.

Since the total number of students remains the same:

$rs = (r + 2)(s - 3)$

$rs = rs - 3r + 2s - 6$

$0 = -3r + 2s - 6$

$3r - 2s = -6$

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Solution (using Elimination Method):

We need to solve the following pair of linear equations:

Subtract equation (2) from equation (1):

$(3r - s) - (3r - 2s) = 3 - (-6)$

$3r - s - 3r + 2s = 3 + 6$

$s = 9$

Substitute the value $s = 9$ into equation (1):

$3r - 9 = 3$

$3r = 3 + 9$

$3r = 12$

$r = \frac{12}{3}$

$r = 4$

So, the number of rows is 4 and the number of students in each row is 9.

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Finding the Total Number of Students:

Total number of students = $r \times s$

Total number of students = $4 \times 9$

Total number of students = $36$

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Result:

The number of students in the class is 36.

Given:

In $\triangle ABC$, the relationships between the angles are:

We also know the angle sum property of a triangle:

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To Find:

The measures of the three angles: $\angle A$, $\angle B$, and $\angle C$.

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Solution:

From equations (1) and (2), we have:

$3 \angle B = 2 (\angle A + \angle B)$

$3 \angle B = 2 \angle A + 2 \angle B$

$3 \angle B - 2 \angle B = 2 \angle A$

$\angle B = 2 \angle A$

This implies:

Now, substitute equations (1) and (4) into the angle sum property (equation (3)):

$\angle A + \angle B + \angle C = 180^\circ$

$\left( \frac{\angle B}{2} \right) + \angle B + (3 \angle B) = 180^\circ$

Multiply by 2 to clear the fraction:

$\angle B + 2 \angle B + 6 \angle B = 360^\circ$

$9 \angle B = 360^\circ$

$\angle B = \frac{360^\circ}{9}$

$\angle B = 40^\circ$

Now find $\angle C$ using equation (1):

$\angle C = 3 \angle B = 3 \times 40^\circ$

$\angle C = 120^\circ$

Find $\angle A$ using equation (4):

$\angle A = \frac{\angle B}{2} = \frac{40^\circ}{2}$

$\angle A = 20^\circ$

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Result:

The three angles of the triangle are:

$\angle A = 20^\circ$

$\angle B = 40^\circ$

$\angle C = 120^\circ$

Given Equations:

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Finding Points for Graphing:

To draw the graphs, we need coordinates of points satisfying each equation.

For equation (i): $5x - y = 5 \implies y = 5x - 5$

x	y = 5x - 5	Point
0	-5	A(0, -5)
1	0	B(1, 0)
2	5	(2, 5)

Point A(0, -5) is the intersection of line (i) with the y-axis.

For equation (ii): $3x - y = 3 \implies y = 3x - 3$

x	y = 3x - 3	Point
0	-3	C(0, -3)
1	0	B(1, 0)
2	3	(2, 3)

Point C(0, -3) is the intersection of line (ii) with the y-axis.

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Graphical Representation and Triangle Vertices:

Plot the points for each equation on graph paper and draw the lines.

The graph shows the two lines intersecting at point B(1, 0).

The triangle is formed by these two lines and the y-axis (the line $x=0$).

The vertices of this triangle are:

• The point of intersection of the two lines: B(1, 0).
• The point where the first line ($5x - y = 5$) intersects the y-axis: A(0, -5).
• The point where the second line ($3x - y = 3$) intersects the y-axis: C(0, -3).

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Result:

The coordinates of the vertices of the triangle formed by the lines $5x - y = 5$, $3x - y = 3$, and the y-axis are (1, 0), (0, -5), and (0, -3).

(i) $px + qy = p - q$ and $qx - py = p + q$

Given equations:

Using the Elimination Method:

Multiply equation (1) by $p$ and equation (2) by $q$:

Add equation (3) and equation (4):

$(p^2x + pqy) + (q^2x - pqy) = (p^2 - pq) + (pq + q^2)$

$p^2x + q^2x = p^2 + q^2$

$(p^2 + q^2)x = p^2 + q^2$

Assuming $p^2 + q^2 \neq 0$, we get $x = 1$.

Substitute $x = 1$ into equation (1):

$p(1) + qy = p - q$

$p + qy = p - q$

$qy = -q$

Assuming $q \neq 0$, we get $y = -1$.

Solution: $x = 1, y = -1$.

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(ii) $ax + by = c$ and $bx + ay = 1 + c$

Given equations:

Using the Elimination Method:

Multiply equation (1) by $a$ and equation (2) by $b$:

Subtract equation (4) from equation (3):

$(a^2x + aby) - (b^2x + aby) = ac - (b + bc)$

$a^2x - b^2x = ac - b - bc$

$(a^2 - b^2)x = c(a - b) - b$

Assuming $a^2 \neq b^2$, $x = \frac{c(a - b) - b}{a^2 - b^2}$.

Multiply equation (1) by $b$ and equation (2) by $a$:

Subtract equation (5) from equation (6):

$(abx + a^2y) - (abx + b^2y) = (a + ac) - bc$

$a^2y - b^2y = a + ac - bc$

$(a^2 - b^2)y = a + c(a - b)$

Assuming $a^2 \neq b^2$, $y = \frac{a + c(a - b)}{a^2 - b^2}$.

Solution: $x = \frac{c(a - b) - b}{a^2 - b^2}, y = \frac{a + c(a - b)}{a^2 - b^2}$.

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(iii) $\frac{x}{a} - \frac{y}{b} = 0$ and $ax + by = a^2 + b^2$

Given equations:

Using the Substitution Method:

From equation (1): $\frac{x}{a} = \frac{y}{b} \implies x = \frac{ay}{b}$.

Substitute this expression for $x$ into equation (2):

$a \left( \frac{ay}{b} \right) + by = a^2 + b^2$

$\frac{a^2y}{b} + by = a^2 + b^2$

Multiply by $b$ to clear the fraction:

$a^2y + b^2y = b(a^2 + b^2)$

$(a^2 + b^2)y = b(a^2 + b^2)$

Assuming $a^2 + b^2 \neq 0$, we get $y = b$.

Substitute $y = b$ back into $x = \frac{ay}{b}$:

$x = \frac{a(b)}{b}$

$x = a$

Solution: $x = a, y = b$.

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(iv) $(a - b)x + (a + b)y = a^2 - 2ab - b^2$ and $(a + b)(x + y) = a^2 + b^2$

Given equations:

Expand the second equation:

Using the Elimination Method:

Subtract equation (1) from equation (2):

$[(a + b)x + (a + b)y] - [(a - b)x + (a + b)y] = (a^2 + b^2) - (a^2 - 2ab - b^2)$

$(a + b)x - (a - b)x = a^2 + b^2 - a^2 + 2ab + b^2$

$(a + b - a + b)x = 2b^2 + 2ab$

$2bx = 2b(b + a)$

Assuming $b \neq 0$, we get $x = a + b$.

Substitute $x = a + b$ into equation (2):

$(a + b)(a + b) + (a + b)y = a^2 + b^2$

$(a + b)^2 + (a + b)y = a^2 + b^2$

$a^2 + 2ab + b^2 + (a + b)y = a^2 + b^2$

$2ab + (a + b)y = 0$

$(a + b)y = -2ab$

Assuming $a + b \neq 0$, we get $y = \frac{-2ab}{a + b}$.

Solution: $x = a + b, y = \frac{-2ab}{a + b}$.

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(v) $152x - 378y = -74$ and $-378x + 152y = -604$

Given equations:

Using the method of adding and subtracting equations:

Add equation (1) and equation (2):

$(152 - 378)x + (-378 + 152)y = -74 - 604$

$-226x - 226y = -678$

Divide by -226:

Subtract equation (2) from equation (1):

$(152 - (-378))x + (-378 - 152)y = -74 - (-604)$

$(152 + 378)x + (-530)y = -74 + 604$

$530x - 530y = 530$

Divide by 530:

Now solve the simpler system of equations (3) and (4).

Add equation (3) and equation (4):

$(x + y) + (x - y) = 3 + 1$

$2x = 4 \implies x = 2$

Substitute $x = 2$ into equation (3):

$2 + y = 3 \implies y = 1$

Solution: $x = 2, y = 1$.

Given:

A cyclic quadrilateral ABCD with angles:

$\angle A = 4y + 20$

$\angle B = 3y - 5$

$\angle C = -4x$

$\angle D = -7x + 5$

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Property of Cyclic Quadrilateral:

We know that the sum of opposite angles in a cyclic quadrilateral is $180^\circ$.

Therefore,

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Formulating Equations:

Using equation (i):

$(4y + 20) + (-4x) = 180$

$4y + 20 - 4x = 180$

$-4x + 4y = 180 - 20$

$-4x + 4y = 160$

Divide by 4:

Using equation (ii):

$(3y - 5) + (-7x + 5) = 180$

$3y - 5 - 7x + 5 = 180$

$-7x + 3y = 180$

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Solving the Equations:

We have the pair of linear equations:

$-x + y = 40$

$-7x + 3y = 180$

From equation (iii), we can express $y$ in terms of $x$:

$y = x + 40$

Substitute this value of $y$ into equation (iv):

$-7x + 3(x + 40) = 180$

$-7x + 3x + 120 = 180$

$-4x = 180 - 120$

$-4x = 60$

$x = \frac{60}{-4}$

$x = -15$

Substitute the value of $x = -15$ back into $y = x + 40$:

$y = -15 + 40$

$y = 25$

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Calculating the Angles:

Now substitute the values of $x$ and $y$ into the expressions for the angles:

$\angle A = 4y + 20 = 4(25) + 20 = 100 + 20 = 120^\circ$

$\angle B = 3y - 5 = 3(25) - 5 = 75 - 5 = 70^\circ$

$\angle C = -4x = -4(-15) = 60^\circ$

$\angle D = -7x + 5 = -7(-15) + 5 = 105 + 5 = 110^\circ$

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Result:

The angles of the cyclic quadrilateral are:

$\angle A = 120^\circ$

$\angle B = 70^\circ$

$\angle C = 60^\circ$

$\angle D = 110^\circ$