Chapter 3 Pair Of Linear Equations In Two Variables

Given:

The pair of linear equations:

and

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To Check / To Solve:

1. Check graphically whether the pair of equations is consistent.

2. If consistent, solve them graphically.

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Solution:

To check the consistency of the pair of linear equations graphically, we need to plot their graphs on the same coordinate plane. If the lines intersect at a point, the system is consistent and the intersection point gives the unique solution. If the lines are parallel, the system is inconsistent (no solution). If the lines coincide, the system is consistent and has infinitely many solutions.

Let's find at least two points for each equation to draw the lines.

For equation (1): $x + 3y = 6$

We can find points by choosing values for $x$ or $y$ and calculating the other variable.

If we take $x = 0$, then $0 + 3y = 6 \implies 3y = 6 \implies y = 2$. Point: (0, 2).

If we take $y = 0$, then $x + 3(0) = 6 \implies x = 6$. Point: (6, 0).

We can also find a third point for accuracy. If we take $x = 3$, then $3 + 3y = 6 \implies 3y = 3 \implies y = 1$. Point: (3, 1).

Table of points for equation (1):

x	y	Point (x, y)
0	2	(0, 2)
6	0	(6, 0)
3	1	(3, 1)

For equation (2): $2x - 3y = 12$

If we take $x = 0$, then $2(0) - 3y = 12 \implies -3y = 12 \implies y = -4$. Point: (0, -4).

If we take $y = 0$, then $2x - 3(0) = 12 \implies 2x = 12 \implies x = 6$. Point: (6, 0).

If we take $x = 3$, then $2(3) - 3y = 12 \implies 6 - 3y = 12 \implies -3y = 6 \implies y = -2$. Point: (3, -2).

Table of points for equation (2):

x	y	Point (x, y)
0	-4	(0, -4)
6	0	(6, 0)
3	-2	(3, -2)

Now, plot these points on a graph paper. Draw a straight line passing through the points for equation (1). On the same graph paper, draw a straight line passing through the points for equation (2).

Observe the graphs of the two lines. We can see that the lines intersect at the point (6, 0).

Since the graphs of the two equations intersect at a single point, the pair of equations is consistent and has a unique solution.

The solution to the system of equations is given by the coordinates of the point of intersection.

The intersection point is (6, 0).

Therefore, the solution is $x = 6$ and $y = 0$.

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Verification:

Substitute $x = 6$ and $y = 0$ into the original equations.

For equation (1): $x + 3y = 6$

$6 + 3(0) = 6$

$6 + 0 = 6$

For equation (2): $2x - 3y = 12$

$2(6) - 3(0) = 12$

$12 - 0 = 12$

Since the values of $x$ and $y$ satisfy both equations, the solution is correct.

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Answer:

The pair of equations is consistent.

The solution is $x = 6$ and $y = 0$.

Given:

The pair of linear equations:

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To Find:

Graphically find whether the pair of equations has no solution, a unique solution, or infinitely many solutions.

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Solution:

To determine the nature of the solutions graphically, we need to plot the graphs of both equations on the same coordinate plane. The relationship between the lines on the graph tells us about the solutions:

• If the lines intersect at a single point, there is a unique solution (consistent system).
• If the lines are parallel and distinct, there is no solution (inconsistent system).
• If the lines coincide (are the same line), there are infinitely many solutions (consistent system).

Let's find at least two points that lie on the line represented by each equation.

For equation (1): $5x - 8y + 1 = 0$. We can rewrite this equation to express $y$ in terms of $x$: $8y = 5x + 1 \implies y = \frac{5x+1}{8}$.

Let's choose a few values for $x$ and find the corresponding values for $y$:

x	y = $\frac{5x+1}{8}$	Point (x, y)
3	$\frac{5(3)+1}{8} = \frac{15+1}{8} = \frac{16}{8} = 2$	(3, 2)
-5	$\frac{5(-5)+1}{8} = \frac{-25+1}{8} = \frac{-24}{8} = -3$	(-5, -3)
-1	$\frac{5(-1)+1}{8} = \frac{-5+1}{8} = \frac{-4}{8} = -0.5$	(-1, -0.5)

For equation (2): $3x - \frac{24}{5}y + \frac{3}{5} = 0$.

To simplify this equation, we can multiply the entire equation by 5 to remove the fractions:

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

$15x - 24y + 3 = 0$

Now, we can see that all the coefficients (15, -24, 3) are divisible by 3. Let's divide the entire equation by 3:

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

Notice that Equation (3) is exactly the same as Equation (1).

This means that any point that satisfies Equation (1) also satisfies Equation (2), and vice-versa. The two equations represent the same line in the coordinate plane.

Since the graphs of the two equations are the same line, they coincide. Coinciding lines share all their points.

Therefore, the pair of equations has infinitely many solutions.

We can plot the points found for Equation (1) (which are also points for Equation (2)) and draw the line. The graph will show a single line representing both equations.

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Answer:

Graphically, the pair of equations has infinitely many solutions because the two equations represent the same line.

Given:

The problem describes two relationships between the number of pants and skirts Champa bought:

1. Number of skirts = (Twice the number of pants) - 2

2. Number of skirts = (Four times the number of pants) - 4

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To Find:

The number of pants and the number of skirts Champa bought, by solving graphically.

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Solution:

Let the number of pants Champa bought be $x$.

Let the number of skirts Champa bought be $y$.

From the first condition, we get the equation:

From the second condition, we get the equation:

To solve this system of linear equations graphically, we need to plot the graph of each equation on the same coordinate plane. The point where the two lines intersect will be the solution $(x, y)$, representing the number of pants and skirts.

For equation (1): $y = 2x - 2$

Let's find at least two points for this line by choosing values for $x$ and calculating $y$:

x	y = $2x - 2$	Point (x, y)
0	$2(0) - 2 = -2$	(0, -2)
1	$2(1) - 2 = 2 - 2 = 0$	(1, 0)
2	$2(2) - 2 = 4 - 2 = 2$	(2, 2)

For equation (2): $y = 4x - 4$

Let's find at least two points for this line by choosing values for $x$ and calculating $y$:

x	y = $4x - 4$	Point (x, y)
0	$4(0) - 4 = -4$	(0, -4)
1	$4(1) - 4 = 4 - 4 = 0$	(1, 0)
2	$4(2) - 4 = 8 - 4 = 4$	(2, 4)

Now, plot these points on a graph paper. Draw a straight line passing through the points for equation (1). On the same graph paper, draw a straight line passing through the points for equation (2).

Observe the graphs of the two lines. The lines intersect at the point (1, 0).

The intersection point represents the solution to the system of equations, where the x-coordinate is the number of pants and the y-coordinate is the number of skirts.

From the graph, the intersection point is $(1, 0)$.

So, $x = 1$ and $y = 0$.

This means Champa bought 1 pair of pants and 0 skirts.

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Verification:

Check if $x=1$ and $y=0$ satisfy the original conditions:

Condition 1: $y = 2x - 2 \implies 0 = 2(1) - 2 \implies 0 = 2 - 2 \implies 0 = 0$ (True)

Condition 2: $y = 4x - 4 \implies 0 = 4(1) - 4 \implies 0 = 4 - 4 \implies 0 = 0$ (True)

The solution is correct.

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Answer:

By solving graphically, we find that Champa bought 1 pair of pants and 0 skirts.

Solution (i): Mathematics quiz problem

Given:

Total number of students = 10.

Number of girls is 4 more than the number of boys.

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To Formulate and Solve:

1. Formulate a pair of linear equations.

2. Find the solution graphically.

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Formulation of Equations:

Let the number of boys be $x$.

Let the number of girls be $y$.

According to the first condition, the total number of students is 10:

According to the second condition, the number of girls is 4 more than the number of boys:

The pair of linear equations is $x + y = 10$ and $y = x + 4$.

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Graphical Solution:

To solve the system graphically, we need to plot the graphs of Equation (1) and Equation (2) on the same coordinate plane. The point of intersection of the two lines will give the solution $(x, y)$.

For Equation (1): $x + y = 10$. We can rewrite this as $y = 10 - x$. Let's find some points on this line:

x	y = 10 - x	Point (x, y)
0	$10 - 0 = 10$	(0, 10)
10	$10 - 10 = 0$	(10, 0)
3	$10 - 3 = 7$	(3, 7)

For Equation (2): $y = x + 4$. Let's find some points on this line:

x	y = x + 4	Point (x, y)
0	$0 + 4 = 4$	(0, 4)
3	$3 + 4 = 7$	(3, 7)
6	$6 + 4 = 10$	(6, 10)

Plot the points for each table on a graph paper. Draw a line through the points for equation (1) and another line through the points for equation (2) on the same graph.

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

The intersection point $(x, y) = (3, 7)$ is the solution to the system of equations.

So, the number of boys is $x = 3$, and the number of girls is $y = 7$.

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Verification:

Check if the solution (3, 7) satisfies the original conditions:

Total students = Boys + Girls = $3 + 7 = 10$. (Correct)

Number of girls = Number of boys + 4 $\implies 7 = 3 + 4 \implies 7 = 7$. (Correct)

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Answer (i):

Number of boys = 3

Number of girls = 7

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Solution (ii): Pencils and Pens cost problem

Given:

Cost of 5 pencils and 7 pens = $\textsf{₹} 50$.

Cost of 7 pencils and 5 pens = $\textsf{₹} 46$.

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To Formulate and Solve:

1. Formulate a pair of linear equations.

2. Find the solution graphically.

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Formulation of Equations:

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

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

According to the first condition, "5 pencils and 7 pens together cost $\textsf{₹} 50$":

According to the second condition, "7 pencils and 5 pens together cost $\textsf{₹} 46$":

The pair of linear equations is $5x + 7y = 50$ and $7x + 5y = 46$.

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Graphical Solution:

To solve the system graphically, we need to plot the graphs of Equation (3) and Equation (4) on the same coordinate plane. The point of intersection of the two lines will give the solution $(x, y)$.

For Equation (3): $5x + 7y = 50$. We can write this as $7y = 50 - 5x$, so $y = \frac{50 - 5x}{7}$. Let's find some points on this line (choose x-values that give integer or easy y-values if possible):

x	y = $\frac{50 - 5x}{7}$	Point (x, y)
3	$\frac{50 - 5(3)}{7} = \frac{50 - 15}{7} = \frac{35}{7} = 5$	(3, 5)
10	$\frac{50 - 5(10)}{7} = \frac{50 - 50}{7} = \frac{0}{7} = 0$	(10, 0)
-4	$\frac{50 - 5(-4)}{7} = \frac{50 + 20}{7} = \frac{70}{7} = 10$	(-4, 10)

For Equation (4): $7x + 5y = 46$. We can write this as $5y = 46 - 7x$, so $y = \frac{46 - 7x}{5}$. Let's find some points on this line (choose x-values that give integer or easy y-values if possible):

x	y = $\frac{46 - 7x}{5}$	Point (x, y)
3	$\frac{46 - 7(3)}{5} = \frac{46 - 21}{5} = \frac{25}{5} = 5$	(3, 5)
8	$\frac{46 - 7(8)}{5} = \frac{46 - 56}{5} = \frac{-10}{5} = -2$	(8, -2)
-2	$\frac{46 - 7(-2)}{5} = \frac{46 + 14}{5} = \frac{60}{5} = 12$	(-2, 12)

Plot the points for each table on a graph paper. Draw a line through the points for equation (3) and another line through the points for equation (4) on the same graph.

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

The intersection point $(x, y) = (3, 5)$ is the solution to the system of equations.

So, the cost of one pencil is $x = 3$ rupees, and the cost of one pen is $y = 5$ rupees.

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Verification:

Check if the solution (3, 5) satisfies the original conditions:

Condition 1: $5x + 7y = 50 \implies 5(3) + 7(5) = 50 \implies 15 + 35 = 50 \implies 50 = 50$. (Correct)

Condition 2: $7x + 5y = 46 \implies 7(3) + 5(5) = 46 \implies 21 + 25 = 46 \implies 46 = 46$. (Correct)

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Answer (ii):

Cost of one pencil = $\textsf{₹} 3$

Cost of one pen = $\textsf{₹} 5$

The general form of a pair of linear equations is $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. We compare the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$ to determine the nature of the lines.

• If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a point.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.

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Solution (i):

Given:

Pair of linear equations:

$5x – 4y + 8 = 0$

$7x + 6y – 9 = 0$

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To Find:

Whether the lines intersect, are parallel, or are coincident.

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Solution:

Comparing the given equations with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we have:

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

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

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{5}{7}$

$\frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}$

$\frac{c_1}{c_2} = \frac{8}{-9} = -\frac{8}{9}$

Compare the ratios:

$\frac{a_1}{a_2} = \frac{5}{7}$

$\frac{b_1}{b_2} = -\frac{2}{3}$

Since $\frac{5}{7} \neq -\frac{2}{3}$, we have $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

Conclusion: Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines representing the given pair of linear equations intersect at a point.

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Solution (ii):

Given:

Pair of linear equations:

$9x + 3y + 12 = 0$

$18x + 6y + 24 = 0$

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To Find:

Whether the lines intersect, are parallel, or are coincident.

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Solution:

Comparing the given equations with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we have:

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

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

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}$

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

$\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$

Compare the ratios:

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

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

$\frac{c_1}{c_2} = \frac{1}{2}$

Since $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$, we have $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Conclusion: Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines representing the given pair of linear equations are coincident.

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Solution (iii):

Given:

Pair of linear equations:

$6x – 3y + 10 = 0$

$2x – y + 9 = 0$

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To Find:

Whether the lines intersect, are parallel, or are coincident.

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Solution:

Comparing the given equations with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we have:

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

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

Calculate the ratios:

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

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

$\frac{c_1}{c_2} = \frac{10}{9}$

Compare the ratios:

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

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

$\frac{c_1}{c_2} = \frac{10}{9}$

We see that $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ since $3 = 3$.

We see that $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ since $3 \neq \frac{10}{9}$.

Thus, we have $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

Conclusion: Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines representing the given pair of linear equations are parallel.

We are asked to compare the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$ for each pair of linear equations in the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, and determine if the pair is consistent or inconsistent.

• If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect (unique solution). The pair is consistent.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident (infinitely many solutions). The pair is consistent.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel (no solution). The pair is inconsistent.

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Solution (i):

The given equations are $3x + 2y = 5$ and $2x – 3y = 7$.

Writing in standard form $ax + by + c = 0$:

$3x + 2y - 5 = 0$

$2x - 3y - 7 = 0$

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

$a_1 = 3$, $b_1 = 2$, $c_1 = -5$

$a_2 = 2$, $b_2 = -3$, $c_2 = -7$

Calculate the ratios:

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

$\frac{b_1}{b_2} = \frac{2}{-3} = -\frac{2}{3}$

Compare the ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$:

$\frac{3}{2} \neq -\frac{2}{3}$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a point.

Conclusion: The pair of linear equations is consistent.

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Solution (ii):

The given equations are $2x – 3y = 8$ and $4x – 6y = 9$.

Writing in standard form $ax + by + c = 0$:

$2x - 3y - 8 = 0$

$4x - 6y - 9 = 0$

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

$a_1 = 2$, $b_1 = -3$, $c_1 = -8$

$a_2 = 4$, $b_2 = -6$, $c_2 = -9$

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

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

$\frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$

Compare the ratios:

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

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

$\frac{c_1}{c_2} = \frac{8}{9}$

Since $\frac{1}{2} = \frac{1}{2}$ and $\frac{1}{2} \neq \frac{8}{9}$, we have $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

Conclusion: The pair of linear equations is inconsistent.

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Solution (iii):

The given equations are $\frac{3}{2} x + \frac{5}{3} y = 7$ and $9x – 10y = 14$.

Writing in standard form $ax + by + c = 0$:

$\frac{3}{2} x + \frac{5}{3} y - 7 = 0$

$9x - 10y - 14 = 0$

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

$a_1 = \frac{3}{2}$, $b_1 = \frac{5}{3}$, $c_1 = -7$

$a_2 = 9$, $b_2 = -10$, $c_2 = -14$

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{\frac{3}{2}}{9} = \frac{3}{2 \times 9} = \frac{3}{18} = \frac{1}{6}$

$\frac{b_1}{b_2} = \frac{\frac{5}{3}}{-10} = \frac{5}{3 \times -10} = \frac{5}{-30} = -\frac{1}{6}$

Compare the ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$:

$\frac{1}{6} \neq -\frac{1}{6}$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a point.

Conclusion: The pair of linear equations is consistent.

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Solution (iv):

The given equations are $5x – 3y = 11$ and $– 10x + 6y = –22$.

Writing in standard form $ax + by + c = 0$:

$5x - 3y - 11 = 0$

$-10x + 6y + 22 = 0$

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

$a_1 = 5$, $b_1 = -3$, $c_1 = -11$

$a_2 = -10$, $b_2 = 6$, $c_2 = 22$

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}$

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

$\frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}$

Compare the ratios:

$\frac{a_1}{a_2} = -\frac{1}{2}$

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

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

Since $-\frac{1}{2} = -\frac{1}{2} = -\frac{1}{2}$, we have $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Conclusion: The pair of linear equations is consistent (coincident lines).

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Solution (v):

The given equations are $\frac{4}{3} x + 2y = 8$ and $2x + 3y = 12$.

Writing in standard form $ax + by + c = 0$:

$\frac{4}{3} x + 2y - 8 = 0$

$2x + 3y - 12 = 0$

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

$a_1 = \frac{4}{3}$, $b_1 = 2$, $c_1 = -8$

$a_2 = 2$, $b_2 = 3$, $c_2 = -12$

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{\frac{4}{3}}{2} = \frac{4}{3 \times 2} = \frac{4}{6} = \frac{2}{3}$

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

$\frac{c_1}{c_2} = \frac{-8}{-12} = \frac{8}{12} = \frac{2}{3}$

Compare the ratios:

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

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

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

Since $\frac{2}{3} = \frac{2}{3} = \frac{2}{3}$, we have $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Conclusion: The pair of linear equations is consistent (coincident lines).

We will check the consistency of each pair of linear equations by comparing the ratios of the coefficients ($\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$). If a pair is consistent, we will find its solution graphically.

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Solution (i): x + y = 5, 2x + 2y = 10

Given:

Equation 1: $x + y = 5$, which can be written as $x + y - 5 = 0$.

Equation 2: $2x + 2y = 10$, which can be written as $2x + 2y - 10 = 0$.

Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

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

$a_2 = 2, b_2 = 2, c_2 = -10$

Let's calculate the ratios of the coefficients:

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

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

$\frac{c_1}{c_2} = \frac{-5}{-10} = \frac{5}{10} = \frac{1}{2}$

Compare the ratios: $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{1}{2}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines representing the equations are coincident. This means the pair of equations is consistent and has infinitely many solutions.

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Graphical Solution (i):

Since the system is consistent, we find the solution graphically. Both equations represent the same line. We only need to plot points for one of the equations, say $x + y = 5$. We can rewrite this as $y = 5 - x$.

Let's find a few points on this line:

x	y = 5 - x	Point (x, y)
0	$5 - 0 = 5$	(0, 5)
5	$5 - 5 = 0$	(5, 0)
2	$5 - 2 = 3$	(2, 3)

Plot these points on a graph paper and draw the line passing through them. Since both equations represent the same line, this single line is the graph for both equations.

The lines coincide, meaning every point on the line is a solution. The solutions are all pairs $(x, y)$ such that $x + y = 5$.

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Solution (ii): x – y = 8, 3x – 3y = 16

Given:

Equation 1: $x - y = 8$, which can be written as $x - y - 8 = 0$.

Equation 2: $3x - 3y = 16$, which can be written as $3x - 3y - 16 = 0$.

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

$a_1 = 1, b_1 = -1, c_1 = -8$

$a_2 = 3, b_2 = -3, c_2 = -16$

Let's calculate the ratios of the coefficients:

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

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

$\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{8}{16} = \frac{1}{2}$

Compare the ratios: $\frac{a_1}{a_2} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{1}{3}$, $\frac{c_1}{c_2} = \frac{1}{2}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines representing the equations are parallel and distinct. This means the pair of equations has no solution and is inconsistent.

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Graphical Solution (ii):

Since the system is inconsistent, there is no solution that can be obtained graphically. The graphs of the two equations are parallel lines that do not intersect.

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

Given:

Equation 1: $2x + y - 6 = 0$.

Equation 2: $4x - 2y - 4 = 0$.

Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

$a_1 = 2, b_1 = 1, c_1 = -6$

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

Let's calculate the ratios of the coefficients:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

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

Compare the ratios: $\frac{a_1}{a_2} = \frac{1}{2}$ and $\frac{b_1}{b_2} = -\frac{1}{2}$.

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines representing the equations intersect at a unique point. This means the pair of equations is consistent and has a unique solution.

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Graphical Solution (iii):

Since the system is consistent, we find the unique solution graphically. We need to plot points for each equation.

For Equation 1: $2x + y - 6 = 0$. We can rewrite this as $y = 6 - 2x$.

x	y = 6 - 2x	Point (x, y)
0	$6 - 2(0) = 6$	(0, 6)
3	$6 - 2(3) = 0$	(3, 0)
1	$6 - 2(1) = 4$	(1, 4)

For Equation 2: $4x - 2y - 4 = 0$. We can rewrite this as $2y = 4x - 4$, so $y = 2x - 2$.

x	y = 2x - 2	Point (x, y)
0	$2(0) - 2 = -2$	(0, -2)
1	$2(1) - 2 = 0$	(1, 0)
2	$2(2) - 2 = 2$	(2, 2)

Plot the points for each table on a graph paper and draw the lines. The point where the lines intersect is the solution.

Upon plotting, the two lines intersect at the point (2, 2).

The unique solution is $(x, y) = (2, 2)$.

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

Given:

Equation 1: $2x - 2y - 2 = 0$.

Equation 2: $4x - 4y - 5 = 0$.

Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

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

$a_2 = 4, b_2 = -4, c_2 = -5$

Let's calculate the ratios of the coefficients:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$

Compare the ratios: $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{2}{5}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines representing the equations are parallel and distinct. This means the pair of equations has no solution and is inconsistent.

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Graphical Solution (iv):

Since the system is inconsistent, there is no solution that can be obtained graphically. The graphs of the two equations are parallel lines that do not intersect.

Given:

Half the perimeter of a rectangular garden is 36 m.

The length of the garden is 4 m more than its width.

---

To Find:

The dimensions (length and width) of the garden.

---

Solution:

Let the width of the rectangular garden be $w$ meters.

Let the length of the rectangular garden be $l$ meters.

According to the condition that the length is 4 m more than its width, we can write the equation:

The perimeter of a rectangle is given by the formula $P = 2(l + w)$.

Half the perimeter is $\frac{1}{2} P = l + w$.

According to the condition that half the perimeter is 36 m, we write the equation:

We now have a system of two linear equations:

$l = w + 4$

$l + w = 36$

Substitute the expression for $l$ from equation (1) into equation (2):

$(w + 4) + w = 36$

Simplify and solve for $w$:

$2w + 4 = 36$

Subtract 4 from both sides:

$2w = 36 - 4$

$2w = 32$

Divide by 2:

$w = \frac{32}{2}$

$w = 16$

The width of the garden is 16 meters.

Now, substitute the value of $w = 16$ into equation (1) to find the value of $l$:

$l = w + 4$

$l = 16 + 4$

$l = 20$

The length of the garden is 20 meters.

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Answer:

The dimensions of the garden are:

Length = 20 m

Width = 16 m

Given:

The linear equation is $2x + 3y – 8 = 0$.

This equation is in the standard form $a_1x + b_1y + c_1 = 0$, where $a_1 = 2$, $b_1 = 3$, and $c_1 = -8$.

We need to write a second linear equation $a_2x + b_2y + c_2 = 0$ such that the pair of equations satisfies the given graphical conditions.

---

To Write:

Another linear equation to form a pair that represents:

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

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Solution:

The conditions for the graphical representation of a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are based on the ratios of their coefficients:

• For intersecting lines: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
• For parallel lines: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
• For coincident lines: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Given equation: $2x + 3y - 8 = 0$, so $a_1 = 2$, $b_1 = 3$, $c_1 = -8$.

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Part (i): Intersecting lines

We need to choose coefficients $a_2$, $b_2$, and $c_2$ such that $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

Using $a_1 = 2$ and $b_1 = 3$, we need $\frac{2}{a_2} \neq \frac{3}{b_2}$.

A simple way is to pick $a_2$ and $b_2$ that are not proportional to $a_1$ and $b_1$. For example, we can swap the coefficients of $x$ and $y$ and keep the signs different or just choose simple non-proportional values.

Let's choose $a_2 = 3$ and $b_2 = -2$. Then $\frac{a_1}{a_2} = \frac{2}{3}$ and $\frac{b_1}{b_2} = \frac{3}{-2}$. Clearly, $\frac{2}{3} \neq -\frac{3}{2}$.

We can choose any value for $c_2$. Let $c_2 = 1$.

So, a possible second linear equation is $3x - 2y + 1 = 0$.

Verification of condition: $\frac{a_1}{a_2} = \frac{2}{3}$ and $\frac{b_1}{b_2} = \frac{3}{-2} = -\frac{3}{2}$. Since $\frac{2}{3} \neq -\frac{3}{2}$, the lines are intersecting.

One possible equation for intersecting lines is $3x - 2y + 1 = 0$.

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Part (ii): Parallel lines

We need to choose coefficients $a_2$, $b_2$, and $c_2$ such that $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

Using $a_1 = 2$, $b_1 = 3$, $c_1 = -8$, we need $\frac{2}{a_2} = \frac{3}{b_2} \neq \frac{-8}{c_2}$.

To make the first part equal, we can multiply $a_1$ and $b_1$ by the same non-zero number. Let's multiply by 2. So, let $a_2 = 2 \times 2 = 4$ and $b_2 = 3 \times 2 = 6$.

Then $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$ and $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$. So $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{1}{2}$.

Now we need $\frac{c_1}{c_2} \neq \frac{1}{2}$. We have $c_1 = -8$. We need $\frac{-8}{c_2} \neq \frac{1}{2}$. This implies $c_2 \neq -8 \times 2$, so $c_2 \neq -16$.

We can choose any value for $c_2$ except -16. Let $c_2 = 0$.

So, a possible second linear equation is $4x + 6y + 0 = 0$, or $4x + 6y = 0$.

Verification of condition: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-8}{0}$ (undefined). Since $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$ is different/undefined, the lines are parallel.

One possible equation for parallel lines is $4x + 6y = 0$.

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Part (iii): Coincident lines

We need to choose coefficients $a_2$, $b_2$, and $c_2$ such that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Using $a_1 = 2$, $b_1 = 3$, $c_1 = -8$, we need $\frac{2}{a_2} = \frac{3}{b_2} = \frac{-8}{c_2}$.

To satisfy this condition, we must multiply $a_1$, $b_1$, and $c_1$ by the same non-zero number. Let's multiply by 3. So, let $a_2 = 2 \times 3 = 6$, $b_2 = 3 \times 3 = 9$, and $c_2 = -8 \times 3 = -24$.

So, a possible second linear equation is $6x + 9y - 24 = 0$.

Verification of condition: $\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}$, $\frac{c_1}{c_2} = \frac{-8}{-24} = \frac{1}{3}$. Since $\frac{1}{3} = \frac{1}{3} = \frac{1}{3}$, the lines are coincident.

One possible equation for coincident lines is $6x + 9y - 24 = 0$.

Given:

Two linear equations:

The third line forming the triangle is the x-axis.

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To Find:

1. Draw the graphs of the given equations (1) and (2).

2. Determine the coordinates of the vertices of the triangle formed by the lines (1), (2), and the x-axis.

3. Shade the triangular region.

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Solution:

To draw the graph of each linear equation, we need to find at least two points that lie on the line. It is usually helpful to find the points where the line intersects the x-axis (by setting $y=0$) and the y-axis (by setting $x=0$).

For equation (1): $x – y + 1 = 0$. We can rearrange this to make it easier to find points, e.g., $y = x + 1$.

If $x = 0$, $y = 0 + 1 = 1$. Point: (0, 1).

If $y = 0$, $0 = x + 1 \implies x = -1$. Point: (-1, 0).

Let's find one more point. If $x = 2$, $y = 2 + 1 = 3$. Point: (2, 3).

Table of points for equation (1):

x	y	Point (x, y)
0	1	(0, 1)
-1	0	(-1, 0)
2	3	(2, 3)

For equation (2): $3x + 2y – 12 = 0$. We can rearrange this to make it easier to find points, e.g., $2y = 12 - 3x \implies y = \frac{12 - 3x}{2}$.

If $x = 0$, $y = \frac{12 - 3(0)}{2} = \frac{12}{2} = 6$. Point: (0, 6).

If $y = 0$, $3x + 2(0) - 12 = 0 \implies 3x = 12 \implies x = 4$. Point: (4, 0).

Let's find one more point. If $x = 2$, $y = \frac{12 - 3(2)}{2} = \frac{12 - 6}{2} = \frac{6}{2} = 3$. Point: (2, 3).

Table of points for equation (2):

x	y	Point (x, y)
0	6	(0, 6)
4	0	(4, 0)
2	3	(2, 3)

Now, plot these points on a graph paper and draw the straight line passing through the points for equation (1). On the same graph paper, plot the points for equation (2) and draw the straight line passing through them. The x-axis is the horizontal line where $y=0$.

The vertices of the triangle formed by these three lines are the points where any two of these lines intersect.

Vertex 1: Intersection of line (1) and the x-axis ($y=0$).

From the table for equation (1), the point where $y=0$ is $(-1, 0)$. This is the intersection of $x - y + 1 = 0$ and $y = 0$.

Vertex 1 is (-1, 0).

Vertex 2: Intersection of line (2) and the x-axis ($y=0$).

From the table for equation (2), the point where $y=0$ is $(4, 0)$. This is the intersection of $3x + 2y - 12 = 0$ and $y = 0$.

Vertex 2 is (4, 0).

Vertex 3: Intersection of line (1) and line (2).

We look for a common point in the tables for equation (1) and equation (2). The point (2, 3) appears in both tables. This is the intersection point of $x - y + 1 = 0$ and $3x + 2y - 12 = 0$.

Vertex 3 is (2, 3).

The triangle is formed by connecting the vertices (-1, 0), (4, 0), and (2, 3). The side of the triangle that lies on the x-axis connects the points (-1, 0) and (4, 0).

Shade the region bounded by these three vertices to represent the triangular region.

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Answer:

The graphs of the equations $x – y + 1 = 0$ and $3x + 2y – 12 = 0$ and the x-axis form a triangle.

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

The triangular region is shaded in the graph above.

Given:

The pair of linear equations:

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To Solve:

Solve the given pair of equations using the substitution method.

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Solution:

The substitution method involves solving one equation for one variable and substituting that expression into the other equation.

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

$x + 2y = 3$

Now, substitute this expression for $x$ into equation (1):

$7x – 15y = 2$

$7(3 - 2y) – 15y = 2$

Distribute the 7:

$21 - 14y – 15y = 2$

Combine the terms with $y$:

$21 - 29y = 2$

Subtract 21 from both sides:

$-29y = 2 - 21$

$-29y = -19$

Divide by -29 to solve for $y$:

Now that we have the value of $y$, substitute it back into equation (3) to find the value of $x$:

$x = 3 - 2y$

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

$x = 3 - \frac{38}{29}$

To subtract, find a common denominator:

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

$x = \frac{87}{29} - \frac{38}{29}$

$x = \frac{87 - 38}{29}$

So, the solution to the system of equations is $x = \frac{49}{29}$ and $y = \frac{19}{29}$.

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Verification (Optional):

Substitute $x = \frac{49}{29}$ and $y = \frac{19}{29}$ into the original equations.

Check equation (1): $7x – 15y = 2$

$7(\frac{49}{29}) - 15(\frac{19}{29}) = \frac{343}{29} - \frac{285}{29} = \frac{343 - 285}{29} = \frac{58}{29} = 2$

Check equation (2): $x + 2y = 3$

$\frac{49}{29} + 2(\frac{19}{29}) = \frac{49}{29} + \frac{38}{29} = \frac{49 + 38}{29} = \frac{87}{29} = 3$

The solution is correct.

Given:

The problem describes a relationship between Aftab's current age and his daughter's current age based on two different time periods.

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To Represent and Solve:

1. Represent the situation algebraically by forming a pair of linear equations.

2. Solve the system of equations using the substitution method.

3. Represent the situation graphically and use the graph to find the solution.

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Formulation of Algebraic Equations:

Let Aftab's present age be $x$ years.

Let his daughter's present age be $y$ years.

According to the first condition:

Seven years ago, Aftab's age was $x - 7$ years.

Seven years ago, his daughter's age was $y - 7$ years.

The condition states, "Seven years ago, I (Aftab) was seven times as old as you (daughter) were then".

So, we can write the equation:

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

Expand the right side:

$x - 7 = 7y - 49$

Rearrange the terms to form a linear equation in the standard form $ax + by = c$:

$x - 7y = -49 + 7$

According to the second condition:

Three years from now, Aftab's age will be $x + 3$ years.

Three years from now, his daughter's age will be $y + 3$ years.

The condition states, "three years from now, I (Aftab) shall be three times as old as you (daughter) will be".

So, we can write the equation:

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

Expand the right side:

$x + 3 = 3y + 9$

Rearrange the terms to form a linear equation:

$x - 3y = 9 - 3$

The situation is represented algebraically by the pair of linear equations:

$x - 7y = -42$

$x - 3y = 6$

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Solution using Substitution Method (Algebraic):

We can use the substitution method to solve this system of equations. From equation (2), it is easy to express $x$ in terms of $y$:

Now, substitute this expression for $x$ from equation (3) into equation (1):

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

Combine the terms involving $y$:

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

$-4y + 6 = -42$

Subtract 6 from both sides of the equation:

$-4y = -42 - 6$

$-4y = -48$

Divide both sides by -4 to find the value of $y$:

$y = \frac{-48}{-4}$

Now that we have the value of $y$, substitute $y = 12$ back into equation (3) (or equation (1) or (2)) to find the value of $x$:

$x = 3y + 6$

$x = 3(12) + 6$

$x = 36 + 6$

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

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Graphical Representation:

To represent the situation graphically, we plot the graphs of the two linear equations on the same coordinate plane. The point of intersection will give the solution $(x, y)$. The equations are:

Equation 1: $x - 7y = -42$. We can write $7y = x + 42$, so $y = \frac{x + 42}{7}$.

Equation 2: $x - 3y = 6$. We can write $3y = x - 6$, so $y = \frac{x - 6}{3}$.

Let's find some points for plotting the graph of Equation 1 ($y = \frac{x + 42}{7}$):

x	y = $\frac{x+42}{7}$	Point (x, y)
0	$\frac{0+42}{7} = 6$	(0, 6)
7	$\frac{7+42}{7} = \frac{49}{7} = 7$	(7, 7)
-7	$\frac{-7+42}{7} = \frac{35}{7} = 5$	(-7, 5)
42	$\frac{42+42}{7} = \frac{84}{7} = 12$	(42, 12)

Let's find some points for plotting the graph of Equation 2 ($y = \frac{x - 6}{3}$):

x	y = $\frac{x-6}{3}$	Point (x, y)
0	$\frac{0-6}{3} = -2$	(0, -2)
6	$\frac{6-6}{3} = 0$	(6, 0)
3	$\frac{3-6}{3} = -1$	(3, -1)
42	$\frac{42-6}{3} = \frac{36}{3} = 12$	(42, 12)

Plot these points on a graph paper with the x-axis representing Aftab's age and the y-axis representing his daughter's age. Draw a straight line through the points for the first equation and another straight line through the points for the second equation on the same graph. Since age cannot be negative, we are mainly interested in the region where $x \ge 0$ and $y \ge 0$.

The graphs of the two lines will intersect at the point (42, 12). This intersection point is the graphical solution, confirming the algebraic solution that Aftab's age is 42 and his daughter's age is 12.

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Answer:

Algebraic Representation:

The pair of linear equations is:

$x - 7y = -42$

$x - 3y = 6$

By solving graphically (finding the intersection point), we find the solution.

Present age of Aftab = 42 years.

Present age of daughter = 12 years.

Given:

The cost of 2 pencils and 3 erasers is $\textsf{₹} 9$.

The cost of 4 pencils and 6 erasers is $\textsf{₹} 18$.

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To Find:

The cost of each pencil and each eraser.

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Solution:

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

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

According to the first condition, "the cost of 2 pencils and 3 erasers is $\textsf{₹} 9$":

According to the second condition, "the cost of 4 pencils and 6 erasers is $\textsf{₹} 18$":

We have a pair of linear equations:

$2x + 3y = 9$

$4x + 6y = 18$

Let's compare the ratios of the coefficients to determine the nature of the solution.

Rewrite the equations in the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

$2x + 3y - 9 = 0$

$4x + 6y - 18 = 0$

Here, $a_1 = 2$, $b_1 = 3$, $c_1 = -9$, and $a_2 = 4$, $b_2 = 6$, $c_2 = -18$.

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

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

$\frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2}$

Comparing the ratios, we find that:

Since the ratios of the coefficients are equal, the lines representing these equations are coincident. This means the system of equations has infinitely many solutions.

In other words, the second equation is simply a multiple of the first equation ($4x + 6y = 18$ is $2 \times (2x + 3y) = 2 \times 9$). The information provided in the second statement is not independent of the first statement.

Because there are infinitely many pairs of $(x, y)$ that satisfy these equations, we cannot determine a unique cost for each pencil and each eraser based on the given information.

Any pair of values for $x$ and $y$ that satisfies the equation $2x + 3y = 9$ is a valid solution for the cost of a pencil and an eraser according to this problem.

For example, if a pencil costs $\textsf{₹} 3$, then $2(3) + 3y = 9 \implies 6 + 3y = 9 \implies 3y = 3 \implies y = 1$. So, one possibility is that a pencil costs $\textsf{₹} 3$ and an eraser costs $\textsf{₹} 1$.

However, if a pencil costs $\textsf{₹} 0$, then $2(0) + 3y = 9 \implies 3y = 9 \implies y = 3$. So, another possibility is that a pencil costs $\textsf{₹} 0$ and an eraser costs $\textsf{₹} 3$.

Without additional independent information, we cannot find a single specific cost for a pencil and an eraser.

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Answer:

The given pair of linear equations has infinitely many solutions. Therefore, the cost of each pencil and each eraser cannot be uniquely determined from the given information.

Given:

The equations representing two rails:

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To Determine:

Will the rails cross each other? (i.e., do the lines intersect?).

---

Solution:

The rails will cross each other if the lines representing their paths intersect. We can determine this by comparing the ratios of the coefficients of the given linear equations.

The given equations are in the standard form $ax + by + c = 0$.

From equation (1), we have $a_1 = 1$, $b_1 = 2$, and $c_1 = -4$.

From equation (2), we have $a_2 = 2$, $b_2 = 4$, and $c_2 = -12$.

Calculate the ratios of the coefficients:

Ratio of x-coefficients: $\frac{a_1}{a_2} = \frac{1}{2}$

Ratio of y-coefficients: $\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$

Ratio of constant terms: $\frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}$

Now, compare the ratios:

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

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

$\frac{c_1}{c_2} = \frac{1}{3}$

We observe that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{1}{2}$, but $\frac{c_1}{c_2} = \frac{1}{3}$.

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

This condition indicates that the lines representing the given pair of linear equations are parallel and distinct.

Parallel lines never intersect.

Therefore, the rails represented by these equations will not cross each other.

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Answer:

No, the rails will not cross each other because the lines representing their paths are parallel.

Solution (i):

Given:

The pair of linear equations:

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To Solve:

Solve the given pair of equations using the substitution method.

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Solution:

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

Substitute this expression for $x$ into equation (1):

$(4 + y) + y = 14$

Combine the terms with $y$:

$4 + 2y = 14$

Subtract 4 from both sides:

$2y = 14 - 4$

$2y = 10$

Divide by 2:

Now, substitute the value of $y = 5$ into equation (3) to find the value of $x$:

$x = 4 + y$

$x = 4 + 5$

The solution is $x = 9$ and $y = 5$.

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Solution (ii):

Given:

The pair of linear equations:

---

To Solve:

Solve the given pair of equations using the substitution method.

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Solution:

First, clear the denominators in equation (2) by multiplying by the LCM of 3 and 2, which is 6:

$6 \times (\frac{s}{3} + \frac{t}{2}) = 6 \times 6$

$2s + 3t = 36$

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

Substitute this expression for $s$ into equation (3):

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

Distribute the 2:

$6 + 2t + 3t = 36$

Combine the terms with $t$:

$6 + 5t = 36$

Subtract 6 from both sides:

$5t = 36 - 6$

$5t = 30$

Divide by 5:

Now, substitute the value of $t = 6$ into equation (4) to find the value of $s$:

$s = 3 + t$

$s = 3 + 6$

The solution is $s = 9$ and $t = 6$.

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Solution (iii):

Given:

The pair of linear equations:

---

To Solve:

Solve the given pair of equations using the substitution method.

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Solution:

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

$3x - y = 3$

$-y = 3 - 3x$

Substitute this expression for $y$ into equation (2):

$9x – 3(3x - 3) = 9$

Distribute the -3:

$9x – 9x + 9 = 9$

Combine the terms with $x$:

$0x + 9 = 9$

The equation $9 = 9$ is a true statement that does not involve $x$ or $y$. This indicates that the two original equations are dependent, meaning they represent the same line (coincident lines). Thus, there are infinitely many solutions.

Any pair $(x, y)$ that satisfies $3x - y = 3$ (or $9x - 3y = 9$) is a solution.

We can express the solution in terms of $x$ (or $y$). From equation (1), the solution is $y = 3x - 3$, where $x$ is any real number.

Conclusion: The system has infinitely many solutions.

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Solution (iv):

Given:

The pair of linear equations:

---

To Solve:

Solve the given pair of equations using the substitution method.

---

Solution:

First, multiply both equations by 10 to remove the decimals:

$10 \times (0.2x + 0.3y) = 10 \times 1.3 \implies 2x + 3y = 13$

$10 \times (0.4x + 0.5y) = 10 \times 2.3 \implies 4x + 5y = 23$

From equation (3), express $2x$ in terms of $y$:

Substitute this expression for $2x$ into equation (4) (note that $4x = 2 \times (2x)$):

$2(2x) + 5y = 23$

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

Distribute the 2:

$26 - 6y + 5y = 23$

Combine the terms with $y$:

$26 - y = 23$

Subtract 26 from both sides:

$-y = 23 - 26$

$-y = -3$

Now, substitute the value of $y = 3$ into equation (5) to find the value of $x$:

$2x = 13 - 3y$}

$2x = 13 - 3(3)$

$2x = 13 - 9$

$2x = 4$

Divide by 2:

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

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Solution (v):

Given:

The pair of linear equations:

---

To Solve:

Solve the given pair of equations using the substitution method.

---

Solution:

From equation (1), we can express $\sqrt{2}x$ in terms of $y$:

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

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

$\sqrt{3} x - \sqrt{8} y = 0$

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

Simplify the terms. Note that $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$.

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

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

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

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

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

$-3y - 2(2) y = 0$

$-3y - 4y = 0$

Combine the terms with $y$:

$-7y = 0$

Divide by -7:

Now, substitute the value of $y = 0$ into equation (3) to find the value of $x$:

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

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

The solution is $x = 0$ and $y = 0$.

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Solution (vi):

Given:

The pair of linear equations:

---

To Solve:

Solve the given pair of equations using the substitution method.

---

Solution:

First, clear the denominators in both equations.

For equation (1), multiply by the LCM of 2 and 3, which is 6:

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

$3(3x) - 2(5y) = -12$

For equation (2), multiply by the LCM of 3, 2, and 6, which is 6:

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

$2x + 3y = 13$

Now we solve the system of equations (3) and (4).

From equation (4), we can express $x$ in terms of $y$:

$2x = 13 - 3y$

Substitute this expression for $x$ into equation (3):

$9x - 10y = -12$

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

Multiply the entire equation by 2 to clear the denominator:

$2 \times [9(\frac{13 - 3y}{2}) - 10y] = 2 \times (-12)$

$9(13 - 3y) - 20y = -24$

Distribute the 9:

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

Combine the terms with $y$:

$117 - 47y = -24$

Subtract 117 from both sides:

$-47y = -24 - 117$

$-47y = -141$

Divide by -47:

Now, substitute the value of $y = 3$ into equation (5) to find the value of $x$:

$x = \frac{13 - 3y}{2}$

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

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

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

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

Given:

The pair of linear equations:

The relationship $y = mx + 3$.

---

To Find:

1. Solve the given pair of equations.

2. Find the value of $m$ for which $y = mx + 3$.

---

Solution:

We can solve the system of equations using the substitution method or elimination method. Let's use substitution.

From equation (1), express $2x$ in terms of $y$:

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

$2x – 4y = – 24$

$(11 - 3y) – 4y = – 24$

Combine the terms with $y$:

$11 - 7y = -24$

Subtract 11 from both sides:

$-7y = -24 - 11$

$-7y = -35$

Divide by -7 to solve for $y$:

Now, substitute the value of $y = 5$ into equation (3) to find the value of $x$:

$2x = 11 - 3y$}

$2x = 11 - 3(5)$

$2x = 11 - 15$

$2x = -4$

Divide by 2:

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

---

We are given the relationship $y = mx + 3$. We need to find the value of $m$ for which this equation holds true at the solution of the given pair of linear equations.

Substitute the values of $x = -2$ and $y = 5$ into the equation $y = mx + 3$:

$5 = m(-2) + 3$

$5 = -2m + 3$

Subtract 3 from both sides:

$5 - 3 = -2m$

$2 = -2m$

Divide by -2 to solve for $m$:

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

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

---

Answer:

Solution to the pair of equations is $x = -2$, $y = 5$.

The value of $m$ is -1.

Solution for (i):

Let the two numbers be $x$ and $y$, where $x$ is the larger number.

According to the first condition, the difference between the two numbers is 26.

According to the second condition, one number is three times the other.

Now, we use the substitution method. Substitute the value of $x$ from equation (2) into equation (1).

$3y - y = 26$

$2y = 26$

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

$y = 13$

Now, substitute the value of $y$ in equation (2) to find $x$.

$x = 3y$

$x = 3(13)$

$x = 39$

Thus, the two numbers are 39 and 13.

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Solution for (ii):

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

Since the angles are supplementary, their sum is $180^\circ$.

According to the second condition, the larger angle exceeds the smaller by $18^\circ$.

Now, we use the substitution method. Substitute the value of $x$ from equation (2) into equation (1).

$(y + 18^\circ) + y = 180^\circ$

$2y + 18^\circ = 180^\circ$

$2y = 180^\circ - 18^\circ$

$2y = 162^\circ$

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

$y = 81^\circ$

Now, substitute the value of $y$ in equation (2) to find $x$.

$x = y + 18^\circ$

$x = 81^\circ + 18^\circ$

$x = 99^\circ$

Thus, the two supplementary angles are $99^\circ$ and $81^\circ$.

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Solution for (iii):

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

According to the first condition, the coach buys 7 bats and 6 balls for $\textsf{₹}3800$.

According to the second condition, she buys 3 bats and 5 balls for $\textsf{₹}1750$.

Now, we use the substitution method. From equation (2), we can express $x$ in terms of $y$.

$3x = 1750 - 5y$

$x = \frac{1750 - 5y}{3}$

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

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

Multiply the entire equation by 3 to eliminate the denominator.

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

$12250 - 35y + 18y = 11400$

$12250 - 17y = 11400$

$-17y = 11400 - 12250$

$-17y = -850$

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

$y = 50$

Now, substitute the value of $y = 50$ in equation (3) to find $x$.

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

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

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

$x = 500$

Thus, the cost of each bat is $\textsf{₹}500$ and the cost of each ball is $\textsf{₹}50$.

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Solution for (iv):

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

According to the first condition, for a distance of 10 km, the charge is $\textsf{₹}105$.

According to the second condition, for a journey of 15 km, the charge is $\textsf{₹}155$.

Now, we use the substitution method. From equation (1), we can express $x$ in terms of $y$.

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

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

$105 + 5y = 155$

$5y = 155 - 105$

$5y = 50$

$y = \frac{50}{5}$

$y = 10$

Now, substitute the value of $y = 10$ in equation (3) to find $x$.

$x = 105 - 10(10)$

$x = 105 - 100$

$x = 5$

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

Now, we need to find the charge for travelling a distance of 25 km.

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

Total charge $= x + 25y$

Substitute $x=5$ and $y=10$.

Total charge $= 5 + 25(10)$

Total charge $= 5 + 250$

Total charge $= 255$

A person has to pay $\textsf{₹}255$ for travelling a distance of 25 km.

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Solution for (v):

Let the fraction be $\frac{x}{y}$, where $x$ is the numerator and $y$ is the denominator ($y \neq 0$).

According to the first condition, if 2 is added to both numerator and denominator, the fraction becomes $\frac{9}{11}$.

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

Cross-multiply:

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

$11x + 22 = 9y + 18$

$11x - 9y = 18 - 22$

According to the second condition, if 3 is added to both numerator and denominator, the fraction becomes $\frac{5}{6}$.

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

Cross-multiply:

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

$6x + 18 = 5y + 15$

$6x - 5y = 15 - 18$

Now, we use the substitution method. From equation (2), we can 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 the entire equation by 6 to eliminate the denominator.

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

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

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

$y - 33 = -24$

$y = -24 + 33$

$y = 9$

Now, substitute the value of $y = 9$ in equation (3) to find $x$.

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

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

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

$x = 7$

Thus, the numerator is 7 and the denominator is 9. The fraction is $\frac{7}{9}$.

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Solution for (vi):

Let the present age of Jacob be $J$ years and the present age of his son be $S$ years.

Five years hence (after 5 years):

Jacob's age will be $J+5$ years.

Son's age will be $S+5$ years.

According to the first condition, five years hence, Jacob's age will be three times that of his son.

$J+5 = 3(S+5)$

$J+5 = 3S + 15$

$J - 3S = 15 - 5$

Five years ago:

Jacob's age was $J-5$ years.

Son's age was $S-5$ years.

According to the second condition, five years ago, Jacob's age was seven times that of his son.

$J-5 = 7(S-5)$

$J-5 = 7S - 35$

$J - 7S = -35 + 5$

Now, we use the substitution method. From equation (1), we can express $J$ in terms of $S$.

Substitute the value of $J$ from equation (3) into equation (2).

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

$3S - 7S + 10 = -30$

$-4S + 10 = -30$

$-4S = -30 - 10$

$-4S = -40$

$S = \frac{-40}{-4}$

$S = 10$

Now, substitute the value of $S = 10$ in equation (3) to find $J$.

$J = 3(10) + 10$

$J = 30 + 10$

$J = 40$

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

Solution:

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

Let their monthly expenditures be $\textsf{₹}4y$ and $\textsf{₹}3y$ respectively, where $y$ is a constant.

We know that Saving = Income - Expenditure.

According to the problem, each person saves $\textsf{₹}2000$ per month.

For the first person:

For the second person:

We will solve this system of linear equations using the substitution method.

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

$7x - 2000 = 3y$

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

$9x - 4\left(\frac{7x - 2000}{3}\right) = 2000$

Multiply the entire equation by 3 to eliminate the denominator:

$3(9x) - 4(7x - 2000) = 3(2000)$

$27x - 28x + 8000 = 6000$

$-x + 8000 = 6000$

$-x = 6000 - 8000$

$-x = -2000$

$x = 2000$

Now, substitute the value of $x = 2000$ into equation (3) to find the value of $y$ (although finding $y$ is not strictly necessary to find the incomes, we can calculate it to check our work or if expenditures were asked).

$y = \frac{7(2000) - 2000}{3}$

$y = \frac{14000 - 2000}{3}$

$y = \frac{12000}{3}$

$y = 4000$

The monthly incomes are $9x$ and $7x$.

First person's monthly income $= 9x = 9 \times 2000 = \textsf{₹}18000$.

Second person's monthly income $= 7x = 7 \times 2000 = \textsf{₹}14000$.

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

Solution:

The given pair of linear equations is:

To eliminate one variable, we can multiply equation (1) by 2:

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

Now, subtract equation (2) from equation (3):

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

$4x + 6y - 4x - 6y = 9$

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

$0 + 0 = 9$

$0 = 9$

Since we have arrived at a false statement ($0=9$), this indicates that the given pair of linear equations has no solution.

The lines represented by these equations are parallel and distinct.

Solution:

Let the two-digit number be $10t + u$, where $t$ is the tens digit and $u$ is the units digit.

Here, $t$ and $u$ are integers, $t \in \{1, 2, ..., 9\}$ and $u \in \{0, 1, ..., 9\}$.

The number obtained by reversing the digits is $10u + t$.

According to the first condition, the sum of the number and the number obtained by reversing the digits is 66.

$(10t + u) + (10u + t) = 66$

$11t + 11u = 66$

Divide both sides by 11:

According to the second condition, the digits of the number differ by 2. This gives us two possible cases:

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Case 1: The tens digit is greater than the units digit by 2.

We have a system of linear equations:

$t + u = 6$

$t - u = 2$

Using the substitution method, from equation (2), we get $t = u + 2$.

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

$(u + 2) + u = 6$

$2u + 2 = 6$

$2u = 6 - 2$

$2u = 4$

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

$u = 2$

Now substitute the value of $u = 2$ back into $t = u + 2$:

$t = 2 + 2$

$t = 4$

The digits are $t=4$ and $u=2$. Both are valid digits (t is not 0, u is a digit).

The number is $10t + u = 10(4) + 2 = 40 + 2 = 42$.

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Case 2: The units digit is greater than the tens digit by 2.

This can be rewritten as:

We have a system of linear equations:

$t + u = 6$

$t - u = -2$

Using the substitution method, from equation (4), we get $t = u - 2$.

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

$(u - 2) + u = 6$

$2u - 2 = 6$

$2u = 6 + 2$

$2u = 8$

$u = \frac{8}{2}$

$u = 4$

Now substitute the value of $u = 4$ back into $t = u - 2$:

$t = 4 - 2$

$t = 2$

The digits are $t=2$ and $u=4$. Both are valid digits (t is not 0, u is a digit).

The number is $10t + u = 10(2) + 4 = 20 + 4 = 24$.

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Both cases result in a valid two-digit number that satisfies the given conditions.

The possible numbers are 42 and 24.

There are two such numbers.

(i) Solve: $x + y = 5$ and $2x – 3y = 4$

Elimination Method:

Given equations:

Multiply equation (1) by 3 to make the coefficient of $y$ equal to the coefficient of $y$ in equation (2) (with opposite sign):

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

Add equation (3) and equation (2):

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

$3x + 3y + 2x - 3y = 19$

$(3x + 2x) + (3y - 3y) = 19$

$5x + 0 = 19$

$5x = 19$

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

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

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

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

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

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

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

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

So, the solution is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

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Substitution Method:

Given equations:

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

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

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

$10 - 2y - 3y = 4$

$10 - 5y = 4$

$-5y = 4 - 10$

$-5y = -6$

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

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

Substitute the value of $y = \frac{6}{5}$ into equation (3):

$x = 5 - \frac{6}{5}$

$x = \frac{5 \times 5}{5} - \frac{6}{5}$

$x = \frac{25}{5} - \frac{6}{5}$

$x = \frac{25 - 6}{5}$

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

So, the solution is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

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(ii) Solve: $3x + 4y = 10$ and $2x – 2y = 2$

Elimination Method:

Given equations:

Multiply equation (2) by 2 to make the coefficient of $y$ equal to the coefficient of $y$ in equation (1) (with opposite sign):

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

Add equation (1) and equation (3):

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

$3x + 4y + 4x - 4y = 14$

$(3x + 4x) + (4y - 4y) = 14$

$7x + 0 = 14$

$7x = 14$

$x = \frac{14}{7}$

$x = 2$

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

$2(2) - 2y = 2$

$4 - 2y = 2$

$-2y = 2 - 4$

$-2y = -2$

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

$y = 1$

So, the solution is $x = 2$ and $y = 1$.

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Substitution Method:

Given equations:

From equation (2), we can simplify it by dividing by 2:

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

$x - y = 1$

Express $x$ in terms of $y$ from the simplified equation:

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

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

$3y + 3 + 4y = 10$

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

$7y + 3 = 10$

$7y = 10 - 3$

$7y = 7$

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

$y = 1$

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

$x = 1 + 1$

$x = 2$

So, the solution is $x = 2$ and $y = 1$.

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

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

Elimination Method:

Multiply equation (1) by 3 to make the coefficient of $x$ equal to the coefficient of $x$ in equation (2):

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

Subtract equation (3) from equation (2):

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

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

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

$0 + 13y = -5$

$13y = -5$

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

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

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

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

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

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

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

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

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

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

$x = \frac{27}{39}$

$x = \frac{\cancel{27}^9}{\cancel{39}_{13}}$

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

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

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Substitution Method:

Given equations:

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

$3x = 5y + 4$

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

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

$\cancel{9}^3\left(\frac{5y + 4}{\cancel{3}_1}\right) - 2y = 7$

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

$15y + 12 - 2y = 7$

$(15y - 2y) + 12 = 7$

$13y + 12 = 7$

$13y = 7 - 12$

$13y = -5$

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

Substitute the value of $y = -\frac{5}{13}$ into equation (3):

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

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

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

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

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

$x = \frac{\cancel{27}^9}{\cancel{39}_{13}}$

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

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

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

First, clear the denominators in both equations.

For the first equation, multiply by the LCM of 2 and 3, which is 6:

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

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

$\cancel{6}^3 \times \frac{x}{\cancel{2}_1} + \cancel{6}^2 \times \frac{2y}{\cancel{3}_1} = -6$

$3x + 4y = -6$

For the second equation, multiply by 3:

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

$3 \times x - 3 \times \frac{y}{3} = 9$

$3x - \cancel{3} \times \frac{y}{\cancel{3}} = 9$

$3x - y = 9$

We now solve the system:

$3x + 4y = -6$

$3x - y = 9$

Elimination Method:

Given equations:

Subtract equation (2) from equation (1) to eliminate $x$:

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

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

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

$0 + 5y = -15$

$5y = -15$

$y = \frac{-15}{5}$

$y = -3$

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

$3x - (-3) = 9$

$3x + 3 = 9$

$3x = 9 - 3$

$3x = 6$

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

$x = 2$

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

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Substitution Method:

Given equations:

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

$3x - 9 = y$

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

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

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

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

$15x - 36 = -6$

$15x = -6 + 36$

$15x = 30$

$x = \frac{30}{15}$

$x = 2$

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

$y = 3(2) - 9$

$y = 6 - 9$

$y = -3$

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

(i) Fraction Problem

Let the fraction be $\frac{x}{y}$, where $x$ is the numerator and $y$ is the denominator ($y \neq 0$).

According to the first condition:

If we add 1 to the numerator and subtract 1 from the denominator, the fraction becomes 1.

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

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

$x+1 = y-1$

$x - y = -1 - 1$

According to the second condition:

It becomes $\frac{1}{2}$ if we only add 1 to the denominator.

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

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

$2x = y+1$

$2x - y = 1$

Now, we solve the pair of linear equations using the elimination method.

Subtract equation (1) from equation (2):

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

$2x - y - x + y = 1 + 2$

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

$x + 0 = 3$

$x = 3$

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

$3 - y = -2$

$-y = -2 - 3$

$-y = -5$

$y = 5$

The numerator is 3 and the denominator is 5. The fraction is $\frac{3}{5}$.

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(ii) Ages Problem

Let the present age of Nuri be $N$ years and the present age of Sonu be $S$ years.

Five years ago:

Nuri's age = $N-5$ years

Sonu's age = $S-5$ years

According to the first condition:

Five years ago, Nuri was thrice as old as Sonu.

$N-5 = 3(S-5)$

$N-5 = 3S - 15$

$N - 3S = -15 + 5$

Ten years later:

Nuri's age = $N+10$ years

Sonu's age = $S+10$ years

According to the second condition:

Ten years later, Nuri will be twice as old as Sonu.

$N+10 = 2(S+10)$

$N+10 = 2S + 20$

$N - 2S = 20 - 10$

Now, we solve the pair of linear equations using the elimination method.

Subtract equation (1) from equation (2):

$(N - 2S) - (N - 3S) = 10 - (-10)$

$N - 2S - N + 3S = 10 + 10$

$(N - N) + (-2S + 3S) = 20$

$0 + S = 20$

$S = 20$

Substitute the value of $S = 20$ into equation (2):

$N - 2(20) = 10$

$N - 40 = 10$

$N = 10 + 40$

$N = 50$

Thus, 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 two-digit number be $10t + u$, where $t$ is the tens digit and $u$ is the units digit.

Here, $t \in \{1, 2, ..., 9\}$ and $u \in \{0, 1, ..., 9\}$.

The number obtained by reversing the digits is $10u + t$.

According to the first condition:

The sum of the digits is 9.

According to the second condition:

Nine times this number is twice the number obtained by reversing the order of the digits.

$9(10t + u) = 2(10u + t)$

$90t + 9u = 20u + 2t$

$90t - 2t = 20u - 9u$

$88t = 11u$

Divide both sides by 11:

$\frac{88t}{11} = \frac{11u}{11}$

$8t = u$

Rearranging in standard form:

Now, we solve the pair of linear equations using the elimination method.

Add equation (1) and equation (2) to eliminate $u$:

$(t + u) + (8t - u) = 9 + 0$

$t + u + 8t - u = 9$

$(t + 8t) + (u - u) = 9$

$9t + 0 = 9$

$9t = 9$

$t = \frac{9}{9}$

$t = 1$

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

$1 + u = 9$

$u = 9 - 1$

$u = 8$

The tens digit is 1 and the units digit is 8. ($t=1$ is not 0, $u=8$ is a digit).

The number is $10t + u = 10(1) + 8 = 10 + 8 = 18$.

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(iv) Bank Withdrawal Problem

Let the number of $\textsf{₹}50$ notes Meena received be $f$.

Let the number of $\textsf{₹}100$ notes Meena received be $h$.

According to the first condition:

Meena got 25 notes in all.

According to the second condition:

The total amount withdrawn is $\textsf{₹}2000$.

(Value of $\textsf{₹}50$ notes) + (Value of $\textsf{₹}100$ notes) = Total amount

$50f + 100h = 2000$

Divide the equation by 50 to simplify:

$\frac{50f}{50} + \frac{100h}{50} = \frac{2000}{50}$

$f + 2h = 40$

Now, we solve the pair of linear equations using the elimination method.

Subtract equation (1) from equation (2) to eliminate $f$:

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

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

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

$0 + h = 15$

$h = 15$

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

$f + 15 = 25$

$f = 25 - 15$

$f = 10$

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

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(v) Lending Library Problem

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

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

According to the first condition:

Saritha paid $\textsf{₹}27$ for a book kept for seven days.

Number of extra days beyond the first three = $7 - 3 = 4$ days.

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

$27 = F + 4D$

According to the second condition:

Susy paid $\textsf{₹}21$ for the book she kept for five days.

Number of extra days beyond the first three = $5 - 3 = 2$ days.

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

$21 = F + 2D$

Now, we solve the pair of linear equations using the elimination method.

Subtract equation (2) from equation (1) to eliminate $F$:

$(F + 4D) - (F + 2D) = 27 - 21$

$F + 4D - F - 2D = 6$

$(F - F) + (4D - 2D) = 6$

$0 + 2D = 6$

$2D = 6$

$D = \frac{6}{2}$

$D = 3$

Substitute the value of $D = 3$ into equation (2):

$F + 2(3) = 21$

$F + 6 = 21$

$F = 21 - 6$

$F = 15$

Thus, the fixed charge for the first three days is $\textsf{₹}15$ and the additional charge for each extra day is $\textsf{₹}3$.