Chapter 3 Pair Of Linear Equations In Two Variables

Given:

A pair of linear equations in two variables:

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To Find:

1. Whether the pair is consistent.

2. The graphical solution if they are consistent.

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Solution:

To check for consistency and solve graphically, we need to find at least two points for each equation to plot them on a graph.

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

We can express $y$ in terms of $x$:

If $x = 0$, $y = \frac{6 - 0}{3} = 2$.

If $x = 6$, $y = \frac{6 - 6}{3} = 0$.

Table of values for equation (1):

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

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

We can express $y$ in terms of $x$:

If $x = 0$, $y = \frac{0 - 12}{3} = -4$.

If $x = 6$, $y = \frac{12 - 12}{3} = 0$.

Table of values for equation (2):

$x$	$y = \frac{2x - 12}{3}$	Point $(x, y)$
$0$	$-4$	$(0, -4)$
$6$	$0$	$(6, 0)$

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Graphical Representation:

Now, we plot the points $A(0, 2)$ and $B(6, 0)$ to draw the line for equation (1). Similarly, we plot the points $C(0, -4)$ and $B(6, 0)$ to draw the line for equation (2).

From the graph, we observe that the two lines intersect at the point $(6, 0)$.

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Conclusion:

1. Since the two lines intersect at a point, the pair of equations has a unique solution. Therefore, the pair is consistent.

2. The coordinates of the point of intersection are $(6, 0)$. Thus, the solution is:

$x = 6$, $y = 0$

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Verification:

Substituting $x = 6$ and $y = 0$ in equation (1): $6 + 3(0) = 6 \Rightarrow 6 = 6$ (True).

Substituting $x = 6$ and $y = 0$ in equation (2): $2(6) - 3(0) = 12 \Rightarrow 12 = 12$ (True).

Given:

The pair of linear equations is:

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To Find:

Whether the pair of equations has no solution, unique solution, or infinitely many solutions using the graphical method.

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Solution:

To plot the graphs, we find the coordinates of points satisfying these equations.

For Equation (1): $5x - 8y + 1 = 0$

We can write $y$ in terms of $x$:

Let's find some points for equation (1):

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

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

Multiplying equation (2) by $\frac{5}{3}$:

We observe that Equation (3) is identical to Equation (1). This means the two lines coincide.

Since the lines are the same, every point on the line is a solution.

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Final Answer:

Since the lines coincide, the pair of equations has infinitely many solutions.

Given:

1. The number of skirts is two less than twice the number of pants.

2. The number of skirts is four less than four times the number of pants.

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To Find:

The number of pants and skirts Champa bought.

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Solution:

Let the number of pants be $x$ and the number of skirts be $y$.

According to the given conditions, we have:

Now, we find points for these equations to plot the graph.

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

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

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

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

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Graphical Solution:

Plotting the points on the graph, we draw two lines. The lines for equation (1) and equation (2) intersect at the point $(1, 0)$.

The point of intersection is $(1, 0)$, which means $x = 1$ and $y = 0$.

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Final Answer:

Champa bought 1 pant and 0 skirts.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

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Given:

Total number of students = 10

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

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To Find:

The number of boys and girls using the graphical method.

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Solution:

Let the number of boys be $x$.

Let the number of girls be $y$.

According to the first condition (Total students):

According to the second condition (Girls are 4 more than boys):

On transposing $x$ to LHS, we get:

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To solve graphically, let us find the points for both equations.

For Equation (i): $y = 10 - x$

$x$	$y = 10 - x$	Point $(x, y)$
0	10	(0, 10)
5	5	(5, 5)
10	0	(10, 0)

For Equation (ii): $y = x + 4$

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

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Graphical Representation:

From the graph, it is observed that the two lines intersect at the point (3, 7).

Therefore, $x = 3$ and $y = 7$.

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Final Answer:

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

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(ii) 5 pencils and 7 pens together cost $\textsf{₹} 50$, whereas 7 pencils and 5 pens together cost $\textsf{₹} 46$. Find the cost of one pencil and that of one pen.

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Given:

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

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

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To Find:

The cost of one pencil and one pen using the graphical method.

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Solution:

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

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

According to the given conditions, we have the following pair of linear equations:

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Now, let us find points to plot these equations on a graph.

For Equation (i): $5x + 7y = 50$

$y = \frac{50 - 5x}{7}$

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

For Equation (ii): $7x + 5y = 46$

$y = \frac{46 - 7x}{5}$

$x$	$y = \frac{46 - 7x}{5}$	Point $(x, y)$
3	$\frac{46-21}{5} = 5$	(3, 5)
8	$\frac{46-56}{5} = -2$	(8, -2)

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Graphical Representation:

From the graph, it is observed that the two lines intersect at the point (3, 5).

Therefore, $x = 3$ and $y = 5$.

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Final Answer:

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

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.

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To Find:

The dimensions (length and width) of the garden.

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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.

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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:

The pair of linear equations is:

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To Find:

1. Draw the graphs of these equations.

2. Find the coordinates of the vertices of the triangle formed by these lines and the x-axis.

3. Shade the triangular region.

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Solution:

To draw the graph, we first find at least two or three solutions for each equation.

For Equation (i): $x - y + 1 = 0$

We can write $y$ in terms of $x$ as:

Now, let's find some points:

If $x = 0$, then $y = 0 + 1 = 1$.

If $x = -1$, then $y = -1 + 1 = 0$.

If $x = 2$, then $y = 2 + 1 = 3$.

The table for equation (i) is as follows:

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

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For Equation (ii): $3x + 2y - 12 = 0$

We can write $y$ in terms of $x$ as:

Now, let's find some points:

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

If $x = 4$, then $y = \frac{12 - 3(4)}{2} = \frac{0}{2} = 0$.

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

The table for equation (ii) is as follows:

$x$	$y = \frac{12 - 3x}{2}$	Point $(x, y)$
$0$	$6$	$(0, 6)$
$4$	$0$	$(4, 0)$
$2$	$3$	$(2, 3)$

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Graphical Representation:

Now, we plot the points on a graph paper and draw the lines passing through them.

From the graph, we observe the following:

1. The two lines intersect at point $(2, 3)$.

2. The line $x - y + 1 = 0$ intersects the x-axis at $(-1, 0)$.

3. The line $3x + 2y - 12 = 0$ intersects the x-axis at $(4, 0)$.

The triangular region formed by these lines and the x-axis is bounded by these three points.

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Final Answer:

The coordinates of the vertices of the triangle are:

$(-1, 0)$, $(4, 0)$, and $(2, 3)$.

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:

Aftab's statement: “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.”

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To Find:

Represent the situation algebraically and solve it using the method of substitution, as well as show the graphical representation.

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Solution (Algebraic Representation):

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

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

Case I: Seven years ago

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

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

According to the condition:

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

$x - 7 = 7y - 49$

Case II: Three years from now

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

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

According to the condition:

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

$x + 3 = 3y + 9$

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Solving by Method of Substitution:

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

Substituting this value of $x$ in equation (i), we get:

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

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

$-4y = -48$

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

Now, substitute $y = 12$ in equation (iii):

$x = 3(12) + 6$

$x = 36 + 6$

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Graphical Representation:

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

For Equation (i): $y = \frac{x + 42}{7}$

x	y = $\frac{x + 42}{7}$	Point (x, y)
0	6	(0, 6)
7	7	(7, 7)
14	8	(14, 8)

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

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

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Final Answer:

The present age of Aftab is 42 years and the present age of his daughter is 12 years.

Given:

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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 given conditions, we can form the following pair of linear equations:

We shall solve this using the Substitution Method.

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

Substituting this value of $x$ in equation (ii), we get:

On simplifying the term in the bracket:

This statement is true for all values of $y$. However, we do not get a specific value of $y$ as a solution. This situation arises because both the given equations are equivalent.

If we look at the ratios of the coefficients:

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

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

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

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident and have infinitely many solutions.

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Final Answer:

The cost of each pencil and eraser cannot be uniquely determined because the given equations have infinitely many solutions. Any pair of values $(x, y)$ that satisfies the equation $2x + 3y = 9$ will be a valid cost for the pencil and eraser.

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Alternate Solution (Graphical Method):

We find points for the equation $2x + 3y = 9$ to plot the graph:

Cost of pencil x ($\textsf{₹}$)	Cost of eraser y ($\textsf{₹}$) ($y=\frac{9-2x}{3}$)	Point (x, y)
0	3	(0, 3)
3	1	(3, 1)
4.5	0	(4.5, 0)

Since the second equation $4x + 6y = 18$ is just a multiple of the first, it will produce the same table and the same line on the graph.

The lines coincide; thus, there is no unique solution.

Given:

The equations representing two rails:

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To Determine:

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

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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:

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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:

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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:

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To Solve:

Solve the given pair of equations using the substitution method.

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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:

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To Solve:

Solve the given pair of equations using the substitution method.

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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:

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To Solve:

Solve the given pair of equations using the substitution method.

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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 $y = mx + 3$.

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To Find:

1. Solve the given pair of equations.

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

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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 $y = 5$.

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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.

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

---

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 $y = -3$.

---

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 $\textsf{₹}3$.