SEBA Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Solutions – Complete Guide

Exercise 3.1 Solutions – Pair of Linear Equations in Two Variables

Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students 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.

Answer:
Let number of boys be x and number of girls be y.

Total students = 10
\[ i.e x + y = 10 \quad ...(1) \]
also, Girls are 4 more than boys:
\[ so, y = x + 4 \quad ...(2) \]

Table for equation (1):

x	y
2	8
6	4

Table for equation (2):

x	y
1	5
3	7

Graph of x + y = 10 and y = x + 4

From the graph, the lines intersect at (3, 7).

Therefore,
Number of boys = 3
Number of girls = 7

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(ii) 5 pens and 7 pencils together cost Rs 50, whereas 7 pens and 5 pencils together cost Rs 46. Find the cost of one pencil and one pen.

Answer:
Let the cost of one pencil be x and cost of one pen be y.

\[so, as per the question 7x + 5y = 50 \quad ...(1) and \] \[ 5x + 7y = 46 \quad ...(2) \]

Table for equation (1):

x	y
5	3
0	10

Table for equation (2):

x	y
5	3
2	4

Graph of 7x + 5y = 50 and 5x + 7y = 46:

From the graph, the lines intersect at (5, 3).

Therefore,
Cost of pencil = Rs 5
Cost of pen = Rs 3

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Q2. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find whether the lines intersect, are parallel or coincident.

(i) \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\)
\[ \frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3} \] Since, \[ \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] Therefore, the lines are intersecting.

(ii) \(9x + 3y + 12 = 0\) and \(18x + 6y + 24 = 0\)
\[ \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2} \] Since, \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Therefore, the lines are coincident.

(iii) \(6x - 3y + 10 = 0\) and \(2x - y + 9 = 0\)
\[ \frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9} \] Since, \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \] Therefore, the lines are parallel.

(iv) \(6x - y + 9 = 0\) and \(2x - 3y + 10 = 0\)
\[ \frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3} \] Since, \[ \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] Therefore, the lines are intersecting.

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Q3. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the following pair of linear equations is consistent, or inconsistent.

(i) \(3x + 2y = 5\) and \(2x - 3y = 7\)
\[ \frac{a_1}{a_2} = \frac{3}{2}, \quad \frac{b_1}{b_2} = \frac{2}{-3} \] Since, \[ \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] Therefore, the lines are intersecting.
Hence, the system is consistent (unique solution).

(ii) \(2x - 3y = 8\) and \(4x - 6y = 9\)
\[ \frac{a_1}{a_2} = \frac{2}{4}, \quad \frac{b_1}{b_2} = \frac{-3}{-6}, \quad \frac{c_1}{c_2} = \frac{8}{9} \] Since, \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \] Therefore, the lines are parallel.
Hence, the system is inconsistent (no solution).

(iii) \( \frac{2}{3}x + \frac{3}{7}y = 7\) and \(9x - 10y = 14\)
Convert: \[ 14x + 9y = 147,\quad 9x - 10y = 14 \] \[ \frac{a_1}{a_2} = \frac{14}{9}, \quad \frac{b_1}{b_2} = \frac{9}{-10} \] Since, \[ \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] Therefore, the lines are intersecting.
Hence, the system is consistent (unique solution).

(iv) \(5x - 3y = 11\) and \(-10x + 6y = -22\)
\[ \frac{a_1}{a_2} = \frac{5}{-10}, \quad \frac{b_1}{b_2} = \frac{-3}{6}, \quad \frac{c_1}{c_2} = \frac{11}{-22} \] Since, \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Therefore, the lines are coincident.
Hence, the system is consistent (infinitely many solutions).

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(v) \( \frac{3}{4}x + 2y = 8\) and \(2x + 3y = 12\)
Convert: \[ 3x + 8y = 32,\quad 2x + 3y = 12 \] \[ \frac{a_1}{a_2} = \frac{3}{2}, \quad \frac{b_1}{b_2} = \frac{8}{3} \] Since, \[ \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] Therefore, the lines are intersecting.
Hence, the system is consistent (unique solution).

Q4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.

(i) \(x + y = 5\) and \(2x + 2y = 10\)
\[ \frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{5}{10} = \frac{1}{2} \] Since, \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Therefore, the lines are coincident.
Hence, the system is consistent (infinitely many solutions).

Table for \(x + y = 5\)

x	y
1	4
3	2

Graph of \(x + y = 5\) and \(2x + 2y = 10\)

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(ii) \(x - 2y = 4\) and \(2x - 4y = 9\)
\[ \frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{4}{9} \] Since, \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \] Therefore, the lines are parallel.
Hence, the system is inconsistent (no solution).

(iii) \(2x - y - 6 = 0\) and \(4x - 2y - 5 = 0\)
Convert: \[ 2x - y = 6,\quad 4x - 2y = 5 \] \[ \frac{a_1}{a_2} = \frac{2}{4}, \quad \frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{6}{5} \] Since, \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \] Therefore, the lines are parallel.
Hence, the system is inconsistent (no solution).

(iv) \(2x - y + 2 = 0\) and \(4x - y - 5 = 0\)
Convert: \[ 2x - y = -2,\quad 4x - y = 5 \] \[ \frac{a_1}{a_2} = \frac{2}{4}, \quad \frac{b_1}{b_2} = \frac{-1}{-1} = 1 \] Since, \[ \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] Therefore, the lines are intersecting.
Hence, the system is consistent (unique solution).

Table for \(2x - y = -2\)

x	y
0	2
1	4

Table for \(4x - y = 5\)

x	y
1	-1
2	3

Graph of \(2x - y + 2 = 0\) and \(4x - y - 5 = 0\)

Solution point: \( \left(\frac{7}{2}, 9\right) \)

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Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer:
Let width of the garden be x m and length be y m.

Given, half of the perimeter = 36 m
\[ i.e\; x + y = 36 \quad ...(1) \]
also,
Length is 4 m more than width:
\[ so,\; y = x + 4 \quad ...(2) \]

Substituting (2) in (1):
\[ x + (x + 4) = 36 \] \[ 2x + 4 = 36 \] \[ 2x = 32 \Rightarrow x = 16 \]
Now, from (2):
\[ y = 16 + 4 = 20 \]

Therefore,
Width of the garden = 16 m
Length of the garden = 20 m

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Q6. Given the linear equation \(2x + 3y - 8 = 0\), write another linear equation such that the pair of equations formed is:

(i) Intersecting lines
Answer:
Given equation: \[ 2x + 3y - 8 = 0 \]
For intersecting lines,
\[ since\; \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] Choose another equation such that ratios are not equal.
\[ so,\; x - y + 1 = 0 \]

Therefore, the given pair represents intersecting lines.

(ii) Parallel lines
Answer:
Given equation: \[ 2x + 3y - 8 = 0 \]
For parallel lines,
\[ since\; \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \] Multiply first equation by 2:
\[ 4x + 6y - 16 = 0 \] Change constant term:
\[ so,\; 4x + 6y + 5 = 0 \]

Therefore, the lines are parallel.

(iii) Coincident lines
Answer:
Given equation: \[ 2x + 3y - 8 = 0 \]
For coincident lines,
\[ since\; \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Multiply by 2:
\[ so,\; 4x + 6y - 16 = 0 \]
Therefore, both equations represent the same line (coincident lines).

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Q7. Draw the graphs of the equations \(x - y + 1 = 0\) and \(3x + 2y - 12 = 0\). Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer:
Given equations:
\[ x - y + 1 = 0 \quad ...(1) \] \[ 3x + 2y - 12 = 0 \quad ...(2) \]
From (1):
\[ x = y - 1 \; \text{or} \; y = x + 1 \]
From (2):
\[ x = \frac{12 - 2y}{3} \]

Table for equation (1):

x	0	1	2
y	1	2	3

Table for equation (2):

x	4	2	0
y	0	3	6

Graph of \(x - y + 1 = 0\) and \(3x + 2y - 12 = 0\)

Now, find intersection points with x-axis:

For (1):
\[ y = 0 \Rightarrow x = -1 \Rightarrow (-1,0) \]
For (2):
\[ y = 0 \Rightarrow 3x = 12 \Rightarrow x = 4 \Rightarrow (4,0) \]
Now, intersection of both lines:
\[ y = x + 1 \] Put in (2):
\[ 3x + 2(x+1) = 12 \] \[ 5x + 2 = 12 \Rightarrow x = 2,\; y = 3 \]

Therefore, vertices of triangle are:
\[ (-1,0),\; (4,0),\; (2,3) \]

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Q8. For what value of p does the pair of equations have a unique solution?

\[ 4x + py + 8 = 0 \] \[ 2x + 2y + 2 = 0 \]

(a) \(p = 4\)
(b) \(p \ne 4\)
(c) \(p = -4\)
(d) \(p \ne -4\)

Answer: (b) \(p \ne 4\)

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Q9. Which of the following has infinitely many solutions?

(a) \(2x + 3y = 6\) and \(4x + 6y - 12 = 0\)
(b) \(x + y = 4\) and \(x - y - 2 = 0\)
(c) \(x - y = 3\) and \(x + y = 7\)
(d) \(x + y = 5\) and \(2x - y = 3\)

Answer: (a)

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Q10. In the following question a statement of assertion (A) is followed by a statement of reason (R).

Assertion (A): The equation \(2x - 3y = 0\) has an infinite number of solutions.
Reason (R): A linear equation in two variables represents a straight line on the coordinate plane.

(a) Both A and R are true and R is correct explanation
(b) Both A and R are true but R is not correct explanation
(c) A is true but R is false
(d) A is false but R is true

Answer: (a)

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Exercise 3.2 Solutions – Pair of Linear Equations in Two Variables

Q1. Solve the following pair of linear equations by the substitution method.

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

Answer:
Given:
\[ x + y = 14 \quad ...(1) \] \[ x - y = 4 \quad ...(2) \]
From (2),
\[ x = y + 4 \quad ...(3) \]
Substituting the value of x from (3) in (1), we get:
\[ (y + 4) + y = 14 \] \[ 2y + 4 = 14 \] \[ 2y = 10 \Rightarrow y = 5 \]
Now substituting the value of y in (3), we get:
\[ x = 5 + 4 = 9 \]

Therefore, \(x = 9,\; y = 5\)

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(ii) \(5 - x = 3\) and \( \frac{5}{3} + \frac{x}{2} = 6\)

Answer:
Given:
\[ 5 - x = 3 \quad ...(1) \] \[ \frac{5}{3} + \frac{x}{2} = 6 \quad ...(2) \]
From (1),
\[ x = 2 \quad ...(3) \]
Substituting the value of x in (2), we get:
\[ \frac{5}{3} + \frac{2}{2} = 6 \] \[ \frac{5}{3} + 1 = \frac{8}{3} \ne 6 \]
Since both equations are not satisfied together,
Therefore, the system has no solution.

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

Answer:
Given:
\[ 3x - y = 3 \quad ...(1) \] \[ 9x - 3y = 9 \quad ...(2) \]
From (1),
\[ y = 3x - 3 \quad ...(3) \]
Substituting the value of y from (3) in (2), we get:
\[ 9x - 3(3x - 3) = 9 \] \[ 9x - 9x + 9 = 9 \] \[ 9 = 9 \]
This is true for all values of x.

Therefore, the system has infinitely many solutions.

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

Answer:
Given:
\[ 0.2x + 0.3y = 1.3 \quad ...(1) \] \[ 0.4x + 0.5y = 2.3 \quad ...(2) \]
Multiplying both equations by 10, we get:
\[ 2x + 3y = 13 \quad ...(1) \] \[ 4x + 5y = 23 \quad ...(2) \]
From (1),
\[ 2x = 13 - 3y \Rightarrow x = \frac{13 - 3y}{2} \quad ...(3) \]
Substituting the value of x from (3) in (2), we get:
\[ 4\left(\frac{13 - 3y}{2}\right) + 5y = 23 \] \[ 2(13 - 3y) + 5y = 23 \] \[ 26 - 6y + 5y = 23 \] \[ 26 - y = 23 \Rightarrow y = 3 \]
Now substituting the value of y in (3), we get:
\[ x = \frac{13 - 9}{2} = 2 \]

Therefore, \(x = 2,\; y = 3\)

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

Answer:
Given:
\[ \sqrt{2}x + \sqrt{3}y = 0 \quad ...(1) \] \[ \sqrt{3}x - \sqrt{8}y = 0 \quad ...(2) \]
From (1),
\[ \sqrt{2}x = -\sqrt{3}y \Rightarrow x = -\frac{\sqrt{3}}{\sqrt{2}}y \quad ...(3) \]
Substituting the value of x from (3) in (2), we get:
\[ \sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0 \] \[ -\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0 \]
Taking LCM and simplifying, we get:
\[ y = 0 \]
Now substituting the value of y in (3), we get:
\[ x = 0 \]

Therefore, \(x = 0,\; y = 0\)

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

Answer:
Given:
\[ \frac{3x}{2} - \frac{5y}{3} = -2 \quad ...(1) \] \[ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \quad ...(2) \]
Multiplying both equations by 6, we get:
\[ 9x - 10y = -12 \quad ...(1) \] \[ 2x + 3y = 13 \quad ...(2) \]
From (2),
\[ 2x = 13 - 3y \Rightarrow x = \frac{13 - 3y}{2} \quad ...(3) \]
Substituting the value of x from (3) in (1), we get:
\[ 9\left(\frac{13 - 3y}{2}\right) - 10y = -12 \] \[ \frac{117 - 27y}{2} - 10y = -12 \] \[ 117 - 27y - 20y = -24 \] \[ 117 - 47y = -24 \Rightarrow y = 3 \]
Now substituting the value of y in (3), we get:
\[ x = \frac{13 - 9}{2} = 2 \]
Therefore, \(x = 2,\; y = 3\)

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Q2. Solve \(2x + 3y = 11\) and \(2x - 4y = -24\) and hence find the value of ‘m’ for which \(y = mx + 3\).

Answer:
Given:
\[ 2x + 3y = 11 \quad ...(1) \] \[ 2x - 4y = -24 \quad ...(2) \]
From (1),
\[ 2x = 11 - 3y \Rightarrow x = \frac{11 - 3y}{2} \quad ...(3) \]
Substituting the value of x from (3) in (2), we get:
\[ 2\left(\frac{11 - 3y}{2}\right) - 4y = -24 \] \[ 11 - 3y - 4y = -24 \] \[ 11 - 7y = -24 \Rightarrow -7y = -35 \Rightarrow y = 5 \]
Now substituting the value of y in (3), we get:
\[ x = \frac{11 - 15}{2} = -2 \]

Now given: \[ y = mx + 3 \] Substitute values:
\[ 5 = m(-2) + 3 \] \[ 5 = -2m + 3 \Rightarrow -2m = 2 \Rightarrow m = -1 \]

Therefore, \(x = -2,\; y = 5\) and \(m = -1\)

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Q3. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Answer:
Let the two numbers be x and y.

Given:
\[ x - y = 26 \quad ...(1) \] One number is three times the other:
\[ x = 3y \quad ...(2) \]
Substituting (2) in (1), we get:
\[ 3y - y = 26 \] \[ 2y = 26 \Rightarrow y = 13 \]
Now from (2):
\[ x = 3 \times 13 = 39 \]

Therefore, the numbers are 39 and 13.

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(ii) The larger of two supplementary angles exceeds the smaller by 18°. Find them.

Answer:
Let thee smaller angle be x and larger angle be y.

Since angles are supplementary:
\[ x + y = 180 \quad ...(1) \] Given:
\[ y = x + 18 \quad ...(2) \]
Substituting (2) in (1), we get:
\[ x + (x + 18) = 180 \] \[ 2x + 18 = 180 \] \[ 2x = 162 \Rightarrow x = 81 \]
Now from (2):
\[ y = 81 + 18 = 99 \]

Therefore, the angles are 81° and 99°.

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(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, he buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and ball.

Answer:
Let cost of one bat be x and cost of one ball be y.

\[ 7x + 6y = 3800 \quad ...(1) \] \[ 3x + 5y = 1750 \quad ...(2) \]
From (2),
\[ 3x = 1750 - 5y \Rightarrow x = \frac{1750 - 5y}{3} \quad ...(3) \]
Substituting (3) in (1), we get:
\[ 7\left(\frac{1750 - 5y}{3}\right) + 6y = 3800 \] \[ \frac{12250 - 35y}{3} + 6y = 3800 \] \[ 12250 - 35y + 18y = 11400 \] \[ 12250 - 17y = 11400 \Rightarrow 17y = 850 \Rightarrow y = 50 \]
Now from (3):
\[ x = \frac{1750 - 250}{3} = 500 \]

Therefore, cost of one bat = Rs 500 and one ball = Rs 50.

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(iv) The taxi charges in a city consist of a fixed charge together with the charge for distance covered. For a journey of 10 km, the charge paid is Rs 105 and for 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km?

Answer:
Let fixed charge of the taxi be x and charge per km be y.

\[ x + 10y = 105 \quad ...(1) \] \[ x + 15y = 155 \quad ...(2) \]
From (1),
\[ x = 105 - 10y \quad ...(3) \]
Substituting (3) in (2), we get:
\[ 105 - 10y + 15y = 155 \] \[ 105 + 5y = 155 \] \[ 5y = 50 \Rightarrow y = 10 \]
Now from (3):
\[ x = 105 - 100 = 5 \]

Therefore, fixed charge = Rs 5 and charge per km = Rs 10.

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(v) A fraction becomes \( \frac{9}{11} \), if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \( \frac{5}{6} \). Find the fraction.

Answer:
Let the numerator be x and denominator be y.
Given:
\[ \frac{x+2}{y+2} = \frac{9}{11} \quad ...(1) \] \[ \frac{x+3}{y+3} = \frac{5}{6} \quad ...(2) \]
From (1):
\[ 11(x+2) = 9(y+2) \] \[ 11x + 22 = 9y + 18 \] \[ 11x - 9y = -4 \quad ...(3) \]
From (2):
\[ 6(x+3) = 5(y+3) \] \[ 6x + 18 = 5y + 15 \] \[ 6x - 5y = -3 \quad ...(4) \]
From (4):
\[ 6x = 5y - 3 \Rightarrow x = \frac{5y - 3}{6} \quad ...(5) \]
Substituting (5) in (3), we get:
\[ 11\left(\frac{5y - 3}{6}\right) - 9y = -4 \] \[ \frac{55y - 33}{6} - 9y = -4 \] \[ 55y - 33 - 54y = -24 \] \[ y - 33 = -24 \Rightarrow y = 9 \]
Now from (5):
\[ x = \frac{45 - 3}{6} = 7 \]
Therefore, the fraction is \( \frac{7}{9} \).

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(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. Find their present ages.

Answer:
Let the present age of Jacob be x years and his son be y years.
Given:
\[ x + 5 = 3(y + 5) \quad ...(1) \] \[ x - 5 = 7(y - 5) \quad ...(2) \]
From (1):
\[ x + 5 = 3y + 15 \] \[ x = 3y + 10 \quad ...(3) \]
Substituting (3) in (2), we get:
\[ 3y + 10 - 5 = 7(y - 5) \] \[ 3y + 5 = 7y - 35 \] \[ 40 = 4y \Rightarrow y = 10 \]
Now from (3):
\[ x = 3(10) + 10 = 40 \]
Therefore, Jacob’s present age is 40 years and his son’s age is 10 years.

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Q4. If \(x + 3\) is a factor of \(x^3 + ax^2 - bx + 6\) and \(a + b = 7\), find the values of a and b.

Answer:
Since \(x + 3\) is a factor of the given polynomial,
by Factor Theorem, we have:
\[ (-3)^3 + a(-3)^2 - b(-3) + 6 = 0 \] \[ -27 + 9a + 3b + 6 = 0 \] \[ 9a + 3b - 21 = 0 \Rightarrow 3a + b = 7 \quad ...(1) \]
Given:
\[ a + b = 7 \quad ...(2) \]
From (2),
\[ b = 7 - a \quad ...(3) \]
Substituting the value of b from (3) in (1), we get:
\[ 3a + (7 - a) = 7 \] \[ 2a + 7 = 7 \Rightarrow a = 0 \]
Now substituting the value of a in (3), we get:
\[ b = 7 - 0 = 7 \]
Therefore, \(a = 0\) and \(b = 7\).

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Q5. Consider the following pair of linear equations of two variables, then choose the correct option from the given alternatives.

\[ 2x = y \quad \text{and} \quad -5x + 2y - 3 = 0 \]

(i) The graphs of the equations intersect at a point.
(ii) The graphs of the equations are parallel to each other.
(iii) The graphs of the equations coincide.
(iv) The equations have a unique solution.

(a) Both (i) and (ii) are true
(b) Both (i) and (iii) are true
(c) Both (ii) and (iii) are true
(d) Both (i) and (iv) are true

Answer: (d) Both (i) and (iv) are true

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Q6. Match the items of column I with column II and choose the correct option.

Column I
A. \(x - y = 0\)
B. \(2x - 3y = 5\) and \(x - y = 1\)
C. \(x + 2y = 6\) and \(4x + 8y = 24\)
D. \(2x + 3y = 6\) and \(4x + 6y = 10\)

Column II
(i) unique solution
(ii) linear equation in two variables
(iii) no solution
(iv) infinitely many solutions and lines coincide

(a) A → (ii), B → (i), C → (iv), D → (iii)
(b) A → (ii), B → (iv), C → (i), D → (iii)
(c) A → (i), B → (i), C → (ii), D → (iii)
(d) A → (iv), B → (ii), C → (i), D → (iii)

Answer: (a)

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Q7. The value of x for which the pair (x, 4) satisfies the equation \(3x + y = 19\) is:

(a) 6
(b) 5
(c) 3
(d) 4

Answer: (b) 5

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Q8. Two equations are given as \(2x + 4y = 10\) and \(kx + 8y = 20\). For which value of k will the system have infinitely many solutions?

(a) 4
(b) 3
(c) 2
(d) 1

Answer: (a) 4

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Exercise 3.3 Solutions – Pair of Linear Equations in Two Variables

Q1(i). Solve: \(x + y = 5\) and \(2x - 3y = 4\) by substitution and elimination method.

Answer:
Method 1: Substitution Method
Given:
\[ x + y = 5 \quad ...(1) \] \[ 2x - 3y = 4 \quad ...(2) \]
From (1),
\[ x = 5 - y \quad ...(3) \]
Substituting (3) in (2), we get:
\[ 2(5 - y) - 3y = 4 \] \[ 10 - 2y - 3y = 4 \] \[ 10 - 5y = 4 \Rightarrow 5y = 6 \Rightarrow y = \frac{6}{5} \]
Now substituting the value of y in (3), we get:
\[ x = 5 - \frac{6}{5} = \frac{19}{5} \]

Method 2: Elimination Method

Given:
\(x + y = 5 \quad ...(1)\)
\(2x - 3y = 4 \quad ...(2)\)

Multiply (1) by 2:
\(2x + 2y = 10 \quad ...(3)\)

Now subtract (2) from (3):

     \(2x + 2y = 10\)
     \(-\; (2x - 3y = 4)\)
     -------------------------
     \(5y = 6\)
     \(y = \frac{6}{5}\)

Now substitute in (1):
\(x + \frac{6}{5} = 5\)
\(x = \frac{19}{5}\)

Therefore, \(x = \frac{19}{5},\; y = \frac{6}{5}\)

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

Answer:
Method 1: Substitution Method
Given:
\(3x + 4y = 10 \quad ...(1)\)
\(2x - 2y = 2 \quad ...(2)\)
From (2):
\(2x = 2 + 2y \Rightarrow x = 1 + y \quad ...(3)\)
Substituting (3) in (1):
\(3(1 + y) + 4y = 10\)
\(3 + 3y + 4y = 10\)
\(7y = 7 \Rightarrow y = 1\)
Now from (3):
\(x = 1 + 1 = 2\)

Method 2: Elimination Method
Multiply (2) by 2:
\(4x - 4y = 4 \quad ...(3)\)
Now subtract (1) from (3):

     \(4x - 4y = 4\)
     \(-\; (3x + 4y = 10)\)
     -------------------------
     \(x - 8y = -6\)
Put \(y = 1\):
\(x = 2\)

Therefore, \(x = 2,\; y = 1\)

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

Answer:
Method 1: Substitution Method
From second equation:
\(9x = 2y + 7 \Rightarrow x = \frac{2y + 7}{9} \quad ...(1)\)
Substituting in first equation:
\(3\left(\frac{2y + 7}{9}\right) - 5y = 4\)
\(\frac{2y + 7}{3} - 5y = 4\)
\(2y + 7 - 15y = 12\)
\(-13y = 5 \Rightarrow y = -\frac{5}{13}\)
Now substitute in (1):
\(x = \frac{9}{13}\)

Method 2: Elimination Method
Write second equation as:
\(9x - 2y = 7 \quad ...(2)\)
Multiply first equation by 3:
\(9x - 15y = 12 \quad ...(3)\)
Now subtract (2) from (3):

     \(9x - 15y = 12\)
     \(-\; (9x - 2y = 7)\)
     -------------------------
     \(-13y = 5\)
     \(y = -\frac{5}{13}\)
Substitute in first equation:
\(x = \frac{9}{13}\)

Therefore, \(x = \frac{9}{13},\; y = -\frac{5}{13}\)

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

Answer:
Method 1: Substitution Method
From second equation:
\(x = 3 + \frac{y}{3} \quad ...(1)\)
Substitute in first equation:
\[ \frac{2}{3 + y/3} + \frac{2}{y} = -1 \] Solving, we get:
\(y = -3\)
Now from (1):
\(x = 3 - 1 = 2\)

Method 2: Elimination Method
Given equations are not linear in standard form,
so elimination is not convenient here.
We convert and solve using substitution.

Therefore, \(x = 2,\; y = -3\)

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Q2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \( \frac{1}{2} \) if we only add 1 to the denominator. What is the fraction?

Answer:
Let the numerator be x and denominator be y.
Given:
\[ \frac{x+1}{y-1} = 1 \quad ...(1) \]Also, \[ \frac{x}{y+1} = \frac{1}{2} \quad ...(2) \]
From (1):
\[ x+1 = y-1 \Rightarrow x - y = -2 \quad ...(3) \]
From (2):
\[ 2x = y + 1 \Rightarrow 2x - y = 1 \quad ...(4) \]
Now subtract (3) from (4):

     \(2x - y = 1\)
     \(-\; (x - y = -2)\)
     ---------------------
     \(x = 3\)
Substitute in (3):
\(3 - y = -2 \Rightarrow y = 5\)
Therefore, the fraction is \( \frac{3}{5} \).

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(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer:
Let present age of Nuri be x years and Sonu be y years.
Given:
\[ x - 5 = 3(y - 5) \quad ...(1) \]Also, \[ x + 10 = 2(y + 10) \quad ...(2) \]
From (1):
\[ x - 5 = 3y - 15 \Rightarrow x - 3y = -10 \quad ...(3) \]
From (2):
\[ x + 10 = 2y + 20 \Rightarrow x - 2y = 10 \quad ...(4) \]
Now subtract (4) from (3):

     \(x - 3y = -10\)
     \(-\; (x - 2y = 10)\)
     ----------------------
     \(-y = -20\)
     \(y = 20\)
Substitute in (4):
\(x - 40 = 10 \Rightarrow x = 50\)
Therefore, Nuri is 50 years old and Sonu is 20 years old.

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(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the digits of the number. Find the number.

Answer:
Let the tens digit be x and units digit be y.
Number = \(10x + y\)
Reversed number = \(10y + x\)
Given:
\[ x + y = 9 \quad ...(1) \]Also, \[ 9(10x + y) = 2(10y + x) \quad ...(2) \]
From (2):
\[ 90x + 9y = 20y + 2x \] \[ 88x - 11y = 0 \Rightarrow 8x - y = 0 \quad ...(3) \]
Now subtract (3) from (1):

     \(x + y = 9\)
     \(-\; (8x - y = 0)\)
     --------------------
     \(-7x + 2y = 9\)
From (3):
\(y = 8x\)
Substitute in (1):
\(x + 8x = 9 \Rightarrow x = 1\)
\(y = 8\)
Therefore, the number is 18.

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(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of each denomination of Rs 50 and Rs 100 she received.

Answer:
Let number of Rs 50 notes be x and Rs 100 notes be y.
Given:
\[ x + y = 25 \quad ...(1) \]Also, \[ 50x + 100y = 2000 \quad ...(2) \]
Divide (2) by 50:
\[ x + 2y = 40 \quad ...(3) \]
Now subtract (1) from (3):

     \(x + 2y = 40\)
     \(-\; (x + y = 25)\)
     ------------------
     \(y = 15\)
Substitute in (1):
\(x + 15 = 25 \Rightarrow x = 10\)
Therefore, She got 10 notes of Rs 50 and 15 notes of Rs 100.

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(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for a book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:
Let the fixed charge be x and charge per extra day be y.
For 7 days the extra days will be 4
\[ x + 4y = 27 \quad ...(1) \] For 5 days the extra days will be 2
\[ x + 2y = 21 \quad ...(2) \]
Now subtract (2) from (1):

     \(x + 4y = 27\)
     \(-\; (x + 2y = 21)\)
     ------------------
     \(2y = 6\)
     \(y = 3\)
Substitute in (2):
\(x + 6 = 21 \Rightarrow x = 15\)
Therefore, fixed charge = Rs 15 and charge per extra day = Rs 3.

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Q3. The monthly incomes of A and B are in the ratio 9 : 7 and their monthly expenditures are in the ratio 4 : 3. If each saves Rs 1600 per month, find the monthly income of each.

Answer:
Let the incomes of A and B be 9x and 7x.
and expenditures pf Aand B be 4y and 3y.
We know, Savings = Income − Expenditure
Therefore, \[ 9x - 4y = 1600 \quad ...(1) \] \[ 7x - 3y = 1600 \quad ...(2) \]
Now subtract (2) from (1):

     \(9x - 4y = 1600\)
     \(-\; (7x - 3y = 1600)\)
     ----------------------
     \(2x - y = 0 \Rightarrow y = 2x\)
Substitute in (1):
\(9x - 8x = 1600 \Rightarrow x = 1600\)
Income of A = \(9x = 14400\)
Income of B = \(7x = 11200\)
Therefore, incomes are Rs 14400 and Rs 11200.

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Q4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer:
Let number of students in each row be x and number of rows be y.
Total students = xy
\[ (x+3)(y-1) = xy \quad ...(1) \] \[ (x-3)(y+2) = xy \quad ...(2) \]
From (1):
\[ xy - x + 3y - 3 = xy \Rightarrow -x + 3y = 3 \quad ...(3) \]
From (2):
\[ xy + 2x - 3y - 6 = xy \Rightarrow 2x - 3y = 6 \quad ...(4) \]
Now add (3) and (4):

     \(-x + 3y = 3\)
     \(+\; (2x - 3y = 6)\)
     ------------------
     \(x = 9\)
Substitute in (3):
\(-9 + 3y = 3 \Rightarrow y = 4\)
Total students = \(9 \times 4 = 36\)
Therefore, total students = 36.

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Q5. A and B each have certain numbers of mangoes. A says to B, “If you give me 30 of your mangoes, I will have twice as many as left with you.” B replies, “If you give me 10, I will have thrice as many as left with you.” How many mangoes does each have?

Answer:
Let A have x number of mangoes and B have y number of mangoes.
\[ x + 30 = 2(y - 30) \quad ...(1) \] \[ y + 10 = 3(x - 10) \quad ...(2) \]
From (1):
\[ x + 30 = 2y - 60 \Rightarrow x - 2y = -90 \quad ...(3) \]
From (2):
\[ y + 10 = 3x - 30 \Rightarrow 3x - y = 40 \quad ...(4) \]
Multiply (3) by 1:
     \(x - 2y = -90\)
     \(3x - y = 40\)
Solve by elimination:
\[ x = 130,\; y = 110 \]
Therefore, A has 130 mangoes and B has 110 mangoes.

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Q6. If the pair of equations \(x + ay = b\) and \(ax + y = 1\) has infinitely many solutions, then choose the correct option.

(i) \(a = 1, b = 1\)
(ii) \(a = -1, b = -1\)
(iii) \(a = 1, b = -1\)
(iv) \(a = -1, b = 1\)

(a) Both (i) and (iii)
(b) Both (ii) and (iv)
(c) Both (i) and (ii)
(d) Both (iii) and (iv)

Answer: (c)

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Q7. For what value of k, the following pair has infinitely many solutions?

\[ kx + 3y - (k-3) = 0 \] \[ 12x + ky - k = 0 \]

Answer: \(k = 6\)

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Q8. Following are the steps of solving a word problem of linear equations in two variables. Choose the correct sequence:

(i) Represent the unknowns using variables.
(ii) Solve the equations.
(iii) Form the required equations.
(iv) Interpret the result.

(a) (i) → (ii) → (iii) → (iv)
(b) (i) → (iii) → (ii) → (iv)
(c) (iv) → (iii) → (ii) → (i)
(d) (ii) → (i) → (iii) → (iv)

Answer: (b)

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Q9(i). Solve: \( \frac{2}{x} + \frac{3}{y} = 2\) and \( \frac{4}{x} - \frac{9}{y} = -1\)

Answer:
Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \)
Then equations become:
\[ 2a + 3b = 2 \quad ...(1) \] \[ 4a - 9b = -1 \quad ...(2) \]
Multiply (1) by 2:
\(4a + 6b = 4 \quad ...(3)\)
Now subtract (2) from (3):

     \(4a + 6b = 4\)
     \(-\; (4a - 9b = -1)\)
     ----------------------
     \(15b = 5\)
     \(b = \frac{1}{3}\)
Substitute in (1):
\(2a + 1 = 2 \Rightarrow a = \frac{1}{2}\)
\[ x = 2,\; y = 3 \]
Therefore, \(x = 2,\; y = 3\)

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Q9(ii). Solve: \( \frac{1}{2x} + \frac{1}{3y} = 2\) and \( \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}\)

Answer:
Let \( \frac{1}{x} = a \), \( \frac{1}{y} = b \)
\[ \frac{a}{2} + \frac{b}{3} = 2 \quad ...(1) \] \[ \frac{a}{3} + \frac{b}{2} = \frac{13}{6} \quad ...(2) \]
Multiply (1) by 6:
\(3a + 2b = 12 \quad ...(3)\)
Multiply (2) by 6:
\(2a + 3b = 13 \quad ...(4)\)
Multiply (3) by 3 and (4) by 2:
     \(9a + 6b = 36\)
     \(-\; (4a + 6b = 26)\)
     -------------------
     \(5a = 10 \Rightarrow a = 2\)
Substitute in (3):
\(6 + 2b = 12 \Rightarrow b = 3\)
\[ x = \frac{1}{2},\; y = \frac{1}{3} \]
Therefore, \(x = \frac{1}{2},\; y = \frac{1}{3}\)

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Q9(iii). Solve: \( \frac{5}{x-1} + \frac{1}{y-2} = 2\) and \( \frac{6}{x-1} - \frac{3}{y-2} = 1\)

Answer:
Let \( \frac{1}{x-1} = a \), \( \frac{1}{y-2} = b \)
\[ 5a + b = 2 \quad ...(1) \] \[ 6a - 3b = 1 \quad ...(2) \]
Multiply (1) by 3:
\(15a + 3b = 6 \quad ...(3)\)
Now add (2) and (3):

     \(15a + 3b = 6\)
     \(+\; (6a - 3b = 1)\)
     -------------------
     \(21a = 7 \Rightarrow a = \frac{1}{3}\)
Substitute in (1):
\(\frac{5}{3} + b = 2 \Rightarrow b = \frac{1}{3}\)
\[ x = 4,\; y = 5 \]
Therefore, \(x = 4,\; y = 5\)

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Q10. In a two-digit number, the sum of the digits is 9. If the digits are reversed, the number is increased by 9. Find the number.

Answer:
Let tens digit be x and units digit be y.
Then number = \(10x + y\)
and number after reverse = \(10y + x\)
\[ x + y = 9 \quad ...(1) \] \[ 10y + x = 10x + y + 9 \quad ...(2) \]
From (2):
\[ 9y - 9x = 9 \Rightarrow y - x = 1 \quad ...(3) \]
Add (1) and (3):
     \(x + y = 9\)
     \(+\; (y - x = 1)\)
     --------------
     \(2y = 10 \Rightarrow y = 5\)
Then \(x = 4\)
Therefore, number = 45.

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Q11. The sum of the digits of a two-digit number is 15. The number decreased by 27 is equal to the number obtained by reversing the digits. Find the number.

Answer:
Let digits number be x and y.
\[ x + y = 15 \quad ...(1) \] \[ 10x + y - 27 = 10y + x \quad ...(2) \]
From (2):
\[ 9x - 9y = 27 \Rightarrow x - y = 3 \quad ...(3) \]
Add (1) and (3):
     \(x + y = 15\)
     \(+\; (x - y = 3)\)
     --------------
     \(2x = 18 \Rightarrow x = 9\)
Then \(y = 6\)
Therefore, number = 96.

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Q12. The difference between two numbers is 4. Twice the smaller number added to three times the larger number gives 82. Find the two numbers.

Answer:
Let larger number be x and smaller be y.
\[ x - y = 4 \quad ...(1) \] \[ 3x + 2y = 82 \quad ...(2) \]
From (1):
\(x = y + 4 \quad ...(3)\)
Substitute in (2):
\[ 3(y+4) + 2y = 82 \] \[ 3y + 12 + 2y = 82 \] \[ 5y = 70 \Rightarrow y = 14 \]
Then \(x = 18\)
Therefore, the numbers are 18 and 14.

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