CBSE 10th Maths - Pair Of Linear Equations in Two Variables - Exercise 3.2

Solution:

(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.

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

According to the question, the algebraic representation is

x + y = 10

x − y = 4

For x + y = 10,

x = 10 − y

X	5	4	6
Y	5	6	4

For x − y = 4,

x = 4 + y

X	5	4	3
Y	1	0	-1

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (7, 3).

Therefore, the number of girls and boys in the class are 7 and 3 respectively.

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

Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.

According to the question, the algebraic representation is

5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,

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

X	3	10	-4
Y	5	0	10

7x + 5y = 46

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

X	8	3	-2
Y	-2	5	12

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (3, 5).

Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

Solution:

(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0

Given expressions;

5x−4y+8 = 0

7x+6y−9 = 0

Comparing these equations with a$_{1}$x+b$_{1}$y+c$_{1}$ = 0

And a$_{2}$x+b$_{2}$y+c$_{2}$ = 0

We get,

a$_{1}$ = 5, b$_{1}$ = -4, c$_{1}$ = 8

a$_{2}$ = 7, b$_{2}$ = 6, c$_{2}$ = -9

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

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

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

Since, $\dfrac{a_{1}}{a_{2}}$ ≠ $\dfrac{b_{1}}{b_{2}}$

So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

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

Given expressions;

9x + 3y + 12 = 0

18x + 6y + 24 = 0

Comparing these equations with a$_{1}$x+b$_{1}$y+c$_{1}$ = 0

And a$_{2}$x+b$_{2}$y+c$_{2}$ = 0

We get,

a$_{1}$ = 9, b$_{1}$ = 3, c$_{1}$ = 12

a$_{2}$ = 18, b$_{2}$ = 6, c$_{2}$ = 24

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{9}{18}$= $\dfrac{19}{2}$

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{3}{6}$= $\dfrac{19}{2}$

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{12}{24}$ = $\dfrac{19}{2}$

Since, $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ = $\dfrac{c_{1}}{c_{2}}$

So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0

Given Expressions;

6x – 3y + 10 = 0

2x – y + 9 = 0

Comparing these equations with a$_{1}$x+b$_{1}$y+c$_{1}$ = 0

And a$_{2}$x+b$_{2}$y+c$_{2}$ = 0

We get,

a$_{1}$ = 6, b$_{1}$ = -3, c$_{1}$ = 10

a$_{2}$ = 2, b$_{2}$ = -1, c$_{2}$ = 9

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

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

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

Since, $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

Solution:

(i) 3x + 2y = 5 ; 2x – 3y = 7

Given : 3x + 2y = 5 or 3x + 2y -5 = 0

and 2x – 3y = 7 or 2x – 3y -7 = 0

Comparing these equations with a$_{1}$x+b$_{1}$y+c$_{1}$ = 0

And a$_{2}$x+b$_{2}$y+c$_{2}$ = 0

We get,

a$_{1}$ = 3, b$_{1}$ = 2, c$_{1}$ = -5

a$_{2}$= 2, b$_{2}$ = -3, c$_{2}$ = -7

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

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

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{-5}{-7}$ = $\dfrac{5}{7}$

Since, $\dfrac{a_{1}}{a_{2}}$ ≠$\dfrac{b_{1}}{b_{1}}$

So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(ii) 2x – 3y = 8 ; 4x – 6y = 9

Given 2x – 3y = 8 and 4x – 6y = 9

Therefore,

a$_{1}$ = 2, b$_{1}$ = -3, c$_{1}$ = -8

a$_{2}$ = 4, b$_{2}$ = -6, c$_{2}$ = -9

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

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

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

Since, $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠$\dfrac{c_{1}}{c_{2}}$

So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

(iii)(3/2)x+(5/3)y = 7; 9x – 10y = 14

Given (3/2)x + (5/3)y = 7 and 9x – 10y = 14

Therefore,

a$_{1}$ = 3/2, b$_{1}$= 5/3, c$_{1}$= -7

a$_{2}$= 9, b$_{2}$ = -10, c$_{2}$ = -14

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{3}{2 \times 9}$ = $\dfrac{1}{6}$

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{5}{3 \times -10}$ = $\dfrac{-1}{6}$

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{-7}{-14}$ = $\dfrac{1}{2}$

Since, $\dfrac{a_{1}}{a_{2}}$ ≠ $\dfrac{b_{1}}{b_{2}}$

So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent.

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

Given, 5x – 3y = 11 and – 10x + 6y = –22

Therefore,

a$_{1}$= 5, b$_{1}$ = -3, c$_{1}$= 11

a$_{2}$ = -10, b$_{2}$ = 6, c$_{2}$ = -22

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

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

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

Since, $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b1}{b2}$ = $\dfrac{c_{1}}{c_{2}}$

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

(v)(4/3)x+2y = 8 ; 2x + 3y = 12

Given,(4/3)x+2y = 8 ; 2x + 3y = 12

Therefore,a$_{1}$ = 4/3 , b$_{1}$= 2 , c$_{1}$ = -8

a$_{2}$ = 2, b$_{2}$ = 3 , c$_{2}$= -12

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{4}{3 \times2 9}$ = $\dfrac{4}{6}$ = $\dfrac{2}{3}$

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

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

Since, $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ = $\dfrac{c_{1}}{c_{2}}$

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

Solution:

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

Given, x + y = 5 and 2x + 2y = 10

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{1}{2}$

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{1}{2}$

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{1}{2}$

Since, $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ = $\dfrac{c_{1}}{c_{2}}$

∴The equations are coincident and they have infinite number of possible solutions.

So, the equations are consistent.

For, x + y = 5 or x = 5 – y

X	4	3	2
Y	1	2	3

For 2x + 2y = 10 or x = (10-2y)/2

X	4	3	2
Y	1	2	3

So, the equations are represented in graphs as follows:

Therefore, the equations have infinite possible solutions.

(ii) x – y = 8, 3x – 3y = 16

Given, x – y = 8 and 3x – 3y = 16

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

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

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{8}{16}$= $\dfrac{1}{2}$

Since, $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0

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

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{1}{-2}$

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{-6}{-4}$= $\dfrac{3}{2}$

Since, $\dfrac{a_{1}}{a_{2}}$ ≠ $\dfrac{b_{1}}{b_{2}}$

The given linear equations are intersecting each other at one point and have only one solution.

Hence, the pair of linear equations is consistent.

Now, for 2x + y – 6 = 0 or y = 6 – 2x

X	0	1	2
Y	6	4	2

And for 4x – 2y – 4 = 0 or y = (4x-4)/2

X	1	2	3
Y	0	2	4

So, the equations are represented in graphs as follows:

From the graph, it can be seen that these lines are intersecting each other at only one point,(2,2).

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0

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

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{-2}{-4}$ = $\dfrac{1}{2}$

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{2}{5}$

Since, $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

Solution:

Let us consider.

The width of the garden is x and length is y.

Now, according to the question, we can express the given condition as;

y – x = 4

and

y + x = 36

Now, taking y – x = 4 or y = x + 4

X	0	8	12
Y	4	12	16

For y + x = 36, y = 36 – x

X	0	36	16
Y	36	0	20

The graphical representation of both the equation is as follows;

From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

Solution:

(i) Intersecting lines

Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;

$\dfrac{a_{1}}{a_{2}}$ ≠ $\dfrac{b_{1}}{b_{2}}$

Thus, another equation could be 2x – 7y + 9 = 0, such that;

$\dfrac{a_{1}}{a_{2}}$ =$\dfrac{2}{2}$ = 1 and $\dfrac{b_{1}}{b_{2}}$ = $\dfrac{3}{-7}$

Clearly, you can see another equation satisfies the condition.

(ii) Parallel lines

Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

Thus, another equation could be 6x + 9y + 9 = 0, such that;

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

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{3}{9}$ = $\dfrac{1}{3}$

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

Clearly, you can see another equation satisfies the condition.

(iii) Coincident lines

Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ = $\dfrac{c_{1}}{c_{2}}$

Thus, another equation could be 4x + 6y – 16 = 0, such that;

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

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

$\dfrac{c_{1}}{c_{2}}$ = -$\dfrac{-8}{-16}$ = $\dfrac{1}{2}$

$\dfrac{a_{1}}{a_{2}}$ = 2/4 = 1/2 ,$\dfrac{b_{1}}{b_{2}}$ = 3/6 = 1/2, $\dfrac{c_{1}}{c_{2}}$ = -8/-16 = 1/2

Clearly, you can see another equation satisfies the condition.

Solution:

Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.

For, x – y + 1 = 0 or x = -1+y

X	0	1	2
Y	1	2	3

For, 3x + 2y – 12 = 0 or x = (12-2y)/3

X	4	2	0
Y	0	3	6

Hence, the graphical representation of these equations is as follows;

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).