RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(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.
(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.
Solution:
(i) Let the number of boys in the quiz be x and the number of girls in the quiz be y.
Total number of students taking part in the quiz = 10
∴ x + y = 10
or x + y – 10 = 0
According to the question
y = x + 4
or x – y = 4
Now let us draw the graph of the linear equations
x + y = 10
and x – y + 4 = 0
x + y = 10
or x = 10 – y …(1)
Putting y = 0 in (1)
x = 10 – 0 = 10
Putting y – 7 in (1)
x = 10 – 7 = 3
Putting y = 10 in (1)
x = 10 – 10 = 0

x	10	3	0
y	0	7	10

Plotting the points A (10, 0), B (3, 7), C (0, 10) and drawing a line joining them, we get the graph of the equation x + y = 10
x – y + 4=0
or x = y – 4 ……(2)
Putting y = 0 in (2)
x = 0 – 4 = -4
Putting y = 7 in (2)
x = 7 – 4 = 3
Putting y = 4 in (2)
x = 4 – 4 = 0

x	– 4	3	0
y	0	7	4

Plotting the points D (- 4, 0). B (3, 7), E (0, 4) and drawing a line joining them, we get the graph of the equation. x – y + 4 = 0. From the graph, it is clear that both the lines meet at the point B (3, 7).
∴ Point B (3, 7) is the graphical situation of the solution.
Therefore number of boys in the quiz = 3 and number of girls in the quiz= 7

(ii) Let the cost of a pencil = ₹ x
and the cost of a pen = ₹ y
According to the first condition
5x + 7y = 50
According to the second condition
7x + 5y = 46
∴ Pair of linear equations is
5x + 7y = 50
7x + 5y = 46
Now let us draw the graph of these linear equations
5x + 7y = 50
or 5x= 50 – 7y ….(1)
or x = \(\frac{50-7 y}{5}\)
Putting y = 0 in (1)
x = \(\frac{50-7 \times 0}{5}\) = \(\frac{50}{5}\)
x = 10
Putting y = 5 in (1)
x = \(\frac{50-7 \times 5}{5}\) = \(\frac{50-35}{5}\)
= \(\frac{15}{5}\) = 3
Putting y = 10 in (1)
x = \(\frac{50-7 \times 10}{5}\) = \(\frac{50-70}{5}\)
= \(\frac{-20}{5}\) = -4

x	10	3	– 4
y	0	5	10

Plotting the points A (10, 0), B (3, 5), C (- 4, 10) and draw ing a line joining them, we get the graph of the equation
5x + 7y = 50
Now from second equation
7x + 5y = 46
or 7x = 46 – 5y

x	– 2	3	8
y	12	5	– 2

Plotting the points E (- 2. 12), B (3, 5), F (8, – 2) and drawing a line joining them, we get the graph of the equation 7x + 5 y = 46
From graph it is clear that both the lines meet at the point B (3, 5)
∴ Point B (3, 5) is the graphical situation of the solution.
Therefore,Cost of one pencil = ₹ 3
Cost of one pen = ₹ 5

Question 2.
On Comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution:
(i) The given pair of linear equations is
5x – 4y + 8 = 0
and 7x + 6y – 9 = 0
Comparing the above equations with
a₁x + b₁y + c₁ = 0
and a₂x + b₂y + c₂ = 0
Here a₁ = 5, b₁ = – 4, c₁ = 8
a₂ = 1, b₂ = 6, c₂ = – 9


Hence the given pair of linear equations intersect at a point.

(ii) The given pair of linear equations is
9x + 3y + 12 = 0
and 18x + 6y + 24 = 0
Comparing the above equations with
a₁x + b₁y + c₁ = 0
and a₂x + b₂y + c₂ = 0
Here a₁ = 9, b₁ = 3, c₁ = 12
a₂ = 18, b₂ = 6,c₂ = 24

Hence the given pair of linear equations are coincident.

(iii) The given pair of linear equations is
6x – 3y + 10 = 0
and 2x – y + 9 = 0
Comparing the above equations with,
a₁x + b₁y + c₁ = 0
and a₂x + b₂y + c₂ = 0
Here a₁ = 6, b₁ = – 3, c₁ = 10
a₂ = 2, b₂ = – 1, c₂ = 9

Hence the given pair of linear equations are parallel to each other.

Question 3.
On comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y = 8; 4x – 6y = 9
(iii) \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11; – 10x + 6y = – 22
(v) \(\frac{4}{3}\)x + 2y = 8 ; 2x + 3y = 12
Solution:
(i) The given pair of linear equations is
3x + 2y = 5
and 2x – 3y = 7
or 3x + 2y – 5 = 0
and 2x – 3y – 7 = 0
Here a₁ = 3, b₁ = 2, c₁ = – 5
a₂ = 2, b₂ = -3, c₂ = -7

Therefore, the pair of linear equations has unique solution. So, the equations are consistent.

(ii) The given pair of linear equations is
2x – 3y = 8
and 4x – 6y = 9
or 2x – 3y – 8 = 0
4x – 6y – 9 = 0
Here a₁ = 2, b₁ = -3, c₂ = – 8
a₂ = 4, b₂ = – 6, c₂ = -9

Hence the given pair of linear equations is consistent.

(iii) The given pair of linear equations is
\(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7
and 9x – 10y = 14

Hence the given pair of linear equations is consistent.

(iv) The given pair of linear equation is
5x – 3y = 11
and – 10x + 6y = – 22
or 5x – 3y – 11 = 0
and – 10x + 6y + 22 = 0
Here a₁ = 5, b₁ = – 3, c₁ = – 11
a₂ = -10, b₂ = 6, c₂ = 22

Therefore, the given linear equations are coincident lines. So, the equation are consistent.

(v) The given pair of linear equations is \(\frac{4}{3}\)x + 2y = 8
and 2x + 3y = 12
or \(\frac{4}{3}\)x + 2y – 8 = 0
and 2x + 3y – 12 = 0
Here a₁ = \(\frac{4}{3}\), b₁ = 2, c₁ = – 8
a₂ = 2, b₂ = 3, c₂ = – 12

Therefore, the given linear equations are coincident lines. So, the equation are consistent.

Question 4.
Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Solution:
(i) The given pair of linear equations is
x + y = 5
and 2x + 2y = 10
or x + y – 5 = 0
and 2x + 2y – 10 = 0
Comparing the above pair of linear equations with the pair of equations
a₁x + b₁y + c₁ = 0
and a₂x + b₂y + c₂ = 0
Here a₁ = 1, b₁ = 1, c₁ = -5
a₂ = 2, b₂ = 2, c₂ = -10

∴ The lines represented by the pair of equations will be consistent.
Hence the given pair of linear equations is consistent.
∴ Let us draw the graph of the given pair of linear equations
x + y = 5
⇒ x = 5 – y …(1)
Putting y = 0 in (1 )x = 5 – 0 = 5
Putting y = 3 in (1 )x = 5 – 3 = 2
Putting y = 5 in (1)x = 5 – 5 = 0

x	5	2	0
y	0	3	5

Plotting the points A(5, 0), B(2, 3), C(0, 5) and drawing a line joining them, wfe get the graph of the equation x + y = 5.
Again 2x + 2y = 10
or 2 (x + y) = 10
or x + y = 5
or x = 5 – y …(2)
Putting y = 0 in (1)
x = 5 – 0 = 5
Putting y = 2 in (2)
x= 5 – 2 = 3
Putting y = 5 in (2)
x = 5 – 5 = 0

x	5	3	0
y	0	2	5

Plotting the points A (5, 0), D (3, 2), C (0, 5) and drawing a line joining them we get the graph of the equation 2x + 2y = 10. From graph it is clear that the given pair of linear equations represents coincident lines. Therefore these equations have infinitely many solutions.

(ii) The given pair of linear equations is
x – y = 8
and 3x – 3y = 16
or x – y – 8 = 0
and 3x – 3y – 16 = 0
Comparing the above pair of equations with the pair of equations
a₁x + b₁y + c₁ = 0
and a₂x + b₂y + c₂ = 0
Here a₁ = 1, b₁ = -1, c₁ = – 8
a₂ = 3, b₂ = -3, c₂ = – 16

∴ The given pair of linear equations will have no solution.
Hence the given pair of linear equations is inconsistent.

(iii) The given pair of linear equations is
2x + y – 6 = 0
and 4x – 2y – 4 = 0
Here a₁ = 2, b₁ = 1, c₁ = – 6
a₂ = 4, b₂ =- 2, c₂ = -4

∴ The given pair of linear equations will have a unique solution.
The given pair of linear equations is consistent.
Let us draw the graph of these linear equations.
2x + y – 6 = 0
or 2x = 6 – y
or x = \(\frac{6-y}{2}\) …(1)
Putting y = 0 in (1)
x = \(\frac{6-0}{2}\) = \(\frac{6}{2}\) = 3
Putting y = 2 in (1)
x = \(\frac{6-2}{2}\) = \(\frac{4}{2}\) = 2
Putting y = – 2 in (1)
x = \(\frac{6-(-2)}{2}\) = \(\frac{6+2}{2}\) = 3
= \(\frac{8}{2}\) = 4

x	3	2	4
y	0	2	– 2

Plotting the points A (3, 0), B (2, 2), C (4, – 2) and drawing a line joining them, we get the graph of the equation 2x + y – 6 = 0
Again 4x – 2y – 4 = 0
or 2 [2x – y – 2] = 0
or 2x – y – 2 = 0
or 2x = y + 2
or x = \(\frac{y+2}{2}\) …(2)
Putting y = 0 in (2),
x = \(\frac{0+2}{2}\) = \(\frac{2}{2}\) = 1
Putting y = 2 in (2),
x = \(\frac{2+2}{2}\) = \(\frac{4}{2}\) = 2
Putting y = – 2 in (2),
x = \(\frac{-2+2}{2}\) = \(\frac{0}{2}\) = 0

x	1	2	0
y	0	2	– 2

Plotting the points D (1, 0), B (2, 2), E (0, – 2) and drawing a line joining them, we get the graph of the equation 4x – 2y – 4 = 0
From the graph it is clear that the given lines meat at the point B (2, 2)
Hence the given pair of linear equations is consistent.

(iv) The given pair of linear equations is
2x – 2y – 2 = 0
and 4x – 4y – 5 = 0
Here a₁ = 2, b₁ = -2, c₂ = -2
a₂ = 4, b₂ = – 4, c₂ = – 5

∴ The given pair of equations have no solution.
Hence the given pair of equations is inconsistent.

Question 5.
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.
Solution:
Let the length of the garden = x m
and the breadth of the garden = y m
∴ Perimeter of the garden
= 2 [x + y] m
∴ Half the perimeter of the garden
= (x + y) m
According to the first condition of the problem
x = y + 4
According to the second condition of the problem
x + y = 36
∴ Pair of linear equations is
x = y + 4
and x + y = 36
x = y + 4 ….(1)
Putting y = 0 in (1),
x = 0 + 4 = 4
Putting y = – 4 in (1),
x = -4 + 4 = 0
Putting y = 16 in (1),
x = 16 + 4 = 20

x	4	0	20
y	0	– 4	16

Plotting the points A (4, 0), B (0, – 4), C (20, 16) and drawing a line joining them, we get the graph of the equation x = y + 4
Again x + y = 36
x = 36 – y …(2)
Putting y = 12 in (2),
x = 36 – 12 = 24
Putting y = 24 in (2),
x= 36 – 24 = 12
Putting y = 16 in (2),
x = 36 – 16 = 20

x	24	12	20
y	12	24	16

Plotting the points D (24, 12), E (12, 24), C (20, 16) and drawing a line joining them, we get the graph of the equation x + y = 36. From graph it is clear that the pair of linear equations meets at point C (20, 16).
∴ C (20, 16), i.e., x = 20 and y = 10 is the solution of the pair of linear equations.
Hence, Length of the garden = 20 m
Width of the garden = 16m

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Solution:
The given linear equation is
2x + 3y – 8 = 0
Comparing this equation with general linear equation
a₁x + b₁y + c₂ = 0.
a₁ = 2, b₁ = 3, c₁ = – 8
(i) When the pair of equations represents the intersecting lines, then the condition is

i.e., the value of a₂ should not be 2 or 0 and the value of b₂ should not be 3 or 0
i.e., a₂ ≠ 2 or b₂ ≠ 3 and a₂ ≠ 0, b₂ ≠ 0 Hence the possible linear equation will be of the type
3x + 2y – 7 = 0

(ii) When the pair of equations represents parallel lines then the condition is

\(\frac{a_{2}}{b_{2}}\) = \(\frac{2}{3}\)
i.e., the ratio of a₂ and b₂ should be 2 : 3 Let a₂ = 2k and b₂ = 3 k, where k is a constant.

Hence the required linear equation is
2kx + 3ky – mk = 0,
m ≠ -8
Hence the possible linear equations will be of the type
2x + 3y – 12 = 0.

(iii) When the pair of equations represents coincident lines then the condition is

Hence the required equation is 2kx + 3ky – 8k = 0, where k is a propotional constant.
Hence the possible linear equation will be of the type 4x + 6y – 16 = 0.

Question 7.
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.
Solution:
Taking pair of linear equations
x – y + 1 = 0
and 3x + 2y – 12 = 0
x – y + 1 = 0
or x = y – 1 …(1)
Putting y = 0 in (1), x = 0 – 1 = – 1
Putting y = 3 in (1), x = 3 – 1 = 2
Putting y = 1 in (1), x = 1 – 1 = 0

x	-1	2	0
y	0	3	1

Plotting the points A (-1, 0), B (2, 3), C (0, 1) and drawing a line joining them, we get the graph of the equation x – y + 1 = 0

3x + 2y – 12 = 0
or 3x = 12 – 2y

x	4	2	0
y	0	3	6

Plotting the points D (4, 0), B (2, 3), E (0, 6) and drawing a line joining them, we get the graph of linear equations 3x – 2y – 12 = 0
The triangular rejoin formed by the pair of linear equations and x-axis has been shaded in the graph. AABD is the triangle fonned thus.
The coordinates of △ABD are : A(- 1, 0), B (2, 3) and D (4, 0)