AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.2

Question 1.
Find three different solutions of the each of the following equations.
i) 3x + 4y = 7
Solution:
Given equation is 3x + 4y = 7

 Choice of  value of x Or y Simplification for y or x Solution x = 0 3 x 0 + 4y = $$\frac{7}{4}$$ (0, $$\frac{7}{4}$$ ) y = 0 3x + 4(0) = 7 ⇒ x = $$\frac{7}{3}$$ ($$\frac{7}{3}$$ ,0) x = 1 3(1) + 4y = 7 ⇒ y = $$\frac{7-3}{-4}$$ = 1 (1, 1)

Choice of x or y Simplification for y or x
Solution

ii) y = 6x
Solution: Given equation is y = 6x ⇒ 6x – y = 0

 Choice of  value of x Or y Simplification for y or x Solution x = 0 6(0) – y = 0 ⇒ y = 0 (0,0) y = 0 6x – 0 = 0 ⇒ x = 0 (0,0) x = 1 6(1) – y = 0 ⇒ y = 6 (1,6) Y = 1 6x – 1 = 0 ⇒ 6x = 1 ⇒ x = $$\frac{1}{6}$$ ($$\frac{1}{6}$$,1)

iii) 2x – y = 7
Solution:
Given equation is 2x – y = 7

 Choice of  value of x Or y Simplification for y or x Solution x = 0 2(0) – y = 7 ⇒ y = -7 (0, -7) y = 0 2x – 0 = 7 ⇒ x = $$\frac{7}{2}$$ ($$\frac{7}{2}$$ , 0) x = 1 2(1) – y = 7 ⇒ -y = 7 – 2  ⇒ y = -5 (1, -5)

iv) 13x – 12y = 25
Solution:
Given equation is 13x – 12y = 25

 Choice of  value of x Or y Simplification for y or x Solution: x = 0 13(0) – 12y = 25 ⇒ y = $$-\frac{25}{12}$$ (0, $$-\frac{25}{12}$$ ) y = 0 13x – 12(0) = 25 ⇒ y = $$\frac{25}{13}$$ ($$\frac{25}{13}$$ ,0) x = 1 13(1) – 12y = 25 ⇒ -12y = 25 – 13 y = $$\frac{12}{-12}$$ = -1 (1, -1)

v) 10x + 11y = 21
Solution:
Given equation is 10x + 11y = 21

 Choice of  value of x Or y Simplification for y or x Solution x = 0 10(0) + 11y = 21 ⇒ y = $$\frac{21}{11}$$ (0, $$\frac{21}{11}$$) y = 0 10x +11(0) = 21 ⇒ x = $$\frac{21}{10}$$ ($$\frac{21}{10}$$ , 0) x = 1 10(1) + 11y = 21 ⇒ 11y = 21 – 10  ⇒ y = $$\frac{11}{11}$$ = 1 (1, 1)

vi) x + y = 0
Solution:
Given equation is x + y = 0

 Choice of a value of x or y Simplification Solution x = 0 0 + y = 0 ⇒ y = 0 (0, 0) x = 1 1 + y = 0 ⇒ y = -1 (1,-1) y = 1 x + 1 = 0 ⇒ x = -1 (- 1, 1)

Question 2.
If (0, a) and (b, 0) are the solutions of the following linear equations. Find a and b.
8x – y = 34
Solution:
Given that (0, a) and (b, 0) are the solutions of 8x – y = 34
∴ 8(0) – a = 34 and 8(b) – 0 = 34
⇒ a = -34 and b = $$\frac{34}{8}=\frac{17}{4}$$

ii) 3x = 7y – 21
Solution:
Given that (0, a) and (b, 0) are the solutions of 3x = 7y – 21
⇒ 3x – 7y = – 21
∴ 3(0) – 7(a) = -21 and 3(b) – 7(0) = -21
⇒ a = $$\frac{-21}{-7}$$ and b = $$\frac{-21}{3}$$
⇒ a = 3 and b = -7

iii) 5x – 2y + 3 = 0
Given that (0, a) and (b, 0) are the solutions of 5x – 2y = – 3
i. e., 5(0) – 2a = – 3 and 5 (b) – 2(0) = – 3
⇒ -2a = -3 and 5b = -3
⇒ a = $$\frac{3}{2}$$ and b = $$\frac{-3}{5}$$

Question 3.
Check which of the following is solution of the equation 2x – 5y = 10.
(i) (0, 2) (ii) (0,-2) (iii)(5, 0) (iv) (2√3, -√3) (v) ($$\frac{1}{2}$$ , 2)
Solution:
i) (0, 2)
The given equation is 2x – 5y = 10
On substituting (0, 2), the L.H.S becomes
2(0)-5(2) = 0-10 = -10
R.H.S = 10
L.H.S ≠ R.H.S
∴ (0, 2) is not a solution.

ii) (0. – 2)
Substituting (0, – 2) in the L.H.S of 2x – 5y = 10, we get
2(0) – 5 (- 2) = 0 + 10 = 10 = R.H.S
∴ (0, – 2) is a solution.

iii) (5, 0)
Substituting (5, 0) in the L.H.S of 2x – 5y = 10, we get
2(5)-5(0) = 10-0 = 10 = R.H.S
∴ (5, 0) is a solution.

iv) 2√3, -√3
On substituting (2√3, -√3 ), the L.H.S becomes
2(2√3) – 5(-√3) = 4√3 + 5√3 = 9√3 ≠ R.H.S
∴ (2√3, -√3) is not a solution.

v) ($$\frac{1}{2}$$, 2)
Given equation is 2x – 5y = 10
Put x = $$\frac{1}{2}$$ and y = 2 in the given equation.
Then 2($$\frac{1}{2}$$) – 5(2) =10
1-10 = 10
-9 = 10 false
∴ ($$\frac{1}{2}$$ , 2) is not a solution.

Question 4.
Find the value of k, if x = 2; y = 1 is a solution of the equation 2x + 3y = k. Find two more solutions of the resultant equation. £3 Each
Solution:
Given that x = 2 and y = 1 is a solution of 2x + 3y = k .
∴ 2(2) + 3(1) = k ⇒ 4 + 3 = k ⇒ k = .7 ’
∴ The equation becomes 2x + 3y = 7

 x 0 1 y 2(0) + 3y = 7 3y = 7 y = $$\frac{7}{3}$$ 2(1) + 3y = 7 3y = 7 – 2= y = $$\frac{5}{3}$$ (x, y) (0,$$\frac{7}{3}$$) (1, $$\frac{5}{3}$$)

Two more solutions are (0, $$\frac{7}{3}$$) and (1, $$\frac{5}{3}$$)

Question 5.
If x = 2 – α and y = 2 + α is a solution of the equation 3x – 2y + 6 = 0, find the value of ‘α’. Find three more solutions of the resultant equation.
Solution:
Given that x = 2 – α and y = 2 + α is a solution of 3x – 2y + 6 = 0
Thus 3 (2 – α) – 2 (2 + α) + 6 = 0
⇒ 6 – 3α – 4 – 2α + 6 = 0
⇒ -5α + 8 = 0
⇒ -5α = -8
∴ α = $$\frac{8}{5}$$
Three more solutions are

 x 0 3x – 2y = -6 3x – 2(0) = -6 y = $$\frac{-6}{3}$$ = -2 1 y 3x – 2y = -6 3(0) – 2y = -6 y = 3 0 3x – 2y = -6 3(1) – 2y = -6 y = $$\frac{9}{2}$$ Solutions (0, 3) (-2, 0) (1, $$\frac{9}{2}$$)

Question 6.
If x = 1;y = 1 is a solution of the equation 3x + ay = 6, find the value of ‘a’.
Solution:
Given that x = 1; y = 1 is a solution of 3x + ay = 6.
Thus 3(1) + a(1) = 6
⇒ 3 + a = 6 ⇒ a = 6 – 3 = 3

Question 7.
Write five different linear equations in two variables and find three solutions for
each of them.
Solution:
i) Let the equations are 2x – 4y = 10

 x 0 1 2 y 2(0) – 4y = 10 y = $$\frac{-5}{2}$$ 2x – 4y = 10 2(1)  – 4y =10y = -2 2x – 4y = 10 2(2) – 4y = 10 y = –$$\frac{-3}{2}$$) Solutions (0, –$$\frac{-5}{2}$$) (1 , -2) (2, $$\frac{-3}{2}$$)

ii) 5x + 6y = 15

 x 0 5x + 6(0) = 15 x = 3 1 y 5(0) + 6y = 15 y = $$\frac{15}{6}$$ = $$\frac{5}{2}$$ 0 5(1) + 6y = 15 y = $$\frac{5}{3}$$ (x, y) (0, $$\frac{5}{2}$$) (3, 0 ) (1, $$\frac{5}{3}$$)

iii) 3x-4y = 12

 x 0 4 1 y -3 0 $$\frac{-9}{4}$$ (x, y) (0, -3) (4, 0 ) (1, $$\frac{-9}{4}$$)

iv) 2x – 7y = 9

 x 0 $$\frac{9}{2}$$ 1 y $$\frac{-9}{7}$$ 0 -1 (x, y) (0, $$\frac{-9}{7}$$) ( $$\frac{9}{2}$$ , 0 ) (1, -1)

v) 7x- 5y = 3

 x 0 $$\frac{3}{7}$$ 1 y $$\frac{-3}{5}$$ 0 $$\frac{4}{5}$$ (x, y) (0, $$\frac{-3}{5}$$ ) ($$\frac{3}{7}$$, 0 ) (1, $$\frac{4}{5}$$)