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 = \(\frac{73}{4}\) = 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) = 010 = 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) = 100 = 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
110 = 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) 3x4y = 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, \(\frac{4}{5}\)) 