Telangana SCERT Solution Class IX (9) Math Chapter 6 Linear Equations in Two Variables Exercise 6.2
(1.) Find three different solutions of the each of the following equations.
i) 3x + 4y = 7 ii) y = 6x iii) 2x − y = 7 iv) 13x − 12y = 25 v) 10x + 11y = 21 vi) x + y = 0
Solution:
(1)(i) Given,
3x + 4y = 7
For, x = 0 we get x,y = (0, 7/4)
3 x 0 + 4y = 7
Or, y = 7/4
For,
y = 0
3 x X + 4 x 0 = 7 ∴ x,y = ( ,0)
Or, x =7/3
For x = 1,
3x 1 + 4y = 7 ∴ x,y = (1, 1)
Or, 4y = 4
Or, y = 1
∴ The three different solutions are (0, 7/4), (7/3 ,0), (1, 1)
(ii) Given,
Y = 6x
For, y = 0,
6x = 0 ∴ x, y = (0, 0)
x = 0
For, x = 0
y = 6 x 0 ∴ x, y = (0, 0)
= 0
For, x = 1
y = 6 x 1 ∴ (1, 6) = x, y
= 6
∴ Three different Solutions are (0, 0) (0, 0), (1, 6)
(iii) given,
2x – y = 7
For, x = 0 x, y = (0, – 7)
2 x 0 – y = 7
Or, y = -7
For, y = o x, y = ( , 0)
2x – 0 = 7
Or, x = 7/2
For x = 1 x, y = (1, -5)
2 x 1 – y = 7
Or, y = -5
Three different Solutions are (0, -7) (7/2, 0), (1, -5)
(iv) 13x – 12y = 25
For x = 0,
13 x 0 – 12 x Y = 25 (x, y) = (0, -25/12)
Or, y = – 25/12
For y = 0,
13X – 12 x 0 = 25 (x, y) = (25/13, 0)
Or, x = 25/13
For, x = 1,
13 x 1 – 12y = 25 (x, y) = (1, 1)
Or, y = (25-13)/12
= 1
Three different solutions (0, – 25/12), (25/13 , 0), (1, 1)
(v) 10x + 11y = 21
For, x = 0
0 + 11y = 21 (x, y) = (0,
y = 21/11
For y = 0
10x + 0 = 21 (x, y) = (21/10 , 0)
Or, x = 21/10
For, x = 1
10 x 1 + 11y = 21 x, y = (1, 1)
Or, y = (21-10)/11
= 1
Three different Solutions are (0, 21/11), (21/10 , 0), (1, 1)
(vi) x + y = 0
For x = 1, x, y = (1, – 1)
y = – 1
For, y = 1, x, y = (-1, 10)
x = -1
For x = 2 x, y = (2, -2)
y = -2
Three different solutions are (1, – 1), (-1, 10), (2, -2)
(2) If (0, a) and (b, 0) are the solutions of the following linear equations. Find ‘a’ and ‘b’.
i) 8x − y = 34 ii) 3x = 7y − 21 iii) 5x − 2y + 3 = 0
Solution:
(2) (i) Given,
8x – y = 34 Solution (0, a); (b, 0)
For, (0, a)
8 x 0 – a = 34
Or, a = -34
For (b, 0)
8 x b – 0 = 34
Or, b = 34/8
Or, b = 17/4
(ii) 3x = 7y – 21 Solution (0, a); (b, 0)
For, (0, a)
3 x 0 = 7 x a – 21
Or, a = 21/7
a = 3
For (b, 0)
3 x b = 7 x 0 – 21
Or, b = -21/3
Or, b = -7
(iii) Given
5x – 2y + 3 = 0 Solution (0, a); (b, 0)
For, (0, a)
5 x 0 – 2 x a + 3 = 0
Or, – 2a = – 3
Or, a =
For (b, 0)
5 x b – 2 x 0 + 3 = 0
Or, b = -3/5
(3) Check which of the following are solutions of the equation 2x − 5y = 10
i) (0, 2) ii) (0, –2) iii) (5, 0) iv) 2√3, √3 v) (1/2, 2)
Solution:
Given,
2x – 5y = 10
(i) For (0, 2)
LHS RHS
2 x 0 – 5 x 2 10
= – 10
Hence it is not a solution
LHS ≠ RHS
(ii) For (0, – 2)
LHS RHS
2 x 0 – 5 x (-2) 10
= +10
= RHS
Hence (0, -2) is a solution 2x – 5y = 10
(iii) For (5, 0)
LHS RHS
2 x 0 – 5 x 0 10
= 10
= RHS
Hence (5, 0) is a solution of 2x – 5y = 10
(iv) For (2√3, – √3)
LHS RHS
2 x 2√3 – 5 x (- √3) 10
= 4√3 + 5√3
= √3(9)
= 9√3
≠RHS
Hence (2√3, – √3) is not a solution of 2x – 5y = 10
(v) For (1/5, 2)
LHS RHS
2 x 1/5 – 5 x 2 10
= 1 – 10
= – 9
≠RHS
Hence (1/5, 2) is not a solution of 2s – 5y = 10
(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.
Solution:
Given,
2x + 3y = K Solution x = 2, y = 1
2 x 2 + 3 x 1 = K
Or, K = 4 + 3
= 7
The resultant equation is 2x + 3y = 7
Now, For x = 0
2 x 0 + 3 x y = 7
Or, y = 7/3
For x = 1
2 x 1 + 3y = 7
Or, y = 5/3
(0,7/3) and (1,5/3) are two more solution of equation 2x + 3y = 7
(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,
3x – 2y + 6 = 0 x = 2 – ∝, y = 2 + ∝ are two solution of the given equation
3 (2 – ∝) -2 (2 + ∝) + 6 = 0
Or, 6 – 3∝ – 4 – 2∝ + 6 = 0
Or, 8 – 5∝ = 0
Or, 5∝ = 8
Or, ∝ = 8/5
For, x = 0
3 x 0 – 2y + 6 = 0
Or, – 2y = -6
Or, y = 3
For x = 1
3 x 1 – 2y + 6 = 0
Or,- 2y = – 9
Or, y = 9/2
For y = 0
3 x X – 2 x 0 + 6 = 0
Or, 3x = – 6
Or, x = – 2
(0, 3) (1,9/2), (-2, 0) are three more solution of the above given equation.
(6) If x = 1, y = 1 is a solution of the equation 3x + ay = 6, find the value of ‘a’.
Solution:
Given, 3x + ay = 6, x = 1, y = 1 are two solution
For, (1, 1)
3 x 1 + a x 1 = 6
Or, a = 6 – 3
= 3
(7.) Write five different linear equations in two variables and find three solutions for each of them?
Solution:
(i) x + y + 1 = 0
For, x = 0
0 + y + 1 = 0
Or, y = – 1
For, x = 1
1 + y + 1 = 0
y = – 2
For y = 0
x + 0 + 1 = 0
x = – 1
Three different solutions are (0, -1) (1, -2) (-1, 0)
(ii) x + y – 1 = 0
For, x = 0
0 + y – 1 = 0
Or, y = 1
For, x = 1
1 + y – 1 = 0
Or, y = 0
For, y = 0
X + 0 – 1 = 0
Or, x = 1
Three different solutions are (0, 1) (1, 0)(1, 0)
(iii) 2x + 3y – 6 = 0
For, x = a
2 x 0 + 3y – 6 = 0
Or, y = 6/3 = 2
For y = 0
2 x X + 3 x 0 – 6 = 0
Or, 2x = 6
Or, x = 3
For x = 1
2 x 1 + 3y – 6 = 0
Or, 3y = 6 – 2
Or, y =
Three different solutions are (0, 2) (3, 0) (1,4/3)
(iv) 2x + 4y – 8 = 0
For x = 0
2 x 0 + 4y = 8
Or, y = 2
For y = 0
2 x + 4 x 0 = 8
Or, x = 4
For x = 1
2 x 1 + 4y = 8
Or, 4y = 6
Or, y = 6/4 = 3/2
Three different solutions are (0, 2) (4, 0) (1,3/2)
(v) 2x + 4y + 8 = 0
For x = 0
2 x 0 + 4y = -8
Or, y = -2
For, x = 1
2 x 1 + 4y = – 8
Or, y = (-10)/4 = (-5)/2
For, y = 0
2 x X + 0 = – 8
Or, x = -4
Three different solutions are (0, -2) (1,-5/2), (-4, 0)
Can you plz add ex 6.3 and 6.4 also.
Hi abhinav how are u
Nikhil yadav
Tq