- Which one of the following options is true, and why? y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions
Solution:
(iiI) y = 3x + 5 ⇒ 3x – y + 5 = 0 has infinitely many solutions. Because it is a linear equation in two variables. Values of y changes as we vary the values of x and vice versa.
- Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i) 2x + y = 7
if x = 1 then 2(1) + y = 7
⇒2 + y = 7
⇒ y = 7 – 2 = 5
If x = 0 then 2(0) + y = 7
⇒ y = 7
If x = 2 then 2(2) + y = 7
⇒ 4+ y = 7
⇒ y = 7 – 4 = 3
⇒ y = 3
If x = -1 then 2(-1) + y = 7
⇒-2 + y = 7
⇒y =7 + 2 = 9
⇒y = 9
(ii) πx + y = 9
If x = 0 then π(0) + y = 9
⇒y = 9
If x = 1/π then π(1) + y = 9
⇒(1/π) + y = 9
⇒y = 9 – 1 = 8
If x =2/π then π(2/π) + y = 9
⇒2 + y = 9
⇒y = 9 – 2 = 7
If x = 9/π then π(9/π) + y = 9
⇒9 + y = 9
⇒y = 9 – 9 = 0
⇒y = 0
(iii) x = 4y
If x = 0 then 4y = 0
⇒y = 0
If x = 1 then 1 = 4y
⇒y = 1/4
If x = 2 then 2 = 4y
⇒y = 2/4 = 1/2
If x = 4 then 4 = 4y
⇒y = 1
- Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)
(ii) (2, 0)
(iii) (4, 0)
(iv) ( 2 , 4 2)
(v) (1, 1)
Solution:
x – 2y = 4 …………(1)
(i) (0, 2)
If x = 0 then eq(1) becomes
0 – 2y = 4
y = –4/2 = -2
Therefore , (0, 2) is not a solution of equation (1)
(ii) (2, 0)
If x = 2 then eq(1) becomes
2 – 2y = 4
-2y = 4 – 2 = 2
y = –2/2 = -1
Therefore , (2, 0) is not a solution of equation (1)
(iii) (4, 0)
If x = 4 then eq(1) becomes
4 – 2y = 4
-2y = 4 – 4 = 0
y = –0/2 = 0
Therefore , (4, 0) is a solution of equation (1)
(iv) (√2 , 4√2)
If x = √2 then eq(1) becomes
√2 – 2y = 4
-2y = 4 – √2
y = 4 – √2/-2 = 4/-2 – √2/-2 = -2 + √2/2
Therefore , (√2 , 4√2) is not a solution of equation (1)
(v) (1, 1)
If x = 1 then eq(1) becomes
1 – 2y = 4
-2y = 4 – 1 = 3
y = 3/-2
Therefore , (1, 1) is not a solution of equation (1)
- Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
If x = 2 and y = 1is a solution of 2x + 3y = k
2(2) + 3(1) = k
4 + 3 = k
k = 7
1 thought on “Linear equations in two variables – Exercise 4.2 – Class IX”
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