Linear equations in two variables – Exercise 4.2 – Class IX

  1. 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.


  1. 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


  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)


  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


 

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