Question

Find two solutions for each of the following equations:
to
(i) 4x + 3y = 12
(ii) 2x+5y=0
(iii) 3y+4=0

Answer 1

Answer:

Given equation is 4x + 3y =12

On putting x = 0 in Eq. (i), we get

4(0) +3y =12 ⇒ 3y =12

⇒ y = 12 / 3 = 4

So, (0, 4) is a solution of given equation.

On putting y = 0 in Eq. (i), we get

4x+ 3 (0) =12 ⇒ 4x =12

x = 12 / 4 =3

So, (3, 0) is a solution of given equation.

On putting x =1 in Eq. (i), we get

4(1) + 3y=12 ⇒ 3y = 12 - 4 = 8 ⇒ y = 8 / 3

So, (1,8 / 3) is another solution of the given equation.

Further, putting x = 2 in Eq. (i), we get

4(2) + 3y = 12 ⇒ 3y = 12 - 8 = 4 ⇒ y = 4 / 3

So, (2, 4 / 3) is also a solution of the given equation.

Hence the four solutions of given equation are (0, 4), (3, 0), (1, 8 / 3) and (2,4 / 3).

Given equation is 3y+4=0

answer in pic