Question

36. Find two solutions for each of the following:
(i) 3x + 4y = 12 (ii) 3x + 5y =0 (iii) 4y +5=0​

Answer 1

Answer:

3x+4y=12

7x=13

X=13/7.

3x+5y=0

8xy=0

xx=8

4y+5=0

4y=5

y=5/4.

Answer 2

Step-by-step explanation:

(i) 3x + 4y = 12

put x = 0

3(0) + 4y = 12

4y = 12

y = 3

put y = 0

3x + 4(0) = 12

3x = 12

x = 4

solution is (4, 0) , (0 , 3)

(ii) 3x + 5y = 0

3x = -5y

put y = 1

3x = -5(1)

3x = -5

x = -5/3

put x = 0

3(0) = -5y

0 = -5y

y = 0

solution is (1, -5/3) , (0 , 0)

(iii) 4y + 5 = 0

4y = -5

y = -5/4