Question

Solve the following pair of simultaneous equations.

3x + y = 2, 2x + 3y = 20

Answer 1

Given equations:

3x + y = 2      ...(1)

2x + 3y = 20      ...(2)

Here, multiplying equation (1) by 3:

3(3x + y) = 3(2)

9x + 3y = 6      ...(3)

Now, subtracting equation (2) from equation (3):

(9x + 3y) − (2x + 3y) = 6 − 20

9x + 3y − 2x − 3y = −14

9x − 2x + 3y − 3y = −14

7x = −14

x = −14 + 7

∴ x = −2

Let’s substitute x = −2 into equation (1):

3(−2) + y = 2

−6 + y = 2

y = 2 + 6

∴ y = 8