# What is the solution (q, r) to this system of linear equations? 12q + 3r = 15 - 4q - 4r = - 44

**Solution:**

12q + 3r = 15 ….. (1)

- 4q - 4r = - 44 ……. (2)

Let us solve the linear equations using the elimination method.

First multiply equation (2) by 3

- 12q - 12r = - 132 ….. (3)

Now add equations (1) and (3)

3r - 12r = 15 - 132

By further calculation

- 9r = - 117

r = 117/9 = 13

Substituting the value of r in equation (2)

- 4q - 4 (13) = - 44

- 4q - 52 = - 44

- 4q = - 44 + 52

So we get

- 4q = 8

Divide both sides by 4

q = - 2

Therefore, the solution (q, r) is (-2, 13).

## What is the solution (q, r) to this system of linear equations? 12q + 3r = 15 - 4q - 4r = - 44

**Summary:**

The solution (q, r) to this system of linear equations 12q + 3r = 15 and - 4q - 4r = - 44 is (-2, 13).