What is the solution (q, r) to this system of linear equations?

12q + 3r = 15

-4q - 4r = - 44

(-18, 29)

(-2, 13)

(8, - 1)

(15, - 44)

(-2, 13)

Step-by-step explanation:

the system can be rewritten as

q = (15 - 3r) / 12

q = (44-4r) / 4

therefore

(15-3r) / 12 = (44-4r) / 4

(15-3r) / 12 = (44-4r) * 3/12 (we can now multiply both sides by 12 to get rid of it)

so 15-3r = (44-4r) * 3

15 - 3r = 132-12r

12r - 3r = 132 - 15

9r = 117

r = 117/9 = 13

now we substitute the value r = 13 in any of the two equation

q = (15 - 3*13) / 12 = (-24) / 12 = - 2

Answer: B