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

B

Step-by-step explanation:

idk

step-by-step explanation:

step-by-step explanation: