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

Divide the first equation, 12q + 3r = 15, by 3 to get 4q + r = 5. Divide the second equation, −4q − 4r = −44, by −4 to get q + r = 11. Subtract the second simplified equation from the first: 3q = −6, so q = −2; substituting gives r = 13. The solution is (−2, 13), choice B.

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