Question

What is the solution (a, c) to this system of linear equations?

2a – 3c = –6

a + 2c = 11

(negative StartFraction 76 over 7 EndFraction, StartFraction 17 over 7 EndFraction)
(3, –4)
(3, 4)
(StartFraction 87 over 7 EndFraction, negative StartFraction 5 over 7 EndFraction)

Answer 1

Substitute a = 11 − 2c into 2a − 3c = −6:

2(11 − 2c) − 3c = −6
22 − 4c − 3c = −6
22 − 7c = −6 → −7c = −28 → c = 4

Then a = 11 − 2(4) = 3.

So (a, c) = (3, 4).