Question

Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.
6, x, plus, y, equals, minus, 3
6x+y=
−3
minus, 28, x, minus, 5, y, equals, 13
−28x−5y=
13

Answer 1

The system is
6x + y = −3
−28x − 5y = 13.

Compute the determinant: det = 6(−5) − 1(−28) = −30 + 28 = −2 ≠ 0, so there is exactly one solution.

Solve by elimination: multiply the first equation by 5: 30x + 5y = −15. Add to the second: (30x − 28x) + (5y − 5y) = −15 + 13 → 2x = −2 → x = −1. Then y = −3 − 6x = −3 − 6(−1) = 3.

Solution: (x, y) = (−1, 3).