## Algebra and Trigonometry 10th Edition

$\begin{bmatrix} 1 & 2 & |7\\ -1 & 1 & |8\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 2 & |7\\ 0 & 3 & |15\\ \end{bmatrix}$ Back-Substitution: 3y = 15 y = 5 x + 2(5) = 7 x + 10 = 7 x = -3 x = -3 y = 5