Chapter 9 - 9.2 - Two-Variable Linear Systems - 9.2 Exercises - Page 646: 19

$(4,-1)$

Multiplying equation 2 by -3, and equation 1 by 2, we get: $6x+4y=20$ and $-6x-15y=-9$ Adding both equations, we get $-11y=11$ and thus $y=-1$ (Elimination) Substituting the value of y in equation 1: $6x-4=20$ This becomes $6x=20+4=24$ and $x=4$ Thus, we get $(4,-1)$ as a solution.