Chapter 9 - 9.2 - Two-Variable Linear Systems - 9.2 Exercises - Page 646: 30

$(7,1)$

Multiplying equation 1 by $6$, we get: $3x+2y=23$ and $x-2y=5$. Adding both equations, we get $4x=28$ and thus $x=\frac{28}{4}=7$ (Elimination) Substituting the value of x in equation 2 : $7-2y=5$ This becomes $2y=7-5=2$ and $y=(2)/2=1$ Thus, we get $(7,1)$ as a solution.