Chapter 9 - 9.2 - Two-Variable Linear Systems - 9.2 Exercises - Page 646: 29

$(5, -2)$

Multiplying equation 1 by $12$ and equation 2 by $4$, we get: $3x+4y=7$ and $8x-4y=48$. Adding both equations, we get $11x=55$ and thus $x=\frac{55}{11}=5$ (Elimination) Substituting the value of x in equation 1: $3\times5+4y=7$ This becomes $4y=7-15=-8$ and $y=(-8)/4=-2$ Thus, we get $(5,-2)$ as a solution.