Chapter 9 - 9.2 - Two-Variable Linear Systems - 9.2 Exercises - Page 646: 16

$(3,\frac{7}{5})$

Multiplying equation 1 by 2, we get: $2x+10y=20$ and $3x-10y=-5$ Adding both equations, we get $5x=15$ and thus $x=3$ (Elimination) Substituting the value of x in equation 1: $2\times 3 + 10y=20$ This becomes $10y=20-6=14$ and $y=\frac{7}{5}$ Thus, we get $(3,\frac{7}{5})$ as a solution.