Chapter 9 - 9.2 - Two-Variable Linear Systems - 9.2 Exercises - Page 646: 13

$(4,1)$

Given that $x+2y=6$ and $x-2y=2$ The first equation becomes: Adding both equations, we get $2x=8$ and thus $x=4$ (Elimination) Substituting the value of x in equation 1: $4+2y=6$ This becomes $2y=2$ and $y=1$ Thus, we get $(4,1)$ as a solution.