## Elementary Linear Algebra 7th Edition

$x_{1}$=5 $x_{2}$=3
Given: (1) $x_{1}$-$x_{2}$=2 (2)$x_{2}$=3 Since we are given that $x_{2}$=3, we can plug this into equation 1 and solve for $x_{1}$: $x_{1}$-3=2 $x_{1}$=5