Chapter 10 - Section 10.5 - Equations of Lines - Exercises - Page 487: 35

$(6, -1)$

$2x + y =11$ $3x + 2y =16$ When we multiply the first equation by 2, the system becomes $4x + 2y = 22$ $3x + 2y = 16$ Subtracting these equations yields $x = 6$. Substituting $x = 6$ into the first equation, we get $12 +y = 11$. Then $y = −1$. So the solution is $(6, −1)$.