## Algebra 2 (1st Edition)

$n=-17$
$8(2n-5)=3(6n-2)\\16n-40=18n-6\\-34=2n\\n=-17$ Checking the solution: $8(2(-17)-5)=8(-34-5)=8(-39)=-312$ and $3(6(-17)-2)=3(-102-2)=3(-104)=-312$. Both sides are equal, thus $n=-17$ is indeed a solution.