## Algebra 1: Common Core (15th Edition)

The solutions are $a=6$ or $a=1$.
$a=\sqrt 7a-6$ Square both sides: $(\sqrt 7a-6)^2=a^2$ $a^2-7a+6=0$ $(a-6)(a-1)=0$ $a=6$ or $a=1$ With $a=6$ then $a=\sqrt 7a-6$ $6=\sqrt 7(6)-6$ $6=6$ With $a=1$ then $a=\sqrt 7a-6$ $1=\sqrt 7(1)-6$ $1=1$ Hence, the solutions are $a=6$ or $a=1$.