## College Algebra (11th Edition)

(a) $x=12$ (b) Use an interval beyond $[-10,10]$ to include 12.
(a) We solve: $-2(x-5)=-x-2$ $-2x+10=-x-2$ $10=x-2$ $x=10+2$ $x=12$ (b) Most standard viewing windows span the interval $[-10,10]$, outside of the solution. We must use an interval around $12$, e.g. $[8,15]$.