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

$y=9$
You are given $4+\sqrt y=7$.Solve for y: $\sqrt y=3$ $\sqrt {(y)^2} = {3}^{2}$ -square both sides- $y=9$ Check solution: $4+\sqrt {9}=7$ $4+3=7$ $7=7$ The solution is $y=9$