Question 46370
Solve and check:
{{{sqrt(2y+7)+4 = y}}} Subtract 4 from both sides.
{{{sqrt(2y+7) = y-4}}} Now square both sides.(Caution! Squaring may introduce an extraneous root)
{{{2y+7 = y^2-8y+16}}} Collect like-terms.
{{{y^2-10y+9 = 0}}} Solve by factoring.
{{{(y-9)(y-1) = 0}}} Apply the zero products principle.
{{{y-9 = 0}}} and/or {{{y-1 = 0}}}

The roots are(?):
y = 9
y = 1

Check y=9
{{{sqrt(2(9)+7)+4 = sqrt(25)+4}}} = {{{5+4 = 9}}} Checks! or {{{-5+4 = -1}}} 
The last solution is an extraneous solution introduced by squaring. Remember that{{{sqrt(25) = +5}}} or {{{-5}}}
Check y=1
{{{sqrt(2(1)+7)+4 = sqrt(9)+4}}} = {{{3+4 = 7}}} or {{{-3+4 = 1}}} The first solution doesn't check. Extraneous solution was introduced when squaring during the solving process.