Question 65696
sqrt2y+7 + 4 = y
{{{sqrt(2y+7)+4=y}}}
{{{sqrt(2y+7)+4-4=y-4}}}
{{{sqrt(2y+7)=y-4}}}
{{{(sqrt(2y+7))^2=(y-4)^2}}}
{{{2y+7=y^2-8y+16}}}
{{{2y-2y+7-7=y^2-8y-2y+16-7}}}
{{{0=y^2-10y+9}}}
0=(y-1)(y-9)
y-1=0 or y-9=0
y-1+1=0+1 or y-9+9=0+9
y=1 or y=9
Check for extraneous solutions:
for y=1
{{{sqrt(2(1)+7)+4=1}}}
{{{sqrt(9)+4=1}}}
{{{3+4=1}}}
{{{7=1}}} False, therefore y=1 is extraneous
For y=9
{{{sqrt(2(9)+7)+4=9}}}
{{{sqrt(18+7)+4=9}}}
{{{sqrt(25)+4=9}}}
{{{5+4=9}}}
{{{9=9}}} True, y=9 is valid.
The only true solution is {{{highlight(y=9)}}}
Happy Calculating!!!