I need help with an algebra problem
Algebra

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√y+5 + √y +13 = 4 (the radical is over y+5 and y+13) Can someone help me please
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Hello. We want to solve for y, correct?
sqrt(y+5) + sqrt(y+13) = 4, is our original equation.
We want to eliminate the radical operations. so we need to square both sides of the equation.
( sqrt(y+5) + sqrt(y+13) )^2 = 4^2
(√(y + 5))^2 + 2*√(y+5) * √(y+13) + (√(y+13))^2 = 16
Expand the left side we get:
where (√(y + 5))^2 = y + 5 and (√(y+13))^2 = y + 13
y + 5 + 2*√(y+5) * √(y+13) + y+13 = 16
The right side 4^2 is equal to 16. Because 4 squared is 16.
We can assume y > 13 so that y + 5 = y + 5 and y + 13 = y + 13.
y + 5 + 2*√(y+5) * √(y+13) + y+13 = 16
Simplify:
2y + 18 + 2*√( (y + 5)*(y + 13) ) = 16
2y +2*√ ( (y + 5)*(y + 13) ) =  2
y + √ ( (y + 5)*(y + 13) ) = 1 this looks like it contradicts the fact that y > 13, So I have to change my assumption.
Assume y < 5 so that y + 5 = (y + 5) and y + 13 = (y+13)
(y+5) + 2*√ ( (y + 5)*(y + 13))  (y + 13) = 16
y  5 + 2*√( (y+5)*(y + 13))  y  13 = 16
2y  18 + 2*√( (y+5)*(y + 13)) = 16
2y + 2*√( (y+5)*(y + 13)) = 34
y + √( (y+5)(y + 13)) = 17 from dividing both sides by 2.
√ ( (y+5)(y + 13) ) = y + 17
(y + 5)(y + 13) = (y + 17)^2
y*y + 18y + 16 = y*y + 34y + 289
16  289 = 16y
y = 273 / 16
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