I need help solving this problem. square root of (2y-3) = square root of (y+7) -2?

Question

the "-2" isn't part of the square root of the second equation

Answer 1

squaring both sides
2y-3=y+7+4-4(y+7)^1/2
y-14=-4(y+7)^1/2
squaring again
y^2-28y+196=16(y+7)
y^2-28y-16y+196-112=0
y^2-44y+84=0
y=[44+/-(44^2-4*84)^1/2]/2
=44+40 /2=42 and 44-40 /2=2
solution y=42 or 2

Answer 2

sqrt(2y - 3) = sqrt(y + 7) - 2

square both sides

(sqrt(2y - 3))^2 = (sqrt(y + 7) - 2)^2
2y - 3 = (sqrt(y + 7) - 2)(sqrt(y + 7) - 2)
2y - 3 = (y + 7) - 2sqrt(y + 7) - 2sqrt(y + 7) + 4
2y - 3 = y + 7 - 4sqrt(y + 7) + 4
2y - 3 = y + 11 - 4sqrt(y + 7)
y - 14 = -4sqrt(y + 7)

same as saying

y - 14 = -sqrt(16(y + 7))

-y + 14 = sqrt(16(y + 7))

14 - y = sqrt(16(y + 7))

square both sides

(14 - y)^2 = 16y + 112
(14 - y)(14 - y) = 16y + 112
196 - 14y - 14y + y^2 = 16y + 112
196 - 28y + y^2 = 16y + 112
y^2 - 28y + 196 = 16y + 112
y^2 - 44y + 84 = 0
(y - 42)(y - 2) = 0

y = 2 or 42

when you plug in 42, the problem works like this

-sqrt(2y - 3) = -sqrt(y + 7) - 2

Answer 3

The answer is Y = 2, because if you plug 2 back into the problem it works. If you plug 42 back into the problem, it doesn't work.