How do you solve #y- 3= \sqrt { - y + 23}#?

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Answer 1 · 2016-12-08T02:53:52

#y = 7# and #y = -2#

First, square each side of the equation to eliminate the square root function and keep the equation balanced:

#(y - 3)^2 = (sqrt(-y + 23))^2#

#(y - 3)(y - 3) = -y + 23#

Expand the right side of the equation by cross multiplying:

#y^2 - 3y - 3y + 9 = -y + 23#

#y^2 - 6y + 9 = -y + 23#

We can now put the quadratic equation in the normal form while keeping the equation balanced:

#y^2 - 6y + 9 + y - 23 = -y + 23 + y - 23#

#y^2 - 5y - 14 = 0#

We can now factor the quadratic equation:

#(y - 7)(y + 2) = 0 #

And finally, we can solve each term for #0#:

#y - 7 = 0#

#y - 7 + 7 = 0 + 7#

#y = 7#

and

#y + 2 = 0#

#y + 2 - 2 = 0 - 2#

#y = -2#