How do you solve $y/(y-3) +6/(y+3)= 1$ ?

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Answer 1 · 2015-10-12T11:37:03

The solution is $y=1$ .

First of all, turn the left member to a single fraction:

$y/(y-3) + 6/(y+3) = (y(y+3) + 6(y-3))/((y+3)(y-3))$

Simplify the numerator, and use the $(a+b)(a-b)=a^2-b^2$ formula for the denominator:

$y(y+3) + 6(y-3) = y^2+3y+6y-18 = y^2 + 9y - 18$

$(y+3)(y-3)=y^2-9$

So, the left member (and the whole equation) become

$(y^2 + 9y - 18)/(y^2-9)=1$

Multiply both members for the denominator:

$cancel(y^2) + 9y - 18 = cancel(y^2)-9$

Isolate $y$ terms and constants on the two sides:

$9y=9$

Solve for $y$ : $y=9/9=1$ .

How do you solve y/(y-3) +6/(y+3)= 1y/(y-3) +6/(y+3)= 1?