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#.