How do you solve #1/y + 1/(y+2) = 1/3#?

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Answer 1 · 2015-11-05T00:28:55

You have two solutions: #y=2pmsqrt(10)#.

Sum the two fractions, obtaining only one denominator:

#1/y+1/(y+2) = ((y+2)+y)/(y(y+2))=(2y+2)/(y(y+2))#

So, the expression becomes

#(2y+2)/(y(y+2)) = 1/3#

Multiply both sides by #y(y+2)# and then by #3#:

#3cancel(y(y+2))(2y+2)/cancel(y(y+2)) = 1/cancel(3) cancel(3) y(y+2)#

The equation becomes

#3(2y+2) = y(y+2)#

Expand both terms:

#6y+6 = y^2+2y#

Bring everything to one side:

#y^2+2y-6y-6=0 iff y^2 -4y -6=0#

Complete the square:

#y^2 -4y -6 = (y^2-4y +4) -10 = (y-2)^2 -10#

So, the equation can be rewritten as

#(y-2)^2 -10 = 0 iff (y-2)^2 =10 \iff y-2 = pmsqrt(10)#

And thus #y=2pmsqrt(10)#.