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

1 Answer
Dec 20, 2016

Answer:

#x=-1+-sqrt(5/2)#.

Explanation:

To add fractions, we need to find a common denominator. For example, to do #1/3+1/5#, we would multiply #1/3# by #5/5# to get #5/15# and #1/5# by #3/3# to get #3/15#, then add them to get #8/15#.

Similarly, #1/x=1/x*(x+3)/(x+3)=(x+3)/(x(x+3))#.

#1/(x+3)=1/(x+3)*x/x=x/(x(x+3))#.

So #1/x+1/(x+3)=(x+3)/(x(x+3))+x/(x(x+3))=(2x+3)/(x(x+3))#.

So now we have #(2x+3)/(x(x+3))=2#.

Multiply both sides by #x(x+3)# to get #2x+3=2x(x+3)#.

Expand the right side to get #2x+3=2x^2+6x#.

Rearrange terms to get #2x^2+4x-3=0#.

This is a quadratic equation with solutions #x=-1+-sqrt(5/2)#.