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

Question

1471 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-10T00:19:41

#x=0# or #y=1#

First rearrange the equation to put the fractions together:

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

Multiply out the denominators. This can be done in one step, but I've done it in two to show exactly what's happening:

Multiply out #x+y#:

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

Multiply out the #y#:

#1(x+y)y=2(x+y)-y(2+x-y)#

Simplify:

#xy+y^2=2x+2y-2y-xy+y^2#

#xy+xy+cancel(y^2)=2x+cancel(2y)-cancel(2y)+cancel(y^2)#

#2xy=2x#

#2xy-2x=0#

#2x(y-1)=0#

and dividing by #2# we get #x(y-1)=0#

as product of #x# and #(y-1)# is #0#. we have

#x=0# or #y=1#