How do you solve #x/(x^2-8)=2/x#?

1 Answer
Jul 27, 2016

Answer:

#x=+-4#

Explanation:

First, move everything from the denominator to the numerator
we do this by multiplying by the LCM of denominators on each side.

#x(x^2-8)* x/(x^2-8)= 2/x *x(x^2-8)#

the #(x^2-8)# on the left side cancels out and the #x# on the right side cancels out

#x(cancel(x^2-8))* x/(cancel(x^2-8))= 2/cancelx *cancelx(x^2-8)#

which leaves us with:

#x*x=2*(x^2-8)#

after this we now have
# x^2=2(x^2-8)#

next we remove the parentheses by multiplying each term by 2
now we have
#x^2=2x^2-16#

next we will move the 16 to the other side to avoid working with negative numbers
#16+x^2=2x^2#

then we will combine like terms by subtracting #x^2#
#16=2x^2-x^2# leaving us with #16=x^2#

then we will get rid of the #x^2# by taking the square root of both sides
#+-sqrt16#=#sqrtx^2#

now we have our final answer of
#x=+-4#