5/(x^2-4)+2/x=2/(x-2) if we factor the leading denominator, it is a "difference of squares" of the form (a^2-b^2) which always factors to (a-b)(a+b) so we have:
5/((x-2)(x+2))+2/x=2/(x-2), to add/subtract fractions we need a common denominator, so in this case we need a common denominator of x(x-2)(x+2) so
5(x)+2(x+2)(x-2)=2(x)(x+2), perform indicated operations...
5x+2(x^2-4)=2x(x+2)
5x+2x^2-8=2x^2+4x subtract 2x from both sides
5x-8=4x subtract 4x from both sides
x-8=0 add 8 to both sides
x=8
check...
5/(x^2-4)+2/x=2/(x-2) when x=8 so
5/(64-4)+2/8=2/(8-2)
5/60+1/4=2/6
5/60+15/60=20/60
20/60=20/60
Simplest form for 1 3/4 would be 7/4
Simplest form for 0.125
-4+2x>-4
2x>0
X> 0
The first answer.
