Question

Solve for x: 5 over quantity x squared minus 4 plus 2 over x equals 2 over quantity x minus 2

Answer 1

$\frac{5}{x^2-4}+ \frac{2}{x}= \frac{2}{x-2} \\ \\ \frac{5}{(x-2)(x+2)}+ \frac{2}{x}= \frac{2}{x-2} \\ \\ \boxed{x \neq 0; \ x \neq б 2;} \\ \\ 5(x)+2(x-2)(x+2)=2(x)(x+2) \\ 5x+2x^2-8=2x^2+4x \\ 5x+2x^2-2x^2-4x=8\\ x=8$

Answer 2

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