can someone help... this is due next block! 4x^2/(x^2-9) -2x/(x+3)=3/(x-3)?

You need to recognize the difference of two squares in denominator.

actually, 4x^2/(x^2-9) -2x/(x+3) = 2x/(x-3) not 3/(x-3)

are you supposed to solve for x? multiply both sides by \[x^2-9\] to clear the fractions you will get \[2x=3\] so \[x=\frac{3}{2}\]

oops

4x^2/(x^2-9) -2x/(x+3)=3/(x-3) => [4x^2-2x(x-3)] / (x^2-9) = 3/(x-3) => [2x^2+6x]/(x^2-9) = 3/(x-3) => 2x(x+3)/(x^2-9) = 3/(x-3) => 2x/(x-3) = 3/(x-3) => x = 3/2

oh not oops. i thought you changed the problem. it is \[\frac{4x^2}{x^2-9}-\frac{2x}{x+3}=\frac{3}{x-3}\] yes? multiply by \[x^2-9=(x+3)(x-3)\] to get \[4x^2-2x(x-3)=3(x+3)\] \[4x^2-2x^2+6x=3x+9\] \[2x^2+3x-9=0\] \[(2x-3)(x+3)=0\] \[x=\frac{3}{2}\] the other solution is not there because it would make the denominator 0