Mathematics OpenStudy (anonymous):

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

You need to recognize the difference of two squares in denominator. OpenStudy (anonymous):

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

are you supposed to solve for x? multiply both sides by $x^2-9$ to clear the fractions you will get $2x=3$ so $x=\frac{3}{2}$ OpenStudy (anonymous):

oops OpenStudy (anonymous):

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 OpenStudy (anonymous):

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

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