Question

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

Answer 1

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

Answer 2

actually, 4x^2/(x^2-9) -2x/(x+3) = 2x/(x-3)

not 3/(x-3)

Answer 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
\[x=\frac{3}{2}\]

Answer 4

oops

Answer 5

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

Answer 6

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