2x/x^2-9 divided by 4x^2/x-3. I know the first step is to factor (x^2-9), which is (x+3)(x-3), and do the reciprocal, which will now be 2x/(x+3)(x-3) * x-3/4x^2, what else do i do next? Please explain:) Thanks.
2006-06-23 16:14:22 · 3 answers · asked by Letrise 5 in Science & Mathematics ➔ Mathematics
[2x/(x+3)(x-3)]*[(x-3)/4(x)]
cancelling 2 and x-3 we will get
=x/(x+3)2(x)
simplify we will get
=x/(2x^2+6x)
2006-06-23 16:26:28 · answer #1 · answered by Anonymous · 0⤊ 0⤋
You're on the right track. Multiply and reduce. You remember how to divide fractions, right? Multiply the dividend by the reciprocal of the divisor. So you get (2x(x-3))/(4(x^2)(x+3)(x-3)). 2x and (x-3) cancel, so the result is 1/(2x(x+3)).
2006-06-23 23:22:34 · answer #2 · answered by anonymous 7 · 0⤊ 0⤋
so after that u cancel (x-3) frm the denominator and numerator and 2x divides the 4x^2= u left with 1/2x(x+3)= 1/2x^2+6
2006-06-23 23:21:35 · answer #3 · answered by dd 4 · 0⤊ 0⤋