Perform indicated operations, simplify: 9x^2/x^2-x-12 * x-4/6x

[Xsan925]

Perform the indicated operations and simplify.

1. 9x^2/x^2-x-12 * x-4/6x

2. x2/3x-15 / 2x^3/x^2-25

 

[Deleted member 4993]

9x^2/x^2-x-12 * x-4/6x


x2/3x-15 / 2x^3/x^2-25

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[HallsofIvy]

9x^2/x^2-x-12 * x-4/6x


x2/3x-15 / 2x^3/x^2-25

Parentheses would help! Is the first one 9(x^2/x^2)- x- 12x- 5/(6x) or is it (9x^2/(x^2- x- 12))((x- 4)/(6x))? Is the second one (x^2/3x- (15/(2x^3))/(x^2- 25) or is it (x^2/(3x- 15))/(2x^3)/(x^2- 15))?

I suspect that the first one is (9x^2/(x^2- x- 12))((x- 4)/(6x)) which is the same as \(\displaystyle \frac{9x^2}{x^2- x- 12}\frac{x- 4}{6x}\). It is important to notice that \(\displaystyle x^2- x- 12= (x- 4)(x+ 3)\) so that product is \(\displaystyle \frac{9x^2}{(x- 4)(x+ 3)}\frac{x- 4}{6x}\). See anything that will cancel?

Assuming that the second one is (x^2/(3x- 15))/(2x^3)/(x^2- 25)), which is the same as \(\displaystyle \frac{\frac{x^2}{3x- 15}}{\frac{2x^3}{x^2- 25}}\). Since you divide by a fraction by inverting it and multiplying, that would be \(\displaystyle \frac{x^2}{3x- 15}\frac{x^2- 25}{2x^3}\). Here it is important to know that \(\displaystyle 3x- 15= 3(x- 5)\) and \(\displaystyle x^2- 25= (x- 5)(x+ 5)\) so that product can be written as \(\displaystyle \frac{x^2}{3(x- 5)}\frac{(x- 5)(x+ 5)}{2x^3}\). Again, cancel "like" factors.