Question

How do you evaluate `\frac{4x}{4x-3}\times \frac{8x-6}{6x^{2}}\div \frac{x+1}{3}`?

Answer 1

Like so:

Explanation:

start with the 2 leftmost terms. factor out 2x from the numerator of the first, and the denominator of the second. This allows you to simplify these:

`(2x(2))/(4x-3) * (8x - 6)/(2x ( 3x)) div (x + 1)/3`

`= 2 /(4x -3) * (8x - 6)/(3x) div (x+1)/3`

...this is a little simpler. Still working with the leftmost 2 terms, we multiply numerators and denominators...

`= (16x - 12)/(12x^2 - 9x) div (x+1)/3`

...now, remember that dividing by a fraction is the same as multiplying by that fractions inverse. So, the line above can be re-written:

`= (16x - 12)/(12x^2 - 9x) * 3/(x+1)`

...once again, multiply numerators and denominators...

`= (48x - 36)/(12x^3 - 9x^2 + 12x^2 - 9x)`

`= (48x -36)/(12x^3 + 3x^2 - 9x)`

...which is correct, but I'm guessing your instructor will want you to factor and simplify, if possible.

...only thing that jumps out to me is that we can factor out 3 from both the numerator & denominator...

`= 3/3 * (16x - 12)/(4x^3 + x^2 - 3x)`

`= (16x - 12)/(4x^3 + x^2 - 3x)`

...can you factor it further?

well, okay, you can factor the denominator:

`= (16x - 12)/(x(4x - 3)(x+1))`

...and you can factor 4 from the numerator:

`=(4(4x - 3))/(x(4x - 3)(x+1))`

...so we have another term in numerator and denominator that will cancel (4x - 3)

`= 4/(x(x+1))`

...which I think may be as simple as we can get it.

GOOD LUCK!