1. ## Rational equations?

Totally lost. ??

2. I assume you want to simplify this expression?

$\displaystyle \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{4x^2 + 6x -4}{4x-28}$

Start off by factorising your polynomial expressions on the right:

$\displaystyle \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{2(x+2)(2x-1)}{4(x-7)}$

Multiply the two fractions:

$\displaystyle \frac{6(x-7)(x+2)(2x-1)}{4(x+2)(2x-1)(x-7)}$

Cancel out factors in denominator and numerator:

$\displaystyle \frac{6}{4} = \frac{3}{2}$

3. Do some factoring on $\displaystyle \frac{4x^2+6x-4}{4x-28}= \frac{2(x^2+3x-2)}{4(x-7)}$

Now have a go at factoring the quadratic in the numerator.

4. Originally Posted by pickslides
Do some factoring on $\displaystyle \frac{4x^2+6x-4}{4x-28}= \frac{2(x^2+3x-2)}{4(x-7)}$

Now have a go at factoring the quadratic in the numerator.
$\displaystyle \frac{2(2x^2+3x-2)}{4(x-7)}$?

5. Originally Posted by rowe
$\displaystyle \frac{2(2x^2+3x-2)}{4(x-7)}$?
Twas careless of me.

6. Originally Posted by pickslides
Twas careless of me.
Sorry, I know you know- but since you highlighted a step my answer didn't, I didn't want our help seeker getting confused!