(3)(x - 7) / (x + 2) (2x - 1) multiplied by (4x^2 + 6x - 4) / (4x - 28)
I could use step by step help on how to simplify this expression.
$\displaystyle \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{4x^2+6x-4}{4x-8}$
just a bit of factoring in each step ...
$\displaystyle \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{2(2x^2+3x-2)}{4(x-2)}$
$\displaystyle \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{2(2x-1)(x+2)}{4(x-2)}$
fully factored, can you finish?
The original expression is $\displaystyle \displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{4x^2+6x-4}{4x-28}$
$\displaystyle \displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{2(2x-1)(x+2)}{4(x-7)}=\frac{2x-1}{2x-1}\ . \frac{x+2}{x+2}\ . \frac{2(3)}{4}\ . \frac{x-7}{x-7}$
Yes, the three fractions containing x, where the numerator and denominators are identical are all just the value 1,
(though strictly speaking, we'd have to say that the function is undefined for values of x giving zero in the denominator).
So the value of the fraction is simply 6/4=1.5.