# Thread: Simplifying Expressions, having trouble with this one

1. ## Simplifying Expressions, having trouble with this one

(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.

2. The second fraction, factor out the 4 in the denominator and 2 in the numerator. Then multiply out the denominator in the first fraction and you should be able to move on from there.

3. Originally Posted by jenbones
(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.
$\frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{4x^2+6x-4}{4x-8}$

just a bit of factoring in each step ...

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

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

fully factored, can you finish?

4. Originally Posted by skeeter
$\frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{4x^2+6x-4}{4x-8}$

just a bit of factoring in each step ...

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

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

fully factored, can you finish?
I am confused one what to do next ...

5. Originally Posted by jenbones
I am confused one what to do next ...
The original expression is $\displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{4x^2+6x-4}{4x-28}$

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

6. Originally Posted by Archie Meade
The original expression is $\displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{4x^2+6x-4}{4x-28}$

$\displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{2(2x-1)(x+2)}{4(x-7)}=\frac{2x-1}{2x-1}\ . \frac{x+2}{x+2}\ . \frac{2(3)}{4}\ . \frac{x-7}{x-7}$
I cancel out (x + 2) / (x + 2) and (x - 7) / (x - 7) next right?

7. Originally Posted by jenbones
I cancel out (x + 2) / (x + 2) and (x - 7) / (x - 7) next right?
Yes, but don't forget about the 3rd set!

8. Originally Posted by Archie Meade
Yes, but don't forget about the 3rd set!
you mean (2x - 1) / (2x -1 ) ?

9. Originally Posted by jenbones
you mean (2x - 1) / (2x -1 ) ?
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.