Simplifying Expressions, having trouble with this one

**jenbones** · August 31st 2010, 10:48 AM

(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.

**wonderboy1953** · August 31st 2010, 11:03 AM

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.

**skeeter** · September 1st 2010, 07:09 AM

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?

**jenbones** · September 1st 2010, 09:14 AM

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 ...

**Archie Meade** · September 1st 2010, 09:29 AM

Originally Posted by jenbones

I am confused one what to do next ...

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

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

**jenbones** · September 1st 2010, 09:34 AM

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?

**Archie Meade** · September 1st 2010, 09:51 AM

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!

**jenbones** · September 1st 2010, 09:53 AM

Originally Posted by Archie Meade

Yes, but don't forget about the 3rd set!

you mean (2x - 1) / (2x -1 ) ?

**Archie Meade** · September 1st 2010, 10:00 AM

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.

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