$\displaystyle \frac{x+2}{2x-3}-\frac{x^2-4}{2x^2-3x}$

The problem is not simplifying correctly, somethings supposed to cross out and I'm not getting it

Printable View

- Nov 9th 2009, 06:34 PMpurplec16Simplifying expressions
$\displaystyle \frac{x+2}{2x-3}-\frac{x^2-4}{2x^2-3x}$

The problem is not simplifying correctly, somethings supposed to cross out and I'm not getting it - Nov 9th 2009, 07:05 PMBacterius
$\displaystyle \frac{x+2}{2x-3}-\frac{x^2-4}{2x^2-3x}$

Put under common denominator :

$\displaystyle \frac{(2x^2-3x)(x+2)}{(2x-3)(2x^2-3x)}-\frac{(2x-3)(x^2-4)}{(2x-3)(2x^2-3x)}$

Expand all this stuff :

$\displaystyle \frac{(2x^2-3x)(x+2) - (2x-3)(x^2-4)}{4x^3 - 6x^2 - 6x^2 + 9x}$

Expand further :

$\displaystyle \frac{(2x^3 + 4x^2 - 3x^2 - 6x) - (2x^3 - 8x - 3x^2 + 12)}{4x^3 - 12x^2 + 9x}$

Again :

$\displaystyle \frac{2x^3 + 4x^2 - 3x^2 - 6x - 2x^3 + 8x + 3x^2 - 12}{4x^3 - 12x^2 + 9x}$

Cancel out terms :

$\displaystyle \frac{4x^2 + 2x - 12}{4x^3 - 12x^2 + 9x}$

Factorize both polynomials :

$\displaystyle \frac{(4x - 6)(x + 2)}{x(4x - 6)(x - \frac{ 3}{2})}$

Cancel out :

$\displaystyle \frac{x + 2}{x(x - \frac{ 3}{2})}$

Finally :

$\displaystyle \frac{x + 2}{x^2 - \frac{ 3}{2}x}$

I think it's right, I haven't checked though ... - Nov 9th 2009, 07:16 PMpurplec16
Thanks so much I checked and I think it's write but I thought the common demonimator means that its suppose to be "2x^2-3x" for the whole equation

- Nov 9th 2009, 07:19 PMBacterius
When adding/substracting fractions, all fractions need to be under the same denominator, otherwise it won't work. To achieve this, we usually multiply both parts of fraction 1 by the denominator of fraction 2, and the other way round. Then, we substract (in your case) the fractions. (we are not substracting the denominators, obviously)

*EDIT : did you see that ? 77 posts, and 7 thanks ... it's my lucky day ...* - Nov 9th 2009, 07:21 PMpurplec16
So aren't you multiplying to get both common denominators in both fractions so that you can minus

- Nov 9th 2009, 07:26 PMBacterius
- Nov 9th 2009, 07:28 PMpurplec16
but I thought in the end when you do that your supposed to have the same denominator which is "2x^2-3x"

- Nov 9th 2009, 07:31 PMBacterius
- Nov 9th 2009, 07:33 PMpurplec16
oh ok thank you I get it now

nite :D - Nov 10th 2009, 03:46 AMHallsofIvy
No, we

**don't**"usually multiply both parts of fraction 1 by the denominator of fraction 2, and the other way round." We usually multiply numerator and denominator of each fraction by whatever is necessary to ge the "least common denominator".

In the original fractions, the denominators are 2x-3 and $\displaystyle 2x^2- 3x= x(2x-3)$. The least common denominator is $\displaystyle x(2x-3)= 2x^2- 3x$. We need to multiply numerator and denominator of the first fraction, $\displaystyle \frac{x+2}{2x-3}$ by x to get $\displaystyle \frac{x^2+ 2x}{x(2x-3)}$. Since the denominator of the second fraction is already x(2x-3), we don't have to change that.

$\displaystyle \frac{x+2}{2x-3}- \frac{x^2- 4}{2x^2- 3x}= \frac{x^2+ 2x}{x(2x-3)}- \frac{x^2-4}{x(2x-3)}$$\displaystyle \frac{x^2+ 2x- x^2+ 4}{x(2x-3)}= \frac{2x+ 4}{x(2x-3)}$

You can, of course, multiply that denominator out to get $\displaystyle \frac{2x+4}{2x^2- 3x}$. - Nov 10th 2009, 10:41 AMBacterius
Ah yes, I didn't think of that.

Surely because when I was in junior I used to multiply the two denominators (typically 7 and 9) together to be sure to get a common divisor, because I wouldn't want to think about it. (Speechless)

I must think of doing this little check ... thanks HallsofIvy