# Thread: I little bit of help needed

1. ## I little bit of help needed

I have a fraction that needs simplifying and I think that I have it right, I just want to make sure my method is right.

[anything that is underlined is canceled]

The fraction is: (2x^2+x-6)/(2x+4)

My simplification is: (3x+3)/2

My method is: (2x^2+x-6)/(2x+4) >> (2x-3)(x-2)/2x+4 >> (2x-3)(x+2)/(2x+4) >> (3x+3)/2

thank you in advance!!!!

Drakmord

2. Originally Posted by drakmord
I have a fraction that needs simplifying and I think that I have it right, I just want to make sure my method is right.

[anything that is underlined is canceled]

The fraction is: (2x^2+x-6)/(2x+4)

My simplification is: (3x+3)/2

My method is: (2x^2+x-6)/(2x+4) >> (2x-3)(x-2)/2x+4 >> (2x-3)(x+2)/(2x+4) >> (3x+3)/2

thank you in advance!!!!

Drakmord
how you make this (2x^2+x-6)/(2x+4) >> to this (2x-3)(x-2)/2x+4 ??

how you cancel like this (2x-3)(x+2)/(2x+4) >> (3x+3)/2

3. Originally Posted by drakmord
I have a fraction that needs simplifying and I think that I have it right, I just want to make sure my method is right.

[anything that is underlined is canceled]

The fraction is: (2x^2+x-6)/(2x+4)

My simplification is: (3x+3)/2

My method is: (2x^2+x-6)/(2x+4) >> (2x-3)(x-2)/2x+4 >> (2x-3)(x+2)/(2x+4) >> (3x+3)/2

thank you in advance!!!!

Drakmord
You can't cancel like that, you cannot cancel where addition is involved.

In this case $2x^2+x-6 = (2x-3)(x+2)$ and $2x+4 = 2(x+2)$

This means your fraction becomes $\frac{2x^2+x-6}{2x+4} = \frac{(2x-3)(x+2)}{2(x+2)} = \frac{2x-3}{2} = x - \frac{3}{2}$

As (x+2) is on the top and the bottom it may be cancelled.

4. Ah.....I see, that is why my answer made so much trouble for me, I did not work it out properly.

Let me make sure I have this straight,what you are saying is:
(2x^2+x-6)/(2x+4) >> (2x-3)(x+2)/2(x+2) NOT "(2x-3)(x-2)/2x+4" like I said before. Therefore, (2x-3)(x+2)/2(x+2) >> (2x-3)(x+2)/2(x+2) >> (2x-3)/2 >> x-3.
Is that correct?

Thanks again.

Drakmord

5. Originally Posted by drakmord
Ah.....I see, that is why my answer made so much trouble for me, I did not work it out properly.

Let me make sure I have this straight,what you are saying is:
(2x^2+x-6)/(2x+4) >> (2x-3)(x+2)/2(x+2) NOT "(2x-3)(x-2)/2x+4" like I said before. Therefore, (2x-3)(x+2)/2(x+2) >> (2x-3)(x+2)/2(x+2) >> (2x-3)/2 >> x-3.
No, your very last step:
$\frac{2x - 3}{2} = x - 3$
is not correct.

It should be this:
$\frac{2x - 3}{2} = \frac{2x}{2} - \frac{3}{2} = x - \frac{3}{2}$.

01

6. Originally Posted by drakmord
Ah.....I see, that is why my answer made so much trouble for me, I did not work it out properly.

Let me make sure I have this straight,what you are saying is:
(2x^2+x-6)/(2x+4) >> (2x-3)(x+2)/2(x+2) NOT "(2x-3)(x-2)/2x+4" like I said before. Therefore, (2x-3)(x+2)/2(x+2) >> (2x-3)(x+2)/2(x+2) >> (2x-3)/2 >> x-3.
Is that correct?

Thanks again.

Drakmord
That's correct, except that $\frac{2x-3}{2}=\frac{2x}{2}-\frac{3}{2}=x-\frac{3}{2}$

Edit* yeongil beat me to it