# (4x-2)/2 * 3x/4 = ?? (Fractions)

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• Jan 4th 2010, 12:10 PM
GrindingxxTeeth
(4x-2)/2 * 3x/4 = ?? (Fractions)
I asked on Yahoo! Answers, but they didn't explain it very well.
Can someone please tell me how to get the answer? The more details the better.

(4x-2)/2 * 3x/4 = ?
• Jan 4th 2010, 12:15 PM
pickslides
Quote:

Originally Posted by GrindingxxTeeth
I asked on Yahoo! Answers, but they didn't explain it very well.
Can someone please tell me how to get the answer? The more details the better.

(4x-2)/2 * 3x/4 = ?

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

Now simplify
• Jan 4th 2010, 12:35 PM
GrindingxxTeeth
Thank you for responding, but how do you get that answer? What do you multiply or add?
• Jan 4th 2010, 12:37 PM
pickslides
Quote:

Originally Posted by GrindingxxTeeth
Thank you for responding, but how do you get that answer? What do you multiply or add?

I multiplied as your question suggested.
• Jan 4th 2010, 12:44 PM
GrindingxxTeeth
But 4x - 2 is 2x. 2x multiplied by 3 is 6x, so where does the 12x2 come from?
• Jan 4th 2010, 12:50 PM
pickslides
Quote:

Originally Posted by GrindingxxTeeth
But 4x - 2 is 2x.

(Doh) No it's not, 4x and -2 are not like terms.
• Jan 4th 2010, 12:55 PM
GrindingxxTeeth
... So then can you show me the steps you used to get that answer?
Sorry if I seem thick here, but I am sooo lost on this problem I don't even know where to begin.
• Jan 4th 2010, 12:56 PM
bigwave
Quote:

Originally Posted by GrindingxxTeeth
But 4x - 2 is 2x. 2x multiplied by 3 is 6x, so where does the 12x2 come from?

($\displaystyle 4x-2)(3x)$

we are distributing the $\displaystyle 3x$ to the terms inside $\displaystyle (4x - 2)$so we have $\displaystyle (4x(3x)- 2(3x))$ which is $\displaystyle 12x^2 - 6x$
• Jan 4th 2010, 12:59 PM
mr fantastic
Quote:

Originally Posted by GrindingxxTeeth
... So then can you show me the steps you used to get that answer?
Sorry if I seem thick here, but I am sooo lost on this problem I don't even know where to begin.

Begin by going back and reviewing the basic algebra that this question depends on. Attempt easier questions as a lead-in to this question. If you do not understand the basic things there is no point trying to understand how they are applied to questions like this one. Your real problem is not this question, it is the underpinning algebra.

By the way, a solution may not seem to be explained very well if you are given guidance on how to solve the question but you do not understand the basic algebra underpinning that guidance.
• Jan 4th 2010, 01:05 PM
GrindingxxTeeth
...
And now I'm off to try math forum #2 since apparently you guys don't like helping people. Why not just show the steps? Is it really that hard?
Whatever. Hopefully a different forum will be more helpful.
• Jan 4th 2010, 01:15 PM
mr fantastic
Quote:

Originally Posted by GrindingxxTeeth
...
And now I'm off to try math forum #2 since apparently you guys don't like helping people. Why not just show the steps? Is it really that hard?
Whatever. Hopefully a different forum will be more helpful.

What we don't like is people who insist on being given a complete solution that shows every single step.

Despite what you might think, it is not helpful to give you an entire solution. Post #2 gave you everything except the last step but you did not understand it. I have said the reasons why.

Do yourself a favor and review the basic material so that you can understand the help that people are trying to give you.

Thread closed.
• Jan 4th 2010, 02:58 PM
Jameson
Quote:

Originally Posted by GrindingxxTeeth
...
And now I'm off to try math forum #2 since apparently you guys don't like helping people. Why not just show the steps? Is it really that hard?
Whatever. Hopefully a different forum will be more helpful.

Bye. Definitely will miss you...