(x+4)/2 = (x-3)/3

I know you need to multiply both sides by the Least common multiple for the denominator, but then I get lost. please help me.

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- Aug 6th 2008, 05:07 PMagentlopezSolving for 'x'
(x+4)/2 = (x-3)/3

I know you need to multiply both sides by the Least common multiple for the denominator, but then I get lost. please help me. - Aug 6th 2008, 05:14 PMDubulus
(x+4)/2 = (x-3)/3

you can cross multiply

3(x+4) = 2(x-3)

3x + 12 = 2x - 6

3x = 2x -18

(3x/2x) = -18

the x cancels out and your left with

3 = -18 which is not true

your original statement is false - Aug 6th 2008, 05:19 PMagentlopez
Okay, but then what would I do for this one when I need to find the least common multiple?

2/3x + 1/2 = 5/6x

I know the LCM is 6, but not sure what goes on after that. I have been doing math all day and really just drawing blanks now. Please explain to me the steps. - Aug 6th 2008, 05:23 PMDubulus
2/3x + 1/2 = 5/6x

No problem.

First, get all the x's in the same multiple. 3 and 6 are similar so this isnt too difficult.

3 * 2 = 6 so write 2/3 as 2(2)/3(2) or 4/6

because 2/3 = 4/6

4/6x + 1/2 = 5/6x - Aug 6th 2008, 05:31 PMagentlopez
Nothing happened to the 1/2?

- Aug 6th 2008, 05:35 PMDubulus
nope, it can stay unless you choose to rewrite it.

heres something you can use

1/2 = n/6

2n = 6

n = 3

1/2 = 3/6 - Aug 9th 2008, 04:49 PMsaranyaJust clarifying
- Aug 9th 2008, 07:28 PMQuick
Before I get going: Saranya is right, dubulus made an error when he divided by 2x (just clearing that up for agentlopez)

$\displaystyle \frac{2}{3}x+\frac{1}{2}=\frac{5}{6}x$

The first thing I like to do is bring all the x's to the same side:

$\displaystyle \frac{1}{2} = \frac{5}{6}x-\frac{2}{3}x$

Then you bring it all together:

$\displaystyle \frac{1}{2} = \left( \frac{5}{6}-\frac{2}{3}\right) x$

Now you need to subtract those two fractions from each other, so you need to find the least common multiple.

$\displaystyle \frac{1}{2} = \left( \frac{5}{6}-\frac{4}{6}\right) x$

Then you can put it all together:

$\displaystyle \frac{1}{2} = \frac{5-4}{6}x \quad\rightarrow\quad \frac{1}{2} = \frac{1}{6}x$

Then to solve for x, you have to divide away that 1/6:

$\displaystyle \frac{1}{2} \div \frac{1}{6}=x$

So now we cross multiply:

$\displaystyle \frac{1}{2} \nearrow\!\!\!\!\!\!\searrow\!\!\!\!\!\!\nwarrow \!\!\!\!\!\!\swarrow \frac{1}{6}=x$

and we get:

$\displaystyle \frac{6}{2}=x\quad\rightarrow\quad\boxed{x=3}$ - Aug 10th 2008, 04:02 PMsaranya
- Aug 10th 2008, 07:00 PMbadgerigar
Just in case agentlopez is trying to figure out why people are giving different answers to the second problem, Quick clearly and correctly got x=3 out of

http://www.mathhelpforum.com/math-he...99185700-1.gif

and Saranya, also correctly, got x = 1/3 out of

$\displaystyle \frac{2}{3x}+\frac{1}{2} = \frac{5}{6x}$

You should make sure you understand the post that solved the problem you were trying to ask. - Aug 10th 2008, 07:48 PMwik_chick88
(x+4)/2 = (x-3)/3

cross multiply, so you get:

3(x+4) = 2(x-3)

then expand the brackets:

3x + 12 = 2x - 6

then take the 2x to the left side and the 12 to the right side:

3x - 2x = -6 - 12

which give:

x = -18