1. ## Solving 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.

2. (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

3. 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.

4. 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

5. Nothing happened to the 1/2?

6. 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

7. ## Just clarifying

Originally Posted by Dubulus
(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

Hi!
I think that upto the third step u r right, after that that +2x has to come for the left side
that is
3x-2x=-18
1x = -18
x = -18
The value of x got.
I hope u may understood now
ok bye.

8. Originally Posted by agentlopez
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.
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:

$\displaystyle \frac{6}{2}=x\quad\rightarrow\quad\boxed{x=3}$

9. Originally Posted by Dubulus
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
hi
just i am trying for u

2/3x + 1/2 = 5/6x
taking lcm on the left side

4 + 3x/6x = 5/6x
then.
4 + 3x/6x - 5/6x = 0
4 + 3x - 5/6x = 0
3x -1/6x = 0
3x -1 = 0
3x = 0 + 1
3x = 1
x = 1/3

10. 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

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.

11. (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