# Need help

##### New member
Both of the problems you ask about have a variable in the denominator. A key step will be to multiply both sides by the denominator, in part to get the variable out of there. There's more to do here than just eliminate a coefficient.

But you are also doing things that don't accomplish what you claim. Does $$\displaystyle \frac{-4}{3k}\cdot-\frac{4}{3} = k$$ as you claim? Write it out this way, rather than in a column, and think carefully about it.
Could you give me an example please?

#### lev888

##### Full Member
Could you give me an example please?
Example of what?
Have you studied fraction multiplication and division?

##### New member
Example of what?
Have you studied fraction multiplication and division?
I just don’t understand what he’s saying

#### lev888

##### Full Member
I just don’t understand what he’s saying
Let's take an easy example. 4=8/x. First, can you figure out what x is without 'solving' the equation?

##### New member
So would I do this...
1. 4*k= (3/k)*k = 4K/3k=8
2.4k/3k divided by 4/3 is k/k and 8 divided by 4/3 = 6

##### New member
I am very confused

#### lev888

##### New member
I didn’t understand what you did with the first step

#### Dr.Peterson

##### Elite Member
Let's look at $$\displaystyle 4 = \frac{8}{x}$$.

We want to solve for x; but it is in the bottom of a fraction, which is not good. In order to undo that division by x, we multiply both sides by x. Since the right side is a fraction, it will be helpful to write x as the fraction x/1:

$$\displaystyle 4\cdot x = \frac{8}{x}\cdot\frac{x}{1}$$​

When we do that multiplication, the x cancels out:

$$\displaystyle 4x = \frac{8}{1}$$​

So we have $$\displaystyle 4x = 8$$.

Now, we want to get x by itself, so we have to undo the multiplication by 4; we do that by dividing by 4:

$$\displaystyle \frac{4x}{4} = \frac{8}{4}$$​

$$\displaystyle x = 2$$​

If you don't follow that, please tell us, again, where you are having trouble, and why.