# I think the idea is to factor out the x's

• Sep 4th 2011, 10:53 AM
David Green
I think the idea is to factor out the x's
I have to equations;

x - 2 / 3, and 2x / 7

I need to solve for x?

First I multiply the denomintors;

3(x - 2) = 7(2x)

next I think I need to factor out the x's?

x(x - 2) from here I am lost?

I am sure I require to get (x......)(x......) but unsure at this point?
• Sep 4th 2011, 10:59 AM
Siron
Re: I think the idea is to factor out the x's
Do you mean:
$\displaystyle \frac{x-2}{3}=\frac{2x}{7}$?
Multiply both sides with 21 which is the LCM of 3 and 7.

Note:
An equation looks like ...=..., so just $\displaystyle \frac{2x}{7}$ is not an equation.
• Sep 4th 2011, 11:12 AM
David Green
Re: I think the idea is to factor out the x's
Quote:

Originally Posted by Siron
Do you mean:
$\displaystyle \frac{x-2}{3}=\frac{2x}{7}$?
Multiply both sides with 21 which is the LCM of 3 and 7.

Note:
An equation looks like ...=..., so just $\displaystyle \frac{2x}{7}$ is not an equation.

I must be doing something wrong because when I check the solutions they don't match?

x - 2 = 2x
3 7

27(x - 2) = 27(2x)

27x - 54 = 54x

27x - 54x = 54

-27x = 54
-27 -27

x = - 2

-2 -2 = - 1.33
3

2 (-2) = - 4
7 7

x = - 0.57???

Something is up I know I can find problems, but I also know I can't find correct solutions?
• Sep 4th 2011, 11:19 AM
Siron
Re: I think the idea is to factor out the x's
If you mutliply both sides with 21 then you get:
$\displaystyle \frac{21}{3}(x-2)=\frac{21}{7}(2x)$
$\displaystyle \Leftrightarrow 7(x-2)=3(2x)$
$\displaystyle \Leftrightarrow 7x-14=6x$
$\displaystyle \Leftrightarrow 14=x$
• Sep 4th 2011, 11:26 AM
David Green
Re: I think the idea is to factor out the x's
Quote:

Originally Posted by Siron
If you mutliply both sides with 21 then you get:
$\displaystyle \frac{21}{3}(x-2)=\frac{21}{7}(2x)$
$\displaystyle \Leftrightarrow 7(x-2)=3(2x)$
$\displaystyle \Leftrightarrow 7x-14=6x$
$\displaystyle \Leftrightarrow 14=x$

Thanks Siron,

I would never had thought about inverting the denominators and numerators to do that?
• Sep 4th 2011, 11:39 AM
Siron
Re: I think the idea is to factor out the x's
You're welcome!
If you have a polynomial equation with fractions in it then it's useful to multiply each side with the LCM of the denominators of the fractions.