1. ## Linear Equations, Fractions

How would I work this problem out if I did not want to cross multiply?

$\displaystyle \frac{18-5x}{3x+2}=\frac{7}{3}$

2. Originally Posted by allyourbass2212
How would I work this problem out if I did not want to cross multiply?

$\displaystyle \frac{18-5x}{3x+2}=\frac{7}{3}$

I don't know why you would want to do it that way?

Multiply both sides by $\displaystyle 3x+2$ and $\displaystyle 3$ and you get rid of both denominators and it becomes simple algebra to solve for x.

3. Originally Posted by allyourbass2212
How would I work this problem out if I did not want to cross multiply?

$\displaystyle \frac{18-5x}{3x+2}=\frac{7}{3}$

$\displaystyle {\text{Cross - Multiply,}} \hfill \\$

$\displaystyle \frac{{18 - 5x}} {{3x + 2}}=\frac{7} {3} \hfill$

$\displaystyle 3(18-5x)=7(3x+2)$

4. Is this what your trying to do with the LCD?

$\displaystyle \frac{18-5x}{3x+2}=\frac{7}{3}$

$\displaystyle \frac{18-5x}{3x+2}-\frac{7}{3}=0$

$\displaystyle \frac{3(18-5x)-7(3x+2)}{3(3x+2)}=0$
Because this still gives you

$\displaystyle 3(18-5x)-7(3x+2)=0$

Which is the same as cross multiplying and dividing which gives

$\displaystyle 3(18-5x)=7(3x+2)$

So you could do it that way, but it's much more work.