# Thread: Help Evaluating Algebraic Problem

1. ## Help Evaluating Algebraic Problem

Express x in terms of y for the expression:

$\frac {x-2y}{3}=\frac{y-2x}{2}$

and hence evaluate:

$\frac {7x-2y}{3x+y}$

So i express x in terms of y in the first expression and get:

$x=\frac {7y}{8}$

And then proceed to evaluate the second part of the question as follows:

$\frac {7\frac{7y}{8}-2y}{3\frac{7y}{8}+y}$

$\frac {\frac {49y}{8}-\frac{2y}{1}}{\frac {21y}{8}+\frac{y}{1}}$

Multiply here by the LCD 8:

$\frac {\frac {49y}{8}-\frac{16y}{8}}{\frac {21y}{8}+\frac{8y}{8}}$

Which gives:

$\frac {\frac {33y}{8}}{\frac {29y}{8}}$

The final answer in the book is:

$\frac {33}{29}$

So do i just cancel the eights and y's in my above answer? Also is my method in evaluating correct? Any help would be great. Thanks!

P.S why does my latex come out so small :?

2. If you have two fractions like that, you can write

$\displaystyle{\frac{\frac{33y}{8}}{\frac{29y}{8}}}$ as

$\displaystyle{\frac{33y}{8}\frac{8}{29y}.}$

Double-click my expressions there to see how I got the bigger fractions. Another command that is useful is the dfrac command.

3. Thanks for the reply! So i can just cancel from there right? My evaluation was okay? Teaching myself this stuff so i don't want to get into bad habits. This forum is great. Everyone is so helpful. Cheers for the latex tip as well mate

4. Yes on all counts. You're welcome! Have a good one.

5. Remember that algebra is essentially arithmetic with symbols.

You learned long ago that "to divide fractions, invert the divisor and multiply":
$\frac{\frac{33y}{8}}{\frac{29y}{8}}}= \frac{33y}{8}\frac{8}{29y}$
which is how Ackbeet wrote your fraction. Now you should be able to see easily that you can cancel the "y" and "8" terms.

6. Tattoo this on your wrist: (a/b) / (c/d) = (a/b) * (d/c)

7. Yeah i was being a bit silly there! I know the rules for arithmetic with fractions. Just didn't look to apply them to the division there :P Thanks for all the tips. It is all very helpful.

Cheers