# Thread: Solving Fractions

1. ## Solving Fractions

(1) $\frac{a+3}{3}+\frac{a-3}{6}=5$

(2) $\frac{3x+2}{2x-3}=\frac{3x-2}{2x-5}
$

(3) $\frac{4m}{m-2}-\frac{13}{3m-6}=\frac{1}{3}$

(4) $\frac{1}{t-5}+\frac{1}{t+5}=\frac{8}{t^2-25}$

(5) $\frac{3}{2b+4}-\frac{4}{b-2}=\frac{3}{2b^2-8}$

(6) $\frac{3y+1}{4}=\frac{13-y}{2}$

Well I just completely failed my math test and i would like if anyone could help me so at least I could have an understanding of how to solve the following equations. I remeber learning how to solve the equations but not with equal signs

2. This is way too many problems for one thread. One question per thread is our policy. These all use the same concepts too, so pick one to start with. Getting the method used to solve these is what you need, not tons of solutions given to you. Let me know which one you wish to focus on.

3. number 2 was the most confusing

4. Originally Posted by purplec16
(2) $\frac{3x+2}{2x-3}=\frac{3x-2}{2x-5}
$
When you have two fractions equal to each other, you use what is called cross multiplication to get rid of both fractions at the same time. Take the numerator of the left fraction and multiply it by the denominator of the right one. Set that equal to the numerator of the right fraction times the denominator of the left. You are essentially multiplying each term by the one diagonal from it.

So $(3x+2)(2x-5)=(3x-2)(2x-3)$ Now you need to FOIL everything out to get rid of the parentheses.

$6x^2-15x+4x-10=6x^2-9x-4x+6 \Rightarrow -11x-10=-13x+6$

The 6x^2 terms canceled and then I grouped the x terms. Solve for x.

5. Thank You, although I'm still having problems working the other questions...

6. Originally Posted by purplec16
Thank You, although I'm still having problems working the other questions...
Make a new thread for each problem then and you will get help.