simplifying Complex Fraction.

**Macmade** · September 17th 2011, 05:01 PM

(4a/a^2-b^2) + (2/a-b) / (3/a-b)-(b/a^2-b^2) help plzz asap

**Quacky** · September 17th 2011, 05:57 PM

Is this your question?

$\frac{4a}{a^2-b^2}+\frac{\frac{2}{a-b}}{\frac{3}{a-b}}-\frac{b}{a^2-b^2}$

As it stands, the problem statement is extremely unclear.

**skeeter** · September 17th 2011, 06:17 PM

... or this?

$\frac{\frac{4a}{a^2-b^2} + \frac{2}{a-b}}{\frac{3}{a-b}-\frac{b}{a^2-b^2}}$

**Macmade** · September 17th 2011, 07:28 PM

its the second one. sorry bout the unclarity, i dont know how to write equations on the internet. lol how fatuous of me. anywayz skeeter guessed it right so anyone care to explain? thanks

**Wilmer** · September 17th 2011, 07:31 PM

Well, how far can you get on your own?
We don't know what you know or been taught.
Is this a problem from math class?

**HallsofIvy** · September 18th 2011, 03:44 AM

$\frac{\frac{4a}{a^2-b^2} + \frac{2}{a-b}}{\frac{3}{a-b}-\frac{b}{a^2-b^2}}$

Since $a^2- b^2= (a- b)(a+ b)$ , $a^2- b^2$ is the least common denominator of all the fractions. Start by multiplying both numerator and denominator by $a^2- b^2$ .

**Wilmer** · September 18th 2011, 04:09 AM

May be easier (less confusing) to first let x = a^2 - b^2 and y = a - b.
Simplify as much as you can with those, then substitute back in;
easier for me, anyway!

**Macmade** · September 18th 2011, 05:45 AM

what i did was i used this rule that a/b + c/d equals ad+cb/bd. I tried to simplify the two fractions seperately but results weren't guud. this is a question for gr.10 math at school.

**skeeter** · September 18th 2011, 06:08 AM

as recommended by Halls of Ivy ...

$\frac{\frac{4a}{a^2-b^2} + \frac{2}{a-b}}{\frac{3}{a-b}-\frac{b}{a^2-b^2}} \cdot \frac{a^2-b^2}{a^2-b^2} = \frac{4a + 2(a+b)}{3(a+b) - b}$

finish it

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