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

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- Sep 17th 2011, 05:01 PMMacmadesimplifying Complex Fraction.
(4a/a^2-b^2) + (2/a-b) / (3/a-b)-(b/a^2-b^2) help plzz asap

- Sep 17th 2011, 05:57 PMQuackyRe: simplifying Complex Fraction.
Is this your question?

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

As it stands, the problem statement is**extremely**unclear. - Sep 17th 2011, 06:17 PMskeeterRe: simplifying Complex Fraction.
... or this?

$\displaystyle \frac{\frac{4a}{a^2-b^2} + \frac{2}{a-b}}{\frac{3}{a-b}-\frac{b}{a^2-b^2}}$ - Sep 17th 2011, 07:28 PMMacmadeRe: simplifying Complex Fraction.
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

- Sep 17th 2011, 07:31 PMWilmerRe: simplifying Complex Fraction.
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? - Sep 18th 2011, 03:44 AMHallsofIvyRe: simplifying Complex Fraction.Quote:

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

- Sep 18th 2011, 04:09 AMWilmerRe: simplifying Complex Fraction.
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! - Sep 18th 2011, 05:45 AMMacmadeRe: simplifying Complex Fraction.
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.

- Sep 18th 2011, 06:08 AMskeeterRe: simplifying Complex Fraction.
as recommended by

**Halls of Ivy**...

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

finish it