[SOLVED] Multiplication of Algeraic Fractions

**waven** · May 18th 2009, 05:13 AM

Q. Simplify: $\frac {a^2 - b^2 + a - b}{a^2 - 4ab + 3b^2} \times \frac {a^2 - 3ab}{a + b + 1}$

this is the furthest i've gotten, im not sure if its even right so far

$\frac {(a + b)(a - b) + (a - b)}{a^2 - 4ab + 3b^2} \times \frac {a(a - 3b)}{a + b + 1}$

Thanks

**Singaporean** · May 18th 2009, 06:03 AM

From where you left off
$\frac {(a-b)((a+b)+1)}{(a-3b)(a-b)} \times \frac {a(a-3b)}{a+b+1}$

Proceeding to cancel similar terms,
$\frac {(a-b)}{(a-3b)(a-b)} \times \frac {a(a-3b)}{1}$

$\frac {(a-b)}{(a-b)} \times {a}$

${a}$

i think this is the correct ans if i am wrong, i hope this at least help u abit

**SENTINEL4** · May 18th 2009, 06:29 AM

I found as a result "a"

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