(6)/(x^2 - 3x) - (5)/(2)(3-x)

**FailInMaths** · March 20th 2011, 03:03 AM

Hi guys, need help on this question

My working:
(6)/(x^2-3x) - (5)/(2)(3-x) = (6)/(x)(x-3) - (5)/(2)(3-x)
= - (6)/(x)(3-x) - (5)/(2)(3-x)
= -6(-2)-5(-x)/2x(3-x)
= (12+5x)/(2x)(3-x)

Answer = (5x+12)/(2x)(x-3)

What is wrong with my working?

P.S sorry that it's a bit messy, I don't know how to make it into that "mathematics" form.

**Plato** · March 20th 2011, 03:32 AM

$\dfrac{6}{x^2-3x}-\dfrac{5}{3-x}=\dfrac{6}{x(x-3)}+\dfrac{5}{2(x-3)}$

**FailInMaths** · March 20th 2011, 03:47 AM

Originally Posted by Plato

$\dfrac{6}{x^2-3x}-\dfrac{5}{3-x}=\dfrac{6}{x(x-3)}+\dfrac{5}{2(x-3)}$

Edited: Thanks a lot Plato, I knew what went wrong

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