Algebra questions, please help!

Hi James,

\(\frac{x}{2(3-x)}-\frac{x^2}{7(x-3)^2}\\ \text{you have to get a common denominator}\\ =\frac{x}{-2(x-3)}-\frac{x^2}{7(x-3)^2}\\ =\frac{-x}{2(x-3)}-\frac{x^2}{7(x-3)^2}\\ =\frac{-x}{2(x-3)}\times \frac{7(x-3)}{7(x-3)}-\frac{x^2}{7(x-3)^2}\times \frac{2}{2}\\ =\frac{-7x(x-3)}{14(x-3)^2}-\frac{2x^2}{14(x-3)^2}\\ =\frac{-7x^2+21x-2x^2}{14(x-3)^2}\\ =\frac{-9x^2+21x}{14(x-3)^2}\\ \)

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I think the other two are easier because you just need to multiply both sides by the lowest common denominator.

With the first on, just multiply both sides by  (x-1)(x+1)

that will get rid of all the fractions and make it much more friendly  :)

See if you can do it :)

2) - Solve for a:
a/(x - 1) - 2/(x + 1) = 4/((x - 1) (x + 1))

Add 2/(x + 1) to both sides:
a/(x - 1) = 2/(x - 1)

Multiply both sides by x - 1:
Answer: |a = 2

Thanks for the help guys 

3) -

Solve for a:
a/(x + 1) + 3/(2 x - 1) = (5 x + 2)/((x + 1) (2 x - 1))

Subtract 3/(2 x - 1) from both sides:
a/(x + 1) = 1/(x + 1)

Multiply both sides by x + 1:
Answer: |a = 1