Question

Simplify each sum or difference.
y 3
___ _ ____
2y=4 y-2

Answer 1

Is that = in the first term supposed to be a minus?

Answer 2

no its minus @Stiwan

Answer 3

\[\frac{y}{2y-4} - \frac{3}{y-2} = \frac{y}{2\cdot (y-2)} - \frac{3}{y-2}\]\[\frac{y - 2\cdot 3}{2\cdot (y-2)} = \frac{y-2 - 4}{2\cdot(y-2)} = \frac{y-2}{2\cdot(y-2)} - \frac{4}{2\cdot(y-2)}\]\[= \frac{1}{2} - \frac{2}{y-2}\]

Answer 4

OMG thank you @Stiwan can u help with two more questions ?

Answer 5

yes, go ahead and ask

Answer 6

thanks and -3x 4
_____ + ______
x^2-9 2x-6 @Stiwan

Answer 7

\[-\frac{3x}{x^2-9} + \frac{4}{2x-6} = -\frac{3x}{(x+3)\cdot(x-3)} + \frac{4}{2\cdot(x-3)}\]\[= \frac{-3x + 2\cdot(x+3)}{(x+3)\cdot(x-3)} =\frac{-x+6}{(x+3)\cdot(x-3)} = \frac{-x+6}{x²-9}\]

Answer 8

I don't think you can get this one simpler than that

Answer 9

omg thank u so much :D @Stiwan what about

2x 4x
---- - -------
x^2-x-2 x^2-3x+2

Answer 10

It's the same as \[\frac{1}{2(x-3)} - \frac{3}{2(x+3)}\] if you like that form better.

Answer 11

so that problem is like that ? wow easy @Stiwan

Answer 12

to clarify, i meant the 2nd problem with that

Answer 13

oh okay thanks :D @Stiwan

Answer 14

for the 3rd:
\[\frac{2x}{x^2 - x - 2} - \frac{4x}{x^2-3x+2} = \frac{2x}{(x-2)\cdot(x+1)} - \frac{4x}{(x-1) \cdot (x-2)}\]\[= \frac{2x(x-1) - 4x(x+1)}{(x-1)\cdot(x+1)\cdot(x-2)} = \frac{(-2x) \cdot (x-3)}{(x-1)\cdot(x+1)\cdot(x-2)}\]

Answer 15

you can use partial fraction expansion for further simplification, if you prefer that