# Thread: operations involving fractions

1. ## operations involving fractions

1. [ (2x/x+3) - (x+2/x) / (x^2 + 3x + 2/x^3+2x^2-3x) ] - (6-9x/ x+2)

2x over x+3 minus x+2 over x etc that's how it goes

2. [(x - 3overx-2 ) ( 1over x+1 - 1)] - 1overx-2 - 1over(x-2)(x-2)

3. [ 1 over x^2-3x+2 + x^2+2xoverx^2-4 ] / x^3+ 1over2x^2-2x-4

2. Originally Posted by eepyej
1. [ (2x/x+3) - (x+2/x) / (x^2 + 3x + 2/x^3+2x^2-3x) ] - (6-9x/ x+2)

2x over x+3 minus x+2 over x etc that's how it goes

...
I try to guess what you mean:

$\dfrac{\dfrac{2x}{x+3} - \dfrac{x+2}{x}}{\dfrac{x^2+3x+2} {x^3+2x^2-3x}} - \dfrac{6-9x}{x+2}$ Factorize the terms in the denominator and add the fractions of the numerator:

$\dfrac{\dfrac{2x^2-(x+2)(x+3)}{x(x+3)}}{\dfrac{(x+2)(x+1)} {x(x+3)(x-1)}} - \dfrac{6-9x}{x+2}$ Expand the bracket in the numerator and collect like terms. Then multiply by the reciprocal of the denominator:

$\dfrac{x^2-5x-6}{x(x+3)} \cdot \dfrac{x(x+3)(x-1)}{(x+2)(x+1)} - \dfrac{6-9x}{x+2}$ Factorize the numerator of the first factor and cancel out like terms:

$\dfrac{(x-6)\rlap{\color{red}====}(x+1)}{\rlap{=====}x(x+3)} \cdot \dfrac{\rlap{=====}x(x+3)(x-1)}{(x+2)\rlap{\color{red}====}(x+1)} - \dfrac{6-9x}{x+2}$ $~=~$ $\dfrac{(x-6)(x-1)}{(x+2)} - \dfrac{6-9x}{x+2} = \dfrac{x^2-7x+6-6+9x}{x+2} = \dfrac{x(x+2)}{x+2} = \boxed{\bold{x}}$

Obviously my guess seems to be correct.

By the way: To prevent any wild guesses you should use parantheses. Your problem should have been written like:

[ (2x/x+3) - (x+2/x) / (x^2 + 3x + 2/x^3+2x^2-3x) ] - (6-9x/ x+2) = (2x/(x+3))-((x+2)/x /((x^2+3x+2)/(x^3+2x^2-3x))) )-(6-9x)/(x+2)

3. could someone help me with the remaining items??

4. Originally Posted by eepyej
...

2. [(x - 3overx-2 ) ( 1over x+1 - 1)] - 1overx-2 - 1over(x-2)(x+1)

...
I assume that there is a typo. I'll do the problem as written above:

$\left(\frac{x-3}{x-2}\right)\left(\frac{1}{x+1} -1 \right)-\frac1{x-2}-\frac1{(x-2)(x+1)}$ Simplify the term in the second paranthesis and the last 2 summands:

$\left(\frac{x-3}{x-2}\right)\left(\frac{1}{x+1} -\frac{x+1}{x+1} \right)-\left(\frac{1\cdot (x+1)}{(x-2)\cdot (x+1)}+\frac1{(x-2)(x+1)}\right)$

$\left(\frac{x-3}{x-2}\right)\left( -\frac{x}{x+1} \right)-\left(\frac{x+2}{(x-2)(x+1)}\right)$ $~=~$ $\frac{-x^2+3x-x-2}{(x-2)(x+1)}$ $~=~$ $\frac{-(x^2-2x+2)}{(x-2)(x+1)}$

And now I'm doing the original version:

$\left(\frac{x-3}{x-2}\right)\left(\frac{1}{x+1} -1 \right)-\frac1{x-2}-\frac1{(x-2)(x-2)}$ Simplify the term in the second paranthesis and the last 2 summands:

$\left(\frac{x-3}{x-2}\right)\left(\frac{1}{x+1} -\frac{x+1}{x+1} \right)-\left(\frac{1\cdot (x-2)}{(x-2)\cdot (x-2)}+\frac1{(x-2)(x-2)}\right)$

$\left(\frac{x-3}{x-2}\right)\left( -\frac{x}{x+1} \right)-\left(\frac{x-1}{(x-2)(x-2)}\right) = \frac{-x^2+3x}{(x+1)(x-2)} - \frac{x-1}{(x-2)(x-2)}$ The common denominator is $(x+1)(x-2)^2$ . Therefore the fractions become:

$\frac{(-x^2+3x)(x-2)}{(x+1)(x-2)^2} - \frac{(x-1)(x+1)}{(x+1)(x-2)^2}$ Expand the brackets and collect like terms:

$\frac{-x(x^2-5x + 6)-(x^2-1)}{(x+1)(x-2)^2} = -\frac{x^3-4x^2+6x-1}{(x+1)(x-2)^2}$