how do I simplify 3x-2/x+2 -2/x-2 = ? I will give you a nice shiny medal if you help me :) Please do show detailed instructions for me please, I really want to memorize the knowledge I have lost. :) Thank you

Question

lacris · Answer

I did multiply (x+2) to x-2 and 2, and I also multiplied (x-2) to x+2 and (3x-2) is that correct?

mathstudent55 · Answer

You need to use parentheses to make the expression above readable.
It is very hard to understand what it is.

mathstudent55 · Answer

Is this what you meant?

$\dfrac{3x-2}{x+2} - \dfrac{2}{x-2 } $

lacris · Answer

yes that's what I mean :D

mathstudent55 · Answer

You are correct.
You need a common denominator.
The common denominator is $(x + 2)(x - 2)$.
That means you multiply the first fraction by $\dfrac{x - 2}{x - 2} $
 and the second fraction by $\dfrac{x + 2}{x + 2} $

lacris · Answer

so for x/x-2 it becomes: $$\frac{2x+4 }{ x^2-4} $$

lacris · Answer

and 3x-2/x+2 becomes: $$\frac{ 3x^2-8x+4 }{ x^2-4 }$$

lacris · Answer

is this correct? :D ;p

mathstudent55 · Answer

$\dfrac{3x-2}{x+2} - \dfrac{2}{x-2 }=$

$= \dfrac{3x-2}{x+2}  \times \dfrac{x - 2}{x-2 } - \dfrac{2}{x-2 } \times \dfrac{x + 2}{x + 2 }$

$= \dfrac{3x^2 - 6x -2x+4}{(x + 2)(x-2) } - \dfrac{2x + 4 }{(x + 2)(x - 2) }$

$= \dfrac{3x^2 - 8x +4}{(x + 2)(x-2) } - \dfrac{2x + 4 }{(x + 2)(x - 2) }$

mathstudent55 · Answer

You are correct so far.

lacris · Answer

so I just subtract them both right? I got $$\frac{ 2x+4 }{ x^2-4 }-\frac{ 3x^2-8x+4 }{ x^2-4 } = \frac{ 3x^2 - 6x}{ x^2-4 }$$

lacris · Answer

Its not one of the answers available, did I do something wrong?

mathstudent55 · Answer

Why did you change the order of the fractions? 
You need to do the first fraction minus the second fraction.
You did the second fraction minus the first fraction.

mathstudent55 · Answer

$= \dfrac{3x^2 - 8x +4}{(x + 2)(x-2) } - \dfrac{2x + 4 }{(x + 2)(x - 2) }$

mathstudent55 · Answer

$= \dfrac{3x^2 - 8x +4}{(x + 2)(x-2) } \color{red}{-} \dfrac{\color{red}{2x + 4} }{(x + 2)(x - 2) }$

$= \dfrac{3x^2 - 8x +4 \color{red}{-2x - 4}}{(x + 2)(x-2) }$

Notice how the minus from the subtraction changes both signs on the numerator of the second fraction (all shown in red above).

lacris · Answer

ooh ahh yes I see what I did wrong, so it becomes $$\frac{ 3x^2+10 }{ x^2-4 }$$ the 4 is cancelled out because 4-4 = 0 and can multiple x+2(x-2) which therefore makes the answer: $$\frac{ x(3x+10)}{ x^2-4 } $$

lacris · Answer

is this correct? :D

mathstudent55 · Answer

$= \dfrac{3x^2 - 8x +4 -2x - 4}{(x + 2)(x-2) }$

$= \dfrac{3x^2 - 10x }{(x + 2)(x-2) }$

You are close, but it is -10x, not +10 in the numerator.

lacris · Answer

oooh I got it wrong because 2x - -8x becomes 2x +8x = 10x?... what was I supposed to do? Im sorry im confused D:

mathstudent55 · Answer

Look at my response above where I used red color for some terms.

lacris · Answer

so the subtraction becomes an addition problem then? :D

mathstudent55 · Answer

In the first line I used red, the fractions already had the common denominator but had not been combined together.
In the second line, the subtraction had been performed.
Notice how the numerator after the subtraction is performed is

$3x^2 - 8x + 4 - 2x - 4$

Once you subtract, you just combine like terms the way they appear. If you see a plus sign, you add, and if you see a minus sign, you subtract.

Here you have $3x^2$ by itself with no other like terms, so it remains as $3x^2$.

Then you have $-8x $ and $-2x$.
Since -8 - 2 = -10, then -8x - 2x = -10x.

Finally, you have 4 - 4, which you already stated above cancel out leaving you 0.

You are left with $3x^2 - 10x$ in the numerator.

lacris · Answer

Ahhh I understand, thank you so much for helping me and being patient! :D

mathstudent55 · Answer

You're welcome.