x+5/x-2=5/x+2+28/x^3-4
no parenthesis ?
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better?
yep, just a sec...
take your time :)
oh sorry, I was doing something else
so you need to find a common denominator
for all 3?
it will work out nicely start with x^2 - 4 ...what can you do to that?
(x-2)(x+2)
do you see what to do yet?
& put everything equal to zero, just for simplicity
honestly im not sure
once i change the x^2-4
Bro what I do ?
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is that a negative in the front?
@tom982
yes
it is equal to zero, like I recommended
Are you still lost? I mean I don't want to solve it for you. These are fun to do
Yeah...I feel pretty confused right now. I never get these....
Uhh once you simplify the denominator you set it to 0...so you subtract the first part to the 3rd...
\[ \frac{x+5}{x-2} = \frac{5}{x+2} + \frac{28}{(x+2)(x-2)} \] \[ \frac{(x+5)(x+2)}{(x-2)(x+2)} = \frac{5(x-2)}{(x+2)(x-2)} + \frac{28}{(x+2)(x-2)} \] \[ (x+5)(x+2) = 5(x-2) + 28 \] \[ x^2 + 7x + 10 = 5x-10+18 \] \[ x^2+2x-8=0 \] Solve for x to give your answer :)
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how do you type so fast @tom982
How do you even type in that font...it's so clear.. @tom982
Far too much practise @adamaero !
equation option
@dtan5457, you can use the Equation tool below to do it for you, but I just type it out myself. Right click on one of the nice text and click Show math as > TeX commands and you'll see the code for that line :)
Anyhow I'm looking further into this .....i have to get this sooner or later....
@tom982 For 28(x+2)(x−2) the denominator just crosses out right? I get how the other 2 the conjugates remove the bottom part
I multiplied both sides by (x-2)(x+2) to get rid of all of the denominators
Actually for the last term I made a typo, it's x^3-4
So the first two get the fraction out...the 3rd term is 28/(x^2+2)(x-2)
How does that cross out exactly?
I'm making this harder than it should be aren't I...
Oh sorry, I didn't see the x^3. I'll redo it all, one sec.
much appreciated
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