How to solve a rational equation
depends largely on the equation
Im going to put up a equation
Do you have an example of such an equation?
ok
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@Mokeira @satellite73
\[\frac{1}{x^4-2x}=\frac{x+5}{x}-2\]?
ick
i would start with \[\frac{1}{x^4-2x}=\frac{-x+5}{x}\] then cross multiply and hope for the best
Cross multiply...let me show you how
Okay @Mokeira and then ill do it to mine and you can tell me if its correct?
Follow what @satellite73 has done and you should get a simplified version of \[x=[(x+5)(x ^{4}-2x)]-[(2x)(x ^{4}-2x)]\]
yea! Lets compare answers
you are going to get a fourth degree polynomial how you are going to solve that is anyone’s guess
wait, I don't know how to cross multiply
i have a feeling there is a typo in the problem this is not really doable without wolfram of something can't really do it with algebra
Well I did make the problem up so it could make like no sense
@_angeliquexoxo please check the problem again and make sure you wrote it exactly right if you did i have no idea how you are supposed to solve it the solutions are complicated
@satellite73 the -2 is a whole number it is not a part of the fraction
oh then nvm forget it
I made it up so that you guys can show me how to do it so then I can do my problem on my own and see if I did it correct @satellite73
a goodgoal, but not very practical in this case why don't you post the actual problem it is probably cooked up to be doable
There are many ways to it. You could cross multiply or remove the denominators so that you remain with like a one-stringed equation to simplify
if you had \[\frac{3}{x+1}=\frac{x+2}{x}\] you would have a chance you could write \[3x=(x+2)(x+1)\] and solve that quadratic
True
but the x on the bottom in my problem has a ^2 and its - 5x
The ^2 just means you square
You can just post your problem then we work it out together step by step
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