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Mathematics 14 Online
OpenStudy (anonymous):

What is the solution to the rational equation

OpenStudy (anonymous):

OpenStudy (anonymous):

+

OpenStudy (anonymous):

OpenStudy (anonymous):

=

OpenStudy (anonymous):

OpenStudy (anonymous):

@phi

OpenStudy (phi):

you could practice using the equation editor

OpenStudy (phi):

I would first factor the denominator of the very first fraction can you do that ?

OpenStudy (mathstudent55):

First factor all denominators. Then multiply both sides by the LCD. This will eliminate all denominators.

OpenStudy (anonymous):

hhaha how do i use the equation editor?

OpenStudy (anonymous):

yes i can!!

OpenStudy (phi):

click on the equation button below the input area in the lower left. you can play with it : learn by doing... meanwhile, can you factor the bottom of the first fraction ?

OpenStudy (anonymous):

thanks and yes

OpenStudy (phi):

what do you get ?

OpenStudy (anonymous):

wait what am i looking for if it tells me what it equals

OpenStudy (phi):

eventually, we will find x = something

OpenStudy (phi):

you should get so far: \[ \frac{5}{(x-2)(x-1) }+ \frac{1}{x-2} = \frac{1}{3(x-1) } \]

OpenStudy (phi):

One good strategy is multiply both sides by (x-2). that means all terms on both sides of the =. what do you get ?

OpenStudy (anonymous):

huh??? what do i multiply that with?

OpenStudy (phi):

do you see how we get \[ \frac{5}{(x-2)(x-1) }+ \frac{1}{x-2} = \frac{1}{3(x-1) } \]?

OpenStudy (phi):

We start with \[ \frac{5}{x^2-3x+2 }+ \frac{1}{x-2} = \frac{1}{3x-3 } \] factor the bottom of the first fraction. that means re-write as(x-2)(x-1) factor out a 3 in the last fraction: 3(x-1) we get our new equation

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