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Mathematics 7 Online
OpenStudy (anonymous):

8x/(x-8)(x-10)=9x+2

OpenStudy (anonymous):

Is this \(\frac{8x}{(x-8)(x-10)} = 9x+2\)?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

The first thing you want to do is get rid of the denominator by multiplying it to the other side.

OpenStudy (anonymous):

so the entire (x-8)(x-10)

OpenStudy (anonymous):

Be careful, you can multiply the denominator to the other side provided the x is not 8 or x is not 10, otherwise we cannot just simply multiply like that

OpenStudy (anonymous):

you can't multiply...

OpenStudy (anonymous):

what do i do then?

OpenStudy (anonymous):

8x/(x-8)(x-10)=9x+2 [8x/(x-8)(x-10)] -(9x+2) = 0 [8x -(9x+2)(x-8)(x-10)] / [(x-8)(x-10)] = 0 for the fraction on the left to be zero, the numerator must be zero. So 8x -(9x+2)(x-8)(x-10) = 0

OpenStudy (anonymous):

That is the explanation, but as you can see, it has the same effect as multiply the denominator to the right side and then bring the right side to the left..

OpenStudy (anonymous):

now i jsut solve>?

OpenStudy (anonymous):

Yep. The key to problems that have variables in the denominator is checking your answers when you solve. If you get that \(x=8\) or \(x=10\) once you solve, you have to say that these aren't valid answers, because they would make the denominator equal zero.

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