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OpenStudy (anonymous):

What is the solution to the rational equation x/x^2-9 - 1/x-3= 1/4x-12?

OpenStudy (anonymous):

x/x^2-9 + 1/x-3 = 1/4x-12 x/(x+3)(x-3)+1/(x-3)=1/4(x-3) LCD: 4(x+3)(x-3) 4x+4(x+3)=(x+3) 4x+4x+12=x+3 7x=-9 x=-9/7

OpenStudy (anonymous):

Thanks

OpenStudy (anonymous):

welcome!

OpenStudy (anonymous):

What is the simplified form of

OpenStudy (anonymous):

(x^2+5x+6/15xy^2)/(2x^2+7x+3/5x^2y)?

hero (hero):

I get something differently for x.

hero (hero):

For the original problem posted.

OpenStudy (anonymous):

What did you get?

OpenStudy (anonymous):

oops copied the problem wrong

hero (hero):

I'm going to post my full solution because my steps are a bit different.

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

I'm using safari

hero (hero):

1. Make sure the denominators are factored: \(\large\frac{x}{(x+3)(x-3)} - \frac{1}{x-3} = \frac{1}{4(x-3)}\) 2. Multiply each fraction by (x-3) to get: \(\large\frac{x}{x+3} - 1 = \frac{1}{4}\) 3. Rewrite 1 as (x+3)/(x+3): \(\large\frac{x}{x+3} - \frac{x+3}{x+3} = \frac{1}{4}\) 4. Combine fractions on the left side to get: \(-\frac{3}{x+3}=\frac{1}{4}\) 5. Cross Multiply to get: \(-12 = x+3\) 6. Subtract 3 from both sides to get: \(-15 = x\)

hero (hero):

That will look messy if you're using anything other than Safari.

OpenStudy (anonymous):

thanks, and can you help me solve the second problem I posted

hero (hero):

Notice I didn't use LCDs to solve that.

OpenStudy (anonymous):

I see what I did wrong, I must have messed up somewhere in the middle of the problem

OpenStudy (anonymous):

What is the simplified form of (x^2+5x+6/15xy^2)/(2x^2+7x+3/5x^2y)?

hero (hero):

You should probably post that as a separate question.

OpenStudy (anonymous):

ok

OpenStudy (inowalst):

Thanks.

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