# Algebra equation with fractions?

$$\frac{10}{r-3}+\frac4{3-r}=6$$

I am not sure how to solve this equation I know both that both denominators on the left side if multiplied by negative 1 equal the other one.

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Do you mean $\frac{10}{r-3}$ or $\frac{10}{r}-3$? You can indicate which you mean by using parentheses –  Brett Frankel Apr 28 '12 at 19:35
Hint: $(r-3)=-(3-r)$ –  Brett Frankel Apr 28 '12 at 19:36
With the hints provided I did this 10(r-3)-4 which then I turned into 10r-30-4=6 which then I got into 10r=40 which then I got my answer as r=4. –  James Del Rio Apr 28 '12 at 19:59
You can accept the answer you found most helpful by clicking the checkmark below the voting arrows. –  Brett Frankel Apr 28 '12 at 20:00

Given your equation: $$\frac{10}{r-3}+\frac4{3-r}=6$$ We can notice that $r-3=-(3-r)$ so $$\frac{10}{r-3}+\frac4{-(r-3)}=\frac{10}{r-3}-\frac4{r-3}=6$$ Let $r-3=t$ then we have $$\frac{10}{t}+\left(-\frac4{t}\right)=\frac{10-4}{t}=\frac{6}{t}=6$$ Then $$\frac{6}{6}=t, 1=t$$ If we substitute $t$ with $r-3$ we have $$1=r-3,r=4$$

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When considering fractions, there are two things you should always consider doing (whether or not you actually use them):

• Finding a common denominator and combining all of the terms in a single fraction
• Clearing denominators (e.g. cross multiplying) by multiplying the equation by something and cancelling as appropriate

So you should never be in a position where you don't know anything to try when faced with an equation involving fractions.

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HINT: Multiply both sides by $r-3$. Note that by your own statement $\dfrac{r-3}{3-r}=\dfrac{-(3-r)}{3-r}=\;$?

This really is very much like the other problem: you can use either $r-3$ or $3-r$ as a common denominator, since each one is a factor of the other.

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