$\frac{10}{r-3}+\frac4{3-r}=6$
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.
$\frac{10}{r-3}+\frac4{3-r}=6$
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.
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.
When considering fractions, there are two things you should always consider doing (whether or not you actually use them):
So you should never be in a position where you don't know anything to try when faced with an equation involving fractions.
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 $1=r-3,r=4$