## andreadesirepen Group Title What is this simplified? one year ago one year ago

2. freewilly922 Group Title

If this expression didn't have variables but just numbers replacing them, you would give each piece a common denom. and then add. Do the same thing with the variables and then add them.

would the denom be x(x-1) ?

4. freewilly922 Group Title

For example $\frac{1}{x+5}+\frac{10}{x^2+5x}\rightarrow\frac{1}{x+5}+\frac{10}{x(x+5)}$ the common Denom. is x(x+5) so all we need to do is multiply the first term by (x/x) $\frac{1}{x+5}+\frac{10}{x(x+5)}\rightarrow\frac{1}{x+5}\frac{x}{x}+\frac{10}{x(x+5)}$\] $\frac{x}{x(x+5)}+\frac{10}{x(x+5)}=\frac{5+10}{x(x+5)}=\frac{15}{x(x+5)}$

5. freewilly922 Group Title

yes!

6. freewilly922 Group Title

be careful with the negative signs though:)

so would it x-1 or x+1 ?

8. freewilly922 Group Title

$x^2+x\rightarrow x(x+1)$