Here's the question you clicked on:
toriebabii
3/x+13/x-11
\[\left[\begin{matrix}3 & 13 \\ x & x-11\end{matrix}\right]\]
are they fractions or are they dividends?
there fractions being added.
sorry forgot to write that
i think it is \[\frac{3}{x}+\frac{13}{x-11}\] but i could be wrong
in any case add fraction the same way you always do \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bc}\]
Thats exactly how you write it
if i am right in the question, you should get \[\frac{3(x-11)+13x}{x(x-11)}\]which you can clean up with some algebra
\[\frac{3}{x}+\frac{13}{x-11}\]\[\frac{3(x-11)+13x}{x^2-11}\] would that be right @satellite73
if you multiply out in the denominator it will be \(x^2-11x\)
Im really confused can you do it step by step?
btw i made a typo above, it should be \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]
@toriebabii did you see what i wrote above about adding fractions? it is the same way with variables as with numbers \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]
in your case \(a=3,b=x,c=13,d=x-11\)
put them directly in to the right hand side of the method for addition \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\] with careful use of parentheses, and you will get the answer
so then like mine said it would be: 3x+13x−11 \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]\[\frac{3}{x}\frac{13}{x-11}\]\[\frac{3(x-11)+13x}{x(x-11)}
\[*\frac{3(x-11)+13x}{x(x-11)}*\]
and thats the answer?
no, now you simplify. \[\frac{3(x-11)+13x}{x(x-11)}\]\[\frac{3x-33+13x}{x^2-11}\]\[\frac{16x-33}{x^2-11}\] do you see anything wlse to do in this equation?
i believe that would be the answer.