toriebabii 2 years ago 3/x+13/x-11

1. toriebabii

$\left[\begin{matrix}3 & 13 \\ x & x-11\end{matrix}\right]$

2. SWAG

3. andriod09

are they fractions or are they dividends?

4. toriebabii

5. toriebabii

sorry forgot to write that

6. satellite73

i think it is $\frac{3}{x}+\frac{13}{x-11}$ but i could be wrong

7. satellite73

in any case add fraction the same way you always do $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bc}$

8. toriebabii

Thats exactly how you write it

9. satellite73

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

10. andriod09

$\frac{3}{x}+\frac{13}{x-11}$$\frac{3(x-11)+13x}{x^2-11}$ would that be right @satellite73

11. satellite73

if you multiply out in the denominator it will be $$x^2-11x$$

12. toriebabii

Im really confused can you do it step by step?

13. satellite73

btw i made a typo above, it should be $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$

14. satellite73

@toriebabii did you see what i wrote above about adding fractions? it is the same way with variables as with numbers $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$

15. satellite73

in your case $$a=3,b=x,c=13,d=x-11$$

16. satellite73

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

17. andriod09

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)} 18. andriod09 \[*\frac{3(x-11)+13x}{x(x-11)}*$

19. toriebabii

20. andriod09

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?

21. andriod09

i believe that would be the answer.