## Spartan_Of_Ares Group Title can some one help me add these rational expressions? one year ago one year ago

1. math>philosophy Group Title

Ok

2. Spartan_Of_Ares Group Title

|dw:1356272011895:dw|

3. math>philosophy Group Title

Find a common denominator

4. Spartan_Of_Ares Group Title

x^2+6?

5. math>philosophy Group Title

I think you need to either review the distributive property or learn FOIL method

6. satellite73 Group Title

$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$ use $$a=8,b=x+2,c=4,d=x+3$$

7. satellite73 Group Title

use parentheses and make a direct substitution

8. Spartan_Of_Ares Group Title

wait so i cross multiply?

9. satellite73 Group Title

there is no such thing as "cross multiply" to add fractions, you compute $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$ that is how it is always done in this case $$a=8$$ so replace $$a$$ by $$8$$ $$b=x+2$$ so replace $$b$$ by $$x+2$$ $$c=4$$ replace $$c$$ by $$4$$ and $$d=x+3$$ replace $$d$$ by $$x+3$$

10. satellite73 Group Title

$\frac{\overbrace{8}^a\times \overbrace{(x+3)}^d+\overbrace{(x+2)}^b\times \overbrace{4}^c}{\overbrace{(x+2)}^b\times \overbrace{(x+3)}^d}$

11. Spartan_Of_Ares Group Title

do i combine the denominators?

12. satellite73 Group Title

you have a choice, you can either leave it in factored form as $(x+2)(x+3)$ or you can multiply out using the distributive law $(x+2)(x+3)=x^2+2x+3x+6=x^2+5x+6$ your choice

13. satellite73 Group Title

your real job is to use the distributive property in the numerator, and then combine like terms, because no one want to look at $8(x+3)+4(x+2)$

14. Spartan_Of_Ares Group Title

well to get a final answer i would need to use $x ^{2}5x+6$ wouldnt I?

15. Spartan_Of_Ares Group Title

so 8x+24 + 4x+8 = 12x+32 ?

16. satellite73 Group Title

your numerator is correct, i would be $$12x+3 2$$

17. satellite73 Group Title

and your denominator is $$x^2+5x+6$$

18. Spartan_Of_Ares Group Title

would i have any restrictions?

19. satellite73 Group Title

yes, the same restrictions you had to begin with the denominator cannot be zero, so you know $$x\neq -2$$ and also $$x\neq -3$$

20. Spartan_Of_Ares Group Title

so it cant be the 2 and 3 from the original denominators?

21. satellite73 Group Title

it cannot be $$-2$$

22. satellite73 Group Title

one denominator is $$x+2$$ so if $$x=-2$$ then you get $$-2+2=0$$ and the denominator cannot be 0

23. Spartan_Of_Ares Group Title

ok that makes sense thanks :)

24. satellite73 Group Title

yw