1. ## Rational Problems

Okay so i dont understand how to do this problem...

x+6 + x+3
x2+x 12 x2-16

(x2=x squared)

Im supposed to simplyfy and state the restrictions on the variables

2. Originally Posted by Ari
Okay so i dont understand how to do this problem...

x+6 + x+3
x2+x-12 x2-16

(x2=x squared)
you are supposed to be simplifying i suppose? please state your questions clearly and specifically.

factor the denominators and then combine the fractions.

do you know how to factor?
do you know how to combine fractions?

3. Im supposed to simplify it and then state the restrictions

4. Originally Posted by Ari
Im supposed to simplify it and then state the restrictions
ok, so try what i said

5. Ok So how would i do

5m(squared) +6m-8 * |20m(squared) +16m
5m(squared) -16m-16 |4-m (squared)

|mean the other fraction

6. Originally Posted by Ari
Ok So how would i do

2 2
5m +6m-8 * 20m +16m
2 2
5m -16m-16 4-m
2
And btw those 2's on top mean squared so 5m mean 5m to the power of 2
i don't understand what you are saying. if you can't use LaTex, please write fractions as follows:

(numerator)/(denominator)

include the brackets

what did you get for the first question

7. I got 2x(squared)+2x-33/(x+4)(x-4)(x-3) for the first one

8. Originally Posted by Ari
I got 2x(squared)+2x-33/(x+4)(x-4)(x-3) for the first one
please rewrite the original question so i can check it. there is a sign missing from the 12 in the original post

9. (x+6)/(x^2+x-12) X (x+3)/(x^2-16)

10. Originally Posted by Ari
(x+6)/(x^2+x-12) X (x+3)/(x^2-16)
if it was $x^2 {\color{red}-}x - 12$ in the denominator of the first fraction, we would be in business. but since it is + x, nothing cancels. we get: $\frac {x + 6}{x^2 + x - 12} \cdot \frac {x + 3}{x^2 - 16} = \frac {(x + 6)(x + 3)}{(x + 4)^2 (x - 4)(x + 3)}$

the restrictions are, we cannot choose the $x$'s that make the denominator zero. what x's are those?

11. Originally Posted by Jhevon
if it was $x^2 {\color{red}-}x - 12$ in the denominator of the first fraction, we would be in business. but since it is + x, nothing cancels. we get: $\frac {x + 6}{x^2 + x - 12} \cdot \frac {x + 3}{x^2 - 16} = \frac {(x + 6)(x + 3)}{(x + 4)^2 (x - 4)(x + 3)}$

the restrictions are, we cannot choose the $x$'s that make the denominator zero. what x's are those?

could you show the steps for that? I thought we are supposed to multyply both bottomt and top when doing the LCD

12. Originally Posted by Ari
could you show the steps for that? I thought we are supposed to multyply both bottomt and top when doing the LCD
you had a multiplication sign, we do not fin any LCD's here. we just multiply the tops to get the new top, and multiply the bottoms to get the new bottom

13. Uggh sorry i messed up its supposed to be a + sign in the middle of the fractions

(x+6)/(x^2+x-12) + (x+3)/(x^2-16)

14. Originally Posted by Ari
Uggh sorry i messed up its supposed to be a + sign in the middle of the fractions

(x+6)/(x^2+x-12) + (x+3)/(x^2-16)
ok.

$\frac {x + 6}{x^2 + x - 12} + \frac {x + 3}{x^2 - 16} = \frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)}$

LCD here is: $(x + 4)(x - 4)(x - 3)$

so we have: $\frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)} = \frac {(x + 6)(x - 4) + (x + 3)(x - 3)}{(x + 4)(x - 4)(x - 3)}$

$= \frac {x^2 + 2x - 24 + x^2 - 9}{(x + 4)(x - 4)(x - 3)}$

$= \frac {2x^2 + 2x - 33}{(x + 4)(x - 4)(x - 3)}$

that cannot be factored any further.

our restrictions are: $x \ne \pm 4$ and $x \ne 3$ since those x's make the denominator zero

15. Originally Posted by Jhevon
ok.

$\frac {x + 6}{x^2 + x - 12} + \frac {x + 3}{x^2 - 16} = \frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)}$

LCD here is: $(x + 4)(x - 4)(x - 3)$

so we have: $\frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)} = \frac {(x + 6)(x - 4) + (x + 3)(x - 3)}{(x + 4)(x - 4)(x - 3)}$

$= \frac {x^2 + 2x - 24 + x^2 - 9}{(x + 4)(x - 4)(x - 3)}$

$= \frac {x^2 + 2x - 33}{(x + 4)(x - 4)(x - 3)}$

that cannot be factored any further.

our restrictions are: $x \ne \pm 4$ and $x \ne 3$ since those x's make the denominator zero

YOUR AWSOME!! but shouldnt it be 2x^2??

Page 1 of 3 123 Last