# Math Help - Rational Problems

1. Originally Posted by Ari
YOUR AWSOME!! but shouldnt it be 2x^2??
correct, i'll change it

2. Ok this is the other problem i dont get(btw i love this forum now)

simplify and state restrictions
x^2+2x-63/7-x

3. Originally Posted by Ari
Ok this is the other problem i dont get(btw i love this forum now)

simplify and state restrictions
x^2+2x-63/7-x
do you mean $\frac {x^2 + 2x - 63}{7 - x}$? if so, you should type (x^2 + 2x - 63)/(7 - x)

4. Originally Posted by Jhevon
do you mean $\frac {x^2 + 2x - 63}{7 - x}$? if so, you should type (x^2 + 2x - 63)/(7 - x)
yeah thats what i meant, BTW how do you type the way you type? i cant get those fractions and stuff properly like you...

5. Originally Posted by Ari
Ok this is the other problem i dont get(btw i love this forum now)

simplify and state restrictions
x^2+2x-63/7-x
$\frac {x^2 + 2x - 63}{7 - x} = \frac {(x + 9)(x - 7)}{-(x - 7)} = -(x + 9)$

the restriction is $x \ne 7$ since we have to consider the domain of the first function as well. in the original question, x = 7 makes the denominator zero

Originally Posted by Ari
yeah thats what i meant, BTW how do you type the way you type? i cant get those fractions and stuff properly like you...
i use LaTex. see here

6. Oh OK so you factor the top and then just cancel them out i see.

(5m^2+6m-8)/(5m^2-16m-16) * (20m^2+16m)/(4-m^2)

$
\frac {5m^2+6m-8}{5m^2-16m-16} \cdot \frac {20m^2+16m}{4-m^2}
$

OMG I DID IT!

7. $
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}
$

$
\frac {(j+3)(j+4)}{(j+3)(j+3)} \div \frac {(j-4)(j+4)}{(j+3)(j+4)}
$

8. I hope you're not letting me do your homework for you... where's that suspicious smiley?
Originally Posted by Ari
Oh OK so you factor the top and then just cancel them out i see.

(5m^2+6m-8)/(5m^2-16m-16) * (20m^2+16m)/(4-m^2)

$
\frac {5m^2+6m-8}{5m^2-16m-16} \cdot \frac {20m^2+16m}{4-m^2}
$
same story here.

$\frac {5m^2 + 6m - 8}{5m^2 - 16m - 16} \cdot \frac {20m^2 + 16m}{4 - m^2} = \frac {(5m - 4) (m + 2)}{(m - 4) (5m + 4)} \cdot \frac {4m (5m + 4)}{(2 - m)(2 + m)}$

$= \frac {4m(5m - 4)}{(m - 4)(2 - m)}$

what do you think the restrictions are? remember, consider the original function as well

OMG I DID IT!
haha, yes you did

not so hard was it?

9. Originally Posted by Ari
$
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}
$
now that you can type with LaTex, try this on your own, and i will correct you if anything

10. Thats what i was planning on doing but i got an email saying you posted so im reading that ATM thanks!

11. m doesnt = 4 or -2 <---is that right?

and im stuck here, do i cancel or recipricate and muiltyply?

$
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}
$

$
\frac {(j+3)(j+4)}{(j+3)(j+3)} \div \frac {(j-4)(j+4)}{(j+3)(j+4)}
$

12. Originally Posted by Ari
m doesnt = 4 or -2 <---is that right?
what about the (2 - x) and the (5m + 4) in the denominators of the original fractions

and im stuck here, do i cancel or recipricate and muiltyply?

$
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}
$

$
\frac {(j+3)(j+4)}{(j+3)(j+3)} \div \frac {(j-4)(j+4)}{(j+3)(j+4)}
$
multiply the first fraction by the reciprocal of the second fraction and see if anything cancels. if not, just multiply

13. so m doesnt = 2 but how do i do the (5m-4)?

$
\frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j+4)}{(j-4)(j+4)}
$

$
\frac {(j+3)(j-3)}{(j+3)(j-4)}
$

did i cancel the right stuff?

14. Originally Posted by Ari
so m doesnt = 2 but how do i do the (5m-4)?
set 5m + 4 = 0, solve for m

$
\frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j+4)}{(j-4)(j+4)}
$

$
\frac {(j+3)(j-3)}{(j+3)(j-4)}
$

did i cancel the right stuff?
no

15. m doesnt = -4/5

then what am i supposed to cancel?

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