Question

Rational expressions help again?

Answer 1

What I've done so far is factor the first one:
z^2
______
(z - 5)(z + 5) and then the LCD is 3(z - 5)(z + 5) so I multipled
2
___
3(z + 5) by (z - 5) and
z + 1/z + 5 by (3)(z - 5)

Then I simplified everything to get

3z^2 2z -5 3z^2 - 12z - 15
_______________ - ______________ = ___________________
3(z - 5)(z + 5) 3(z - 5)(z + 5) 3(z - 5)(z + 5)

Then I factored the right side to get

3(z + 1)(z - 5)
_______________
3(z - 5)(z + 5)

Is that right so far and can I cancel out (2 - 5)?

Answer 2

I would leave the right side as you had it 3z^2 - 12z - 15

on the left, add/subtract the numerators to get
3z^2 -2z + 10

equate the numerators on both sides
3z^2 -2z + 10 = 3z^2 - 12z - 15

now simplify

Answer 3

so first, I'll subtract the 3z^2 from the right and the left side...then add 10 to both sides..then add 12z to both sides...so now I have 10z = -5, divide by 10, z = -1/2?

Answer 4

\[ \cancel{3z^2} -2z + 10 = \cancel{3z^2} - 12z - 15 \]
I would add -10 to both sides
and +12z to both sides

Answer 5

those minus signs are tough to keep track of...

Answer 6

I did add -10 and 12z?

Answer 7

or do you mean + (-10)?

Answer 8

\[ -2z +10 = -12z -15 \\
-2z +10 -10 = -12z -15 -10 \\
-2z = -12z -25 \\
-2z + 12z = -25
10z = -25
\]

Answer 9

but it wasn't +10 it was -10?

Answer 10

I was wondering if you did something wrong.

the second term was
\[ - \frac{2}{3z+15} =\\

- \frac{2}{3(z+5)}
\]
to get a common denominator of 3(z-5)(z+5) multiply top and bottom by (z-5)
\[ - \frac{2(z-5)}{3(z-5)(z+5)} \]
write that as
\[ + \frac{-2(z-5)}{3(z-5)(z+5)} \]
distribute the -2 to get
\[ + \frac{-2z+10}{3(z-5)(z+5)} \]

Answer 11

Ohh I didn't see the "-" >< okay, I'm going to redo it, I have to brb for a while so I'll ttyl, thank you for your help Phi!

Answer 12

@phi is the answer z = -2.5?

Answer 13

yes. you can try z = -2.5 in the original equation... and you will see that it works.

Answer 14

Awesome, thank you!!