Question

needed precal help: see attachment

Answer 1

@Wholock

Answer 2

This one will take a little work, hang with me here.

Answer 3

Alrighty, so, let's start by simplifying the bottom.

We are going to separate it into 2 equations.

\[((x + 1)/x) * -4/1\]
Figure the LCD
\[((x+1)/x) * -4/x\]
Add them together
\[(x + 1 - 4x)/x\]

Add like terms to get
-3x + 1

Answer 4

okay, following..

Answer 5

Now for the top!

\[((3(x + 1))/x) * -1/1\]

Using the same rules, take the LCD:

\[((3(x+1)/x) * (x/x)\]

(3(x + 1) - x)/x

(2x + 3)/x

Answer 6

This makes the equation

(2x + 3)/-(3x + 1)

Answer 7

so 6x + 4?

Answer 8

The domain is all reall numbers except those that make the equation zero..

Answer 9

so 3rd choice?

Answer 10

No! don't worry about simplifying further. Just work the denominator.

Answer 11

okay im so 1/3 then now it's the first choice?

Answer 12

Correct!

Answer 13

Sorry about the wait!

Answer 14

okay, thought so! no problem, i'm so thankful that you're helping! one more :'D

Answer 15

Sure thing!

Answer 16

i will just attach it here

Answer 17

Kay, begin by finding an LCD.

\[(1/2x) + (3/1)\]

\[(1/2x) + ((3 * 2x)/2x)\]

\[(1 + 6x)/2x\]
The LCD now will work for the numerator as well, so now you simply multiply the numerator by the LCD which is 2x.

Answer 18

2/((1+6x)/2x)

2 * 2x = ?

Answer 19

4x

Answer 20

@Wholock

Answer 21

Yes! That's it!

Answer 22

so the first choice?

Answer 23

Ta-da!

Answer 24

thank you soso much for all your help!

Answer 25

No problem! Anytime.