Question

Can someone plz help me get started on these fractions
f(x)=3x-8 div 4x+20

Answer 1

What do you need?

Answer 2

\[\frac{ 3x - 8}{4x + 20}\to \frac{3x-8}{4(x+5)}\]As there is nothing common between denominator and numerator so we cannot simply it.

Answer 3

sorry I just found this.

Answer 4

No worries! :D

Answer 5

How do I find the domain

Answer 6

hi

Answer 7

domain is all real numbers except for -5

Answer 8

No division by zero. So what is x going to be when the denominator is zero?

Answer 9

0

Answer 10

set the denominator equal to zero right? @Akshay_Budhkar

Answer 11

no. x cannot = -5.

Answer 12

Yes @sai

Answer 13

@saifoo.khan

Answer 14

f(x)=3x-8 over 4x+20

Answer 15

@denise9423 Domain means all of the values x can take on.

Here's your domain:

\[x \neq-5\]

Answer 16

If you have a look the function is having real values for all values except for x=5 when it is not defined as the denominator is zero

Answer 17

can you show me how to get 5

Answer 18

4x+20 =0

Solve for x.

x=-5.

Answer 19

Do you know why division by zero is not allowed?

Answer 20

ok,then can i sub it also 3x-8

Answer 21

no!

Answer 22

you only need to worry about the bottom.

Answer 23

no to ur question
ok

Answer 24

Remember this: What is 1 divided by 0? In other words, what times zero can equal 1? 0*2? 0*3? There is nothing you can multiply by zero to get 1.

So whenever you get these fractions and they ask you to find the domain, just think about what x has to be in order for the denominator to equal zero.

Answer 25

If you have an equation in the denominator, then set that equal to zero :-)

Answer 26

*and solve for x.

Answer 27

like this 4x+20=0

Answer 28

Yup! :-)

Answer 29

ok

Answer 30

Or if you have 1/(x^2+2x+3) . If you have something weird like that, then set it equal to zero and use the quadratic formula to solve for your two x-values. Those x-values are what x can't equal. Sorry, I sound like I'm talking in riddles lol.

Answer 31

ok so -7xyy over 28xxxxxy=0

Answer 32

Are you in algebra?

Answer 33

yes and Its rough

Answer 34

I'm gonna be more specific with what I said, sorry my fault:

Or if you have 1/(x^2+2x+3) . If you have something weird like that, then set it equal to zero and use the quadratic formula to solve for your two x-values. Those x-values are what x can't equal. Sorry, I sound like I'm talking in riddles lol.

x^2+2x+3 = 0 (then use the quadratic formula and solve for x)

Answer 35

That's okay!

Answer 36

I don't think they'll ask you to find the domain of something with x's and y's. Or are they?

Answer 37

yes

Answer 38

no sorry

Answer 39

simplify

Answer 40

I can help you on something like that, but first off is that denominator stuff making sense? Really sleep on that and try to remember it b/c it's super important for calculus.

Answer 41

-7xyy/28xxxxxy

Answer 42

Okay well what multiples to 7 and 8 have in common? What can you divide them both by?

Answer 43

I mean 28.

Answer 44

yes it does bc it will help me remember to put=0 so I can solve for x

Answer 45

good :-)

Answer 46

4

Answer 47

-7xyy/28xxxxxy

You can divide the numerator and denominator by 7.

Also, which has the smallest amount of x's, and then which one has the smallest amount of y's?

Answer 48

3x7=28

Answer 49

no 4x7=28

Answer 50

The numerator has the smallest amount of x's. There is 1 of them. The denominator has the smallest amount of y's, which is 1 y.

So we'll divide the numerator and denominator by 7.

-1*x*y*y/4xxxxxy

Answer 51

bottom has 5 x's and the top has 2 y's

Answer 52

We have to cancel out the smallest amount of x's. And we also cancel out the smallest amount of y's.

So we have: -1*y/4xxxx

I canceled out an x on the top and bottom. Then I canceled out a y on the top and bottom.

Answer 53

ok

Answer 54

sorry i have a 911 gotta chk out b back later thank you for all ur help god bless

Answer 55

k

Answer 56

Anyway, -1*y/4xxxx can be re-written as:

-y/4x^4

See Paul, my favorite math teacher: (I swear by his notes)
http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx