Question

Subtracting the rational expressions x + 1/ 3x +1 minus x/ x- 4 and find the restrictions that resist with (or in) the resulting rational expression

Answer 1

@ranga @phi @hartnn

Answer 2

Please use the equation editor (see button below the input area)

Answer 3

\[\frac{ 2x }{ 3x+1} - \frac{ x }{ x-4}\]

Answer 4

the common denominator will be
\[ (3x+1)(x-4) \]
what should you multiply the first fraction by to get this common denominator?

Answer 5

You would have to multiply it by 1, right?

Answer 6

btw, your equation does not match what you posted (x+1) in the top of the first fraction versus 2x ??

Answer 7

I mean't to do 2x, I must of accidentantly hit the one, I'm sorry.

Answer 8

before going on, which problem are we doing ?

Answer 9

The one I wrote with the equation editor. Disregard the x+1 I was typing fast and meant 2x

Answer 10

@phi

Answer 11

ok, so multiply the first fraction by 1, but in the form
\[ \frac{(x-4)}{(x-4)} \]

and the second fraction by
\[ \frac{(3x+1)}{(3x+1)} \]

Answer 12

btw, I would *not* multiply out the denominator (x-4)(3x+1)
we want it in factored form, because we want to find any x's that make the bottom zero (these will be the x's that are not allowed)

Answer 13

Okay, and for (X-4) would it be 1(x-4) with the multiplication (sorry it's not in equation editior for some reason it was messing up for me)

Answer 14

like this
\[ \frac{ 2x }{ (3x+1)}\cdot \frac{(x-4)}{(x-4)} - \frac{ x }{ (x-4)}\cdot \frac{(3x+1)}{(3x+1)} \]

Answer 15

distribute the 2x in
2x*(x-4)

distribute the -x in
-x*(3x+1)

Answer 16

2x-4
-3x +1 ?

Answer 17

For the first fraction
2x * (x-4)
distribute the 2x means 2x times each term inside the parens
you write 2x*x and 2x * -4
then simplify

Answer 18

So it would be 2x and -8x?

Answer 19

almost, except 2*x * x is 2*x*x
(you get x*x in there)
people use the short-hand x^2 for x*x (i.e. \(x^2= x\cdot x\))

Answer 20

So could we use 2x^2

Answer 21

yes, the top of the first fraction becomes
2x^2 -8x

Answer 22

\[\frac{ 2^{x2}-8x }{ (3x+1)(x-4) }\]

Answer 23

now the second fraction
distribute the -x in
-x*(3x+1)
-3x^2 -x

Answer 24

I assume you mean
\[ \frac{ 2x^2-8x }{ (3x+1)(x-4) } \]

Answer 25

Yes, still getting used to equation editior, i'm sorry!

Answer 26

yes, the editor is more tricky than the algebra.

Answer 27

So the numerator for the second equation is -3x^2- x and now we gotta solve, right?

Answer 28

you don't solve fractions. but you can add the tops when the two fractions have the same denominator (as we have here)

Answer 29

in other words we now have
\[ \frac{ 2x^2-8x -3x^2 -x }{(3x+1)(x-4)} \]

Answer 30

using the same rule that allows
\[ \frac{1}{4} + \frac{2}{4} = \frac{1+2}{4} \]

Answer 31

Hmm, okay. So now is it kinda like combining the like terms? so it's 5x^2 - 8x

Answer 32

combine like terms

2 x^2 take away 3 x^2 leaves how many x^2 ?

Answer 33

-8 x's take away another x gives how many x's ?

Answer 34

We have -1x^2 and -9x's?

Answer 35

yes.

Answer 36

So the new numerator now is -1x^2 - (-9x)?

Answer 37

so that is the top


if this is multiple choice, you might have to tweak the answer to match the choices.

**
So the new numerator now is -1x^2 - (-9x)?
**
we already took care of the minus sign when we distributed -x (instead of just x)
so the the top is just
\[ -x^2 -9x \]

Answer 38

It's not multiple choice, it's part of an assigment. THe numerator would be correct.

Answer 39

now for the *excluded values*

we are not allowed to divide by zero

the bottom has (3x+1)(x-4)
if either factor is zero, the whole bottom becomes zero (and we do not want that)
what x makes
x-4 = 0 ?

what x makes
3x+1 = 0
?

Answer 40

-4 for the first one and -3 for the second?

Answer 41

ok on the first. that means x=4 is excluded


3*-3 + 1 is -9+1 = -8 which is not zero.

use algebra for the second problem
3x+ 1 = 0
first add -1 to both sides
3x + 1 -1 = 0-1 and simplify
3x = -1
now divide both sides by 3

Answer 42

*for the first one, I am sure you mean x=4 makes
x -4 = 0

Answer 43

So this means the "restrictions" are x=4 and x=-3x?

Answer 44

how did you get x= -3x ? (which is only true if x is zero)


3x+ 1 = 0
first add -1 to both sides
3x + 1 -1 = 0-1 and simplify
3x = -1
now divide both sides by 3

Answer 45

Well because negative 1 divided by 3 is a decimal.. so I assumed you would do 3x divided by -1 unless it's a decimal for the restriction

Answer 46

3x = -1
now divide both sides by 3

-1 divided by 3 is written -⅓
on the left side 3x/3 is just x (because 3/3 is 1 and 1*x is x)
we ge
x= -⅓
as your other restricted value.

Answer 47

Oh i'm sorry, thanks so much though for your help, I really appreciate it