Question

Use the rational expression below to answer the following questions
a) what values of x are restrictions on the variable
b) simplify the rational expression

Answer 1

here's the rational expression and problems, its for a finals study guide

Answer 2

Values of x that make any denominator 0 (any time throughout the simplification) are restrictions.

Answer 3

Any thoughts on what those might be?

Answer 4

no idea, i took most of this class a year ago, and algebra 1 my freshman year so these concepts are super vague for me

Answer 5

if i get what youre saying right, its any number in the numberator that can make the denominator 0?
i dont remember how to simplify the expressions like this exactly

Answer 6

It's just any number in general that you can plug in for 'x' that make the denominator = 0

\[\large \frac{x^2 + 3x - 4}{5 - x} \div \frac{x^2 + 5x + 4}{x^2 - 2x - 15}\]

What numbers can you plug in here for 'x' would make the denominator of either fraction = 0?

Answer 7

im not sure, would i just plug in the values i have until i get a denominator or 0 with one of the x values?

again its been over a year since ive touched this concept so i'm not sure how to do it

Answer 8

Well I mean yeah that is 1 way to do it

But lets see

The denominator of the first fraction is \(\large 5 - x\) so what can you make 'x' that will make that = 0?

The denominator of the second fraction is \(\large x^2 - 2x - 15\) so what can 'x' be that would make this = 0?

Answer 9

to make the denominator on the left size 0 i guess x=5 so i can check that off as a restriction?
5-5=0

Answer 10

Right...x = 5 is 1 restriction

Answer 11

so on the other side i have to find x^2-2x-15
and whatever x is has to make all of the come out to 0?

Answer 12

Right

Just factor and make it easier

we need 2 numbers that multiply to make -15 and at the same time add to make -2

-5 and 3 come to mind

so the factors of that would be

\[\large (x - 5)(x + 3)\]

And so what can you make 'x' be so that comes out to 0?

Answer 13

so (5-5)(-3-3) would equal (0)(0)

Answer 14

or should x be the same in both the factors?

Answer 15

Nope it can definitely be 2 different numbers

(5 - 5)(anything) = 0 so if x = 5 this = 0
(anything)(-3 - 3) = 0 so if x = -3 this = 0

So we have 1 new restriction to cross off...-3

Answer 16

okay, so those would be the restrictions, 5 and -3 because both of those when plugged in make the denominator 0

Answer 17

Sorry that was a typo...it should have been (-3 + 3)=0

But still correct

Answer 18

And yes, that statement is correct

But that is just so far...as another poster has stated above, restrictions can appear at ANY point during simplification :) so we need to begin simplifying

Answer 19

yeah, that's what i really need help with, I dont remember how to simplify these

Answer 20

Alright

Well first...do you know how I got *above*

\[\large x^2 - 2x - 15 = (x - 5)(x + 3)\]?

Answer 21

i know you factored them
5 and three are both factors of 15 ?
that's about all i know on that though

Answer 22

Okay, well no that's pretty much it :) just wanted to make sure!

So we have

\[\large \frac{x^2 + 3x - 4}{5 -x} \div \frac{x^2 + 5x + 4}{x^2 - 2x - 15}\]

I would begin by turning this into multiplication....so to do that...all we do is flip the second fraction and turn this into multiplication..

So

\[\large \frac{x^2 + 3x - 4}{5 -x} \times \frac{x^2 -2x - 15}{x^2 + 5x + 4}\]

Okay?

Answer 23

alrighty, i see what you did there

Answer 24

Good, so now...that WAS considered a simplification...so we need to check if that introduced any more restrictions!

The first fraction is still the same

But now...the second fraction has a denominator of \(\large x^2 + 5x + 4\)

So what can 'x' be that makes that = 0?

Answer 25

ok so i have to factor that part first right then find the value of x

Answer 26

So I would factor that as well

two numbers that multiply to make 4 and add to make 5.....looks like 4 and 1

So

\[\large x^2 + 5x + 4 = (x + 4)(x + 1)\]

what can 'x' be that makes it = 0?

Answer 27

Oh, yeah sorry I jumped the gun there >.<

Answer 28

so the values of x are -4 and -1 to make that 0
looks like we did the same thing at the same time

Answer 29

Lol well yes both of those are now our new restrictions :)

Answer 30

okay, so the multiplication part, thats conisdered simplified,
and the 5, -1, -4, -3 are the restrictions ?

Answer 31

That is considered the FIRST part of the simplification...we still have more to do :D

And yes all of those are the correct restrictions!

Answer 32

whoops last part was an accident, whats the next part of the simplification?

Answer 33

Now, to simplify this down

lets catch up quick

We had

\[\large \frac{x^2 + 3x - 4}{5 - x} \times \frac{x^2 - 2x - 15}{x^2 + 5x + 4}\]

And we have already done some factoring so lets write those out here too

\[\large x^2 + 3x - 4 = (x+4)(x-1)\]
\[\large x^2 - 2x - 15 = (x+5)(x-3)\]
\[\large x^2 + 5x + 4 = (x+4)(x+1)\]

There...all caught up!

Answer 34

yeah, got all that down on paper!

Answer 35

So now...since we have all the factors...lets just write those into that equation we have up there

\[\large \frac{(x+4)(x-1)}{5-x}\times \frac{(x-5)(x+3)}{(x+4)(x+1)}\]

Answer 36

So...lets just combine these...since this is all just multiplication

\[\large \frac{(x+4)(x-1)(x-5)(x+3)}{(5 - x)(x+4)(x+1)}\]

Answer 37

And now, we just cancel out what we can...whatever is on top and on bottom will cancel

\[\large \frac{\cancel{(x+4)}(x-1)(x-5)(x+3)}{(5 - x)\cancel{(x+4)}(x+1)}\]

Answer 38

alrighty, i get that

Answer 39

Now, there is just 1 more thing we need to do

Is it true that \(\large (5 - x) = -(x - 5)\) ?

Answer 40

im not sure, wouldnt the second one mean you have to multiply the inside by -1 ?
like -(x-5) would be (-x+5)
not entirely sure, i could be reading that wrong

Answer 41

Right you have to multiply the inside by -1

So we HAD

5 - x

And what I'm concluding is that

-(x - 5) is the same

If we distribute the -1....we have

-x + 5

which is the same as 5 - x just written in n another order

Answer 42

okay

Answer 43

So with that little....proof? I guess lol we know that we can now write

\(\large (5 - x)\) as \(\large -(x - 5)\)

So in what we have left of our equation

\[\large \frac{(x-1)(x-5)(x+3)}{(5-x)(x+1)}\]

We can now write that as

\[\large \frac{(x-1)(x-5)(x+3)}{-(x-5)(x+1)}\]

So now we have 1 more factor that cancels

\[\large \frac{(x-1)\cancel{(x-5)}(x+3)}{-\cancel{(x-5)}(x+1)}\]

So finally ALL we have left is

\[\large - \frac{(x-1)(x+3)}{(x+1)}\]

Answer 44

whew! that wasn't too hard to follow thankfully,
do you think you could do the complex fraction underneath too with me?

Answer 45

Sure :)

Answer 46

thanks! fractions are the main thing i really am not good at

Answer 47

Alright..so

\[\huge \frac{\frac{x}{x-3}}{\frac{1}{x} - \frac{1}{x-3}}\]

The first thing we wanna do is get a common denominator on the bottom

Answer 48

okay, so would a common denominator be 3?

Answer 49

Not quite...

So we have *just the bottom part*

\[\large \frac{1}{x} - \frac{1}{x-3}\]

To get a common denominator here...we would multiply \(\large \frac{1}{x}\times \frac{(x-3)}{(x-3)}\) and we would also multiply \(\large \frac{1}{x-3} \times \frac{x}{x}\)

This would make the denominator in both cases x(x-3) right?

Answer 50

yeah, im following i think

Answer 51

Thats just a quick way to find a common denominator...multiply the first fraction by the denominator of the second fraction...and vice-versa

But okay...so once we evaluate all that out and combine them over the common denominator ...we have

\[\large \frac{(x-3) - (x)}{x(x - 3)}\] right?

Answer 52

right, i get that!

Answer 53

Okay good...now that was just the bottom part we focused on...lets bring the top back into this

\[\huge \frac{\frac{x}{x-3}}{\frac{(x-3)-x}{x(x-3)}}\]

Just a quick thing we can obviously see

\[\huge \frac{\frac{x}{x-3}}{\frac{(\cancel{x}-3)\cancel{-x}}{x(x-3)}}\] right?

so we really just have

\[\huge \frac{\frac{x}{x-3}}{\frac{-3}{x(x-3)}}\]

Answer 54

arighty, i see what you canceled out there :)

Answer 55

Great! Okay so now, for this next part...let me just write this as

\[\large \frac{x}{x-3} \div \frac{-3}{x(x-3)}\]

Just so you can see it like that last question

Now, we should turn this into multiplication...how?

Answer 56

okay so you just seperated it into two fractions like we did in the last question,
so to make it multiplication you flip the second fraction right?

Answer 57

Right and right! :D

so we have

\[\large \frac{x}{x-3} \times \frac{x(x-3)}{-3}\]

So then we just combine them again since its just multiplication!

\[\large \frac{x\times x(x-3)}{(x-3) \times -3}\] can you simplify it from there? :)

Answer 58

the (x-3) on the top and bottom cancel out right?

Answer 59

Correct :)

Answer 60

then youd be left with x^2/-3 ?

Answer 61

Perfect! :)

or just \(\large -\frac{x^2}{3}\)

but either way yes correct! :)

Answer 62

awesome! i'm trying to look through the rest of this to make sure there's nothing else on the same topic, i think everything else is sequences, probability, and then trig and functions :(

Answer 63

Thank you so much! i may ask for some help later on if i get stumped with any of the other questions on those topics!

Answer 64

Absolutely shoot me a message if you need to...I'm not gonna be on for a little bit after this but if I get on and you still need me I'll be there :)

Answer 65

thanks!