Question

Rational Functions Help:
A box has a volume of x^3-16x cm^3. Another box has a volume of x^3-4x^2 cm^3. What is the ratio of the volume of the first box to that of the second?

What is the sum 3x/x^2-9 + 3/x^2-6x+9? State any restrictions on the variable.

What is the difference 4x/x^2-3x+2 - 5/6x-12 State any restrictions on the variable.

@whpalmer4

Answer 1

For the first part, since you need a ratio, write the two expressions as a fraction, with the first expression in the numerator, and the second expression in the denominator. Then factor the numerator and denominator.
What do you get?

Answer 2

For all of these, you might as well factor numerator and denominator, as that will ease your job of making a common denominator.

To find restrictions on the variable, find all values of \(x\) that make the denominator equal 0. \(x\) cannot take on those values, as doing so would result in a division by 0, which has an undefined value.

Answer 3

@mathstudent55 After simplifying the ratio and factoring it. I got x(x+4)/x^2. @whpalmer4 How would I begin factoring The sum and the Difference of problems though? :/

Answer 4

@whpalmer4 Can you help me set up the second problem because I am so confused?

Answer 5

What is the sum \[\frac{3x}{x^2-9} + \frac{3}{x^2-6x+9}\]

As I suggested, you need to factor the denominators.

Do you know how to factor the expressions \[x^2-9\]and\[x^2-6x+9\]

Answer 6

(x+3) and (x-3)

(x-3) (x-3)

After that Idk what to do.

Answer 7

Okay, that's good. Now write the fractions as factored:\[\frac{3x}{(x-3)(x+3)} + \frac{3}{(x-3)(x-3)}\]

Now before you go any further, figure out which values of \(x\) will cause the denominator of either fraction to equal 0. Those values of \(x\) are the restricted values.

Answer 8

Next, establish a common denominator for both fractions. Do you know how to do that?

Answer 9

First Part.

Here is the ratio of the two volumes written as a fraction.

\(\dfrac{x^3-16x}{x^3-4x^2} \)

Now we factor the numerator and denominator.

The first step is to factor out a common factor. Both the numerator and the denominator have common factors.

\(=\dfrac{x(x^2-16)}{x^2(x-4)} \)

Now we factor \(x^2 - 16\) in the numerator. As you wrote above, you know this is the difference of two squares and factors into \(x + 4)(x - 4) \).

\(=\dfrac{x(x + 4)(x - 4)}{x^2(x-4)} \)

Now we divide the numerator and denominator by the common terms (which is usually referred to as canceling out common terms.)

\(=\dfrac{\cancel{x}(x + 4)\cancel{(x - 4)}}{\cancel{x}x\cancel{(x-4)}} \)

We are left with

\(= \dfrac{x + 4}{x} \)

Your answer was \(\dfrac{x(x + 4)}{x^2} \).
Your answer is equivalent to the given problem and to the final answer, but you did not reduce the fraction completely because you could have divided the numerator and denominator by x. In other words, your answer is a correct step in trying to find the final answer.

Answer 10

@mathstudent55 Okay thank you so much!! So my final answer would be x+4/x? Do I have to cube the answer to find the ratio of the volumes? Because it asks "What is the ratio of the volume of the first box to that of the second?"


@whpalmer4 So the restrictions would be x cannot equal 3 or -3? And a common denominator I think would be (x+3)(x-3)^2. Right?

Answer 11

No. Don't cube anything. That is the final answer for the first problem. Each expression you were given was already a volume. It just so happened that the expressions, once written as a ratio in a fraction, could be simplified. If you write your answer is a single line, then it should be written as (x + 4)/x (using parentheses around the x + 4.)

Answer 12

Yes, \((x+3)(x-3)^2\) would be the logical common denominator. Convert both fractions to that denominator and combine, then simplify if possible.

Answer 13

Also, those are the correct restrictions on \(x\).

Answer 14

@mathstudent55 Oh okay thank you soo much! How do you award medals on here? you and @whpalmer4 were great help!! :)

Answer 15

Click the blue "Best Response" button on a post that helped you. Only can do one per question, go ahead and give it to @mathstudent55, he typed more :-)

Answer 16

Click on Best Response, but you can only give one medal per post. I gave you one, so give one to whpalmer4.

Answer 17

Alright, thank you to both of you! I am new on here so I am still trying to figure this thing out. How do you convert both fractions and combine? @whpalmer4

Answer 18

A favorite saying of mine: "it's amazing what can be accomplished when no one cares about who gets the credit" :-)

Answer 19

Multiply by a fraction that equals 1, but is written as missing terms / missing terms:

so if you have \[\frac{(x-2)}{(x+3)}\]and you want it over \((x+3)(x-3)\) you would multiply by this:
\[\frac{(x-2)}{(x+3)}*\frac{(x-3)}{(x-3)} = \frac{(x-2)(x-3)}{(x+3)(x-3)} \]
then expand the numerator and add with the other fraction(s)

Answer 20

(just a made-up example, any similarly to the problem at hand is strictly coincidental!)

Answer 21

similarity, not similarly...

Answer 22

really, it's just the same as if you were operating on numbers alone...

Answer 23

Alright since the LCD is (x+3)(x-3)^2? How would I do the square to get it for the denominators of both fractions? I am sorry, I'm just so confused

Answer 24

Square the 3x? so 3x^2/(x+3)(x-3)^2?

Answer 25

\[(x+3)(x-3)^2 = (x+3)(x-3)(x-3)\]

Your left hand fraction is \[\frac{3x}{(x-3)(x+3)} \]so we multiply it by \[\frac{(x-3)}{(x-3)}\]

Your right hand fraction is \[\frac{3}{(x-3)(x-3)}\]so we multiply it by \[\frac{(x+3)}{(x+3)}\]

when you do that and expand the numerators, you should be able to add them together and simplify (might have to factor the numerator again to do so)

Answer 26

in both cases, we multiply by a fraction with the value 1, so we aren't doing anything except changing to a more convenient form.

Answer 27

It's also possible that after you combine them that you won't be able to appreciably simplify, but you won't know until you do the work.

Answer 28

However, as I mentioned, it is important to do the check for restricted values first, before any cancellation of common terms from numerator and denominator, because the restrictions still apply, even if the terms that contributed them appear to go away.

Answer 29

Okay, So i got 3x^2-9/(x-3)(x+3)(x-3) and 3x^2 +9/(x-3)(x+3)(x-3).

is that right?

Answer 30

For example\[\frac{x(x+2)}{(x+2)(x-2)} =\frac{x}{x-2} \]but it still has the restriction \[x \ne -2\] as well as \[x \ne 2\]

Answer 31

Uh, no you need to recheck your work. Also, if you are going to write a fraction on one line like that, you must put parentheses around the numerator and denominator. What you've written is actually \[3x^2 - \frac{9}{(x-3)(x+3)(x-3)} + 3x^2 + \frac{9}{(x-3)(x+3)(x-3)}\]

Answer 32

(3x^2-9)/(x-3)(x+3)(x-3) would be the way to write the left hand fraction on one line

Answer 33

actually, (3x^2-9)/((x-3)(x+3)(x-3)) would be safer.

Answer 34

just doing the numerators:

\[3x*\frac{(x-3)}{(x-3)} = \]

\[3*\frac{(x+3)}{(x+3)} = \]

See your mistakes?

Answer 35

oh so the right hand fraction is 3x+9/(x-3)(x-3)(x+3) I hope that was right :/

Answer 36

Still didn't put parentheses around your numerator! Also, you need to go recheck your work on the left fraction.

Answer 37

Okay so it would be (3x+9)/...... I guess I am just still confused where did I mess up at? You said the left fraction was right, right? Just the right fraction was wrong?

Answer 38

No, I didn't say the left was correct.

Answer 39

Alright sorry about that, I'll re-do these problems again.

Answer 40

Just check your work carefully.

\[3x*\frac{(x-3)}{(x-3)} = \]

Answer 41

(3x^2-9x)/(x-3). If you factor it you get 3x(x-3)/(x-3) and the two (x-3)'s cancel out giving you 3x?

Answer 42

Okay, you don't want simplify here, you're just constructing the terms to add for the numerator. You did spot your mistake, I see, good :-)

\[\frac{3x^2-9x}{(x-3)(x+3)(x-3)} + \frac{3x+9}{(x-3)(x+3)(x-3)} = \]

Answer 43

Alright :) So do I simplify the numerators at this point and start canceling like terms?

Answer 44

Combine the two fractions, then simplify, if possible.

Answer 45

So it would be (3x^2-6x+9)/(x-3)(x+3)(x-3) after combining?

Answer 46

Again, you really need to put parentheses around that denominator if you want it to mean the right thing.

Yes, you have the right combined fraction. Unfortunately, the only factoring that can be done at this point is to factor out a 3 from the numerator, which doesn't really simplify the fraction at all.

Answer 47

What do you mean? I thought I did put parentheses in the denominator? I put (x-3)(x+3)(x-3). So it would be 3(x^2-2x+3)/(x-3)(x+3)(x-3) as my final answer since it cant be further simplified?

Answer 48

No, you need parentheses around all of those terms as well:
\[3(x^2-2x+3)/((x-3)(x+3)(x-3))\]otherwise what you are writing is \[\frac{3(x^2-2x+3)}{(x-3)}(x+3)(x-3)\]

Answer 49

Oh, okay! So 3(x^2-2x+3)/((x-3)(x+3)(x-3)) is the final answer since I cannot simplify anymore? Because i don't see anymore ways to simplify this answer.

Answer 50

Yes. If you wanted to make it a little tidier, you could of course write it as
\[\frac{3(x^2-2x+3)}{(x+3)(x-3)^2}\]or 3(x^2-2x+3)/((x+3)(x-3)^2)

Answer 51

Alright, thank you so much. That is starting to make a little more sense. Don't kill me, but could you help me through the 3rd problem if I tried doing it?

Answer 52

For the 3rd one I got (19x-5)/(6(x-2)(x-1)). Is that right? or do I need to re check my work?

Answer 53

Assuming the 3rd problem is supposed to be \[\frac{4 x}{x^2-3 x+2}-\frac{5}{6 x-12}\]I do not get the same answer. Also, what are your restricted values?

Answer 54

I got x does not equal 2 and 1.

Answer 55

agree on the restricted values. try your fraction algebra again

Answer 56

Okay it should say 4x/((x-2)(x-1)) - 5/(6(x-2)). The LCD would be 6(x-2)(x-1)?

Answer 57

yes

hint: we got the same results, but you had different signs in the numerator.

I'll check back later tonight, but I'm going offline for 5 hours or so.

Answer 58

Okay I will re-do that problem! Thanks again for your help! :)

Answer 59

Would the answer be (19x+5)/(6(x-2)(x-1))? Sorry for responding so late, internet has been down

Answer 60

@whpalmer4

Answer 61

Yes, you found your mistake (or at least didn't repeat it!)

Answer 62

Okay so the final answer is (19x+5)/(6(x-2)(x-1)) with restrictions on 2 and 1?

Answer 63

It is, or my name isn't Mortimer J. Schnicklegruber III :-)

Answer 64

That's a pretty awesome name ;) Thank you so much! In fact, I have a surprise for you! Just follow along with my plan :)