Question

solve the rational equation. check your answer.
2c/c - 4 - 2 = 4/c + 5

Answer 1

I think you need some ( ) or the equation editor.... but to guess, did you mean:
\[\frac{2c}{c - 4} - 2=\frac{4}{c + 5}\]

Answer 2

Yes, that is what I meant.

Answer 3

Have you tried using multiplication to cancel out the fractions?

Answer 4

I know I need to cross multiply, but I don't know which numbers to multiply.

Answer 5

Well, you can't just cross multiply because of that -2. But you can multiply through by each deniminaotr.

Answer 6

I mean you have the distribution to deal with:\[\frac{2c}{c - 4} - 2=\frac{4}{c + 5}\implies \\
( c + 5)\left(\frac{2c}{c - 4} - 2\right)=\left(\frac{4}{c + 5}\right)( c + 5)\implies \\
\frac{( c + 5)2c}{c - 4} - ( c + 5)2=4\]

Answer 7

Once you do the same thing to \(c-4\) you will have it without the fractions. Then you can get it all to one side of the equation, combined like terms, and solve for 0s. However, if any answer you find would invalidate the original equation, it is not a useable answer! You have fractions, so if the divisor would be 0 because of the answer you find....

Answer 8

I know my final answer should be -14, I just need to show my work

Answer 9

Well, keep going the way I started that. You will have all the work.

Answer 10

If there is something about what I did that you do not understand, what is it? That way I can explain it.

Answer 11

I don't know what to do next

Answer 12

You see how I got rid of one denominator, right?

Answer 13

Not really, I'm terrible with math

Answer 14

OK, lets go back to fraction and numbers for a bit. If I multiply \(\frac{1}{2}\) by 2, what happens to it?

Answer 15

It becomes 1

Answer 16

Yes. And there is a process to that.
The 2 goes up on top, and multiplies the 1.
\(\frac{1}{2}\cdot 2=\frac{1\cdot 2}{2}\)
But anything times 1 is itself, so that simplifies:
\(\frac{1\cdot 2}{2}=\frac{2}{2}\)
Then, because there is a 2 on the top and the bottom, it cancels:
\(\frac{2}{2}=\frac{1}{1}\)
Which as you pointed out is just 1.

Now, what if we put a letter in there for a variable? It mmakes it a little different, but it is the same set of steps.

Answer 17

If I multiply \(\frac{3}{x}\) by x, what happens to it?
The x goes up on top, and multiplies the 3.
\(\frac{3}{x}\cdot x=\frac{3\cdot x}{x}\)
Then, because there is an x on the top and the bottom, it cancels:
\(\frac{3x}{x}=\frac{3}{1}\)
Now, 3 over 1 is just 3.

Not that different, right?

Answer 18

yeah, it's just confusing because it's such a big problem. I never know what to multiply first

Answer 19

No problem. It is getting there. I just want to show one last like thing this leads to.

Answer 20

So now that you have seen this with a variable, what about a term, an expression, or what you might call a factor!

What if I have \(\frac{4}{x + 5}\) and I multiply it by \(x + 5\)?
\[\frac{4}{x + 5}\cdot ( x + 5)\implies \frac{4\cdot ( x + 5)}{x + 5}\implies \frac{4}{1}\implies 4\]Do you see how that worked?

Answer 21

Yes

Answer 22

Oops. Pasredin your problem wrong. LOL. Fixing that:

Now that we have reviewed fractions a bit, lets see how that applies to your problem.

So I looked at your problem:\[\frac{2c}{c - 4} - 2=\frac{4}{c + 5}\]And I thought to myself, "There are fractions. Only two ways to deal with fractions. Common denominators or eliminate the fractions."

Personally, I like fractions. They are very useful. However, in this case you need to solve for c in the end. To solve for c, you must eliminate the fractions at some point. So that told me the best path to take is eliminating the fractions.

To eliminate the \( c + 5\) on the right, I can do exactly what I did above! However, anything I multiply on the right, I MUST multiply on the left, so I have to multiply both sides by \(c + 5\).

Answer 23

That was why my first step was:\[\frac{2c}{c - 4} - 2=\frac{4}{c + 5}\implies \\
( c + 5)\left(\frac{2c}{c - 4} - 2\right)=\left(\frac{4}{c + 5}\right)( c + 5)\implies \\
\frac{( c + 5)2c}{c - 4} - ( c + 5)2=4 \]The second step is to do something similar to get rid of \(c-4\). Can you do that second step now?

Answer 24

What do I need to do to get rid of the c - 4?

Answer 25

Well, look at how I multiplied through the c+5. That is what you need to do, but with c-4.

Answer 26

what do I multiply it by?

Answer 27

Well, in every example with fractions above I got rid of the fraction by multiplying by what was on the bottom of the fraction. Does that explain how you choose what to multiply by?

Answer 28

Not really. I would need to see it in the problem

Answer 29

\[\frac{( c + 5)2c}{c - 4} - ( c + 5)2=4\]\(\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\)There is is.

Answer 30

so multiply it by itself on the left and it by (c+5) on the right?

Answer 31

Well, there are three pieces of this equation.
\[\text{One is:}\frac{( c + 5)2c}{c - 4} \text{ The second is}- ( c + 5)2 \text{ And there is the }4\]All three of these need to be multiplied by \(c-4\) to get rid of that fraction.

Answer 32

I don't know how to distribute

Answer 33

Well, did you see how I distributed when I did the part I did?

Answer 34

I keep looking it over but it doesn't really make sense

Answer 35

I still do not understand how to do it with the c-4

Answer 36

Well, see how my pruple arrows go to three places? Same sort of distribution.

Answer 37

I see where the arrows go but I don't really know how to use the information to do it with the c - 4

Answer 38

Same basic procedure:

Answer 39

I don't know how to multiply c- 4 if it were just 4 I could do it

Answer 40

Hmm.... do you know how to do FOIL for things like (x+2)(x+2)?

Answer 41

No

Answer 42

OK.
You saw the arrows, right? On the second distribute picture. That is the path the entire \((c-4)\) nees to follow.

Answer 43

With the ( ) around it. They are important still.

Answer 44

How is that relevant?

Answer 45

Because they let you know that it still needs to be distributed after. ( ) are a way of saying something needs to be multiplied. For example, when I did this:
\[\frac{( c + 5)2c}{c - 4} - ( c + 5)2=4\]
The \((c+5)\) still needs to be multiplied by a few things there.

Answer 46

This is exactly where i am at. Could you please walk me through the rest of this problem so I can try to understand how to finish the rest my my assignment?

Answer 47

That is what I have been trying to do. I was going to take care of the divisors first, then finish multiplying things out.

So, can you use the arrows in the second picture to distribute the \((c-4)\)? The arrows show you where it needs to go.