Rational Expressions

[mlbmaniaco]

Right now in math class we are learning rational expressions. Since I am in an advanced math class, it seems like we learn a new lesson everyday. So if you don't understand something, you pretty much need to teach yourself. I don't really understand rational expressions, so can someone tell me if I am doing these two problems right(If I am doing wrong please tell me how to do them.)

Problem: x/3 = 4/x+4

Answer: 1) First I found a common denominator.
3x+12(x/3) = 3x+12(4/x+4)
2) So I got x+4 = x+3
3) Then the answer would be x=-4, x=-3

Am I right?

[TD]

Watch out with the brackets, I suppose you mean

\frac{x}{3} = \frac{4}{{x + 4}}

Now, multiply both sides with a common denominator to get rid of the denominators, so with for example 3\left( {x + 4} \right)

[Hurkyl]

First off, you need to learn to use parentheses correctly. Whenever an addition or subtraction is supposed to happen before a multiplication or division, you need parentheses.

For example:

4/x+4 = \frac{4}{x} + 4

but

4/(x+4) = \frac{4}{x+4}


(3) is totally wrong -- if x+3 = x+4, then x=-4 and x=-3 are certainly not solutions.

But... (2) is also totally wrong. You skipped a bunch of steps, so I don't know what you're doing wrong. Multiplication by 3x+12 was a reasonable idea, though. Could you post your work?

[mlbmaniaco]

See, I don't know my work. I have no idea what I am doing. This is as far as I got with the book. I am supposed to simplify and check

[TD]

Well, try what I said. By multiplying both sides with 3, you lose the left denominator. Then, multiply both sides with x+4, this will get rid off the right denominator :smile:

[mlbmaniaco]

so would i then have 3(3) = x2 = 4 ?

[mlbmaniaco]

Never Mind, I think I'll just give up. It is way to hard for me to understand

[TD]

I'll show you that first step. We multiply both sides with 3.
At the LHS, the 3 will cancel out with the denominator, as we wanted.
At the RHS, you can simplify it by multiplying it when the nominator.

\frac{x}{3} = \frac{4}{{x + 4}} \Leftrightarrow 3 \cdot \frac{x}{3} = 3 \cdot \frac{4}{{x + 4}} \Leftrightarrow x = \frac{{12}}{{x + 4}}

Now, try losing the right denominator by multiplying both sides with (x+4) in the same way :smile:

[mlbmaniaco]

So would I do this?

x+4 * x = 12/ x+4 * x+4

Then I would get . . .

x(x+4) = 12(x+4)

Right?

If so, what do I do next?

[mlbmaniaco]

wait I made a mistake . . .

It would be x(x=4) = 12
Right...

[mlbmaniaco]

I mean x(x+4) = 12

Right

[Hurkyl]

Yes:

x/3 = 4/(x+4)

implies

x(x+4) = 12

[TD]

I mean x(x+4) = 12

Right
Correct! :smile:

Now bring everything to 1 side and you have a quadratic equation.
Solve with the quadratic formula or by factoring.

[mlbmaniaco]

We are supposed to solve by factoring, so how do I do that?

[TD]

So we have

x\left( {x + 4} \right) = 12 \Leftrightarrow x^2 + 4x - 12 = 0

Personally, I would factor just by finding zeroes :smile:
The divisors of the constant (-12) are 'possible candidates'...