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Rational Expressions

  • Thread starter mlbmaniaco
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  • #1
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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?
 

Answers and Replies

  • #2
TD
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Watch out with the brackets, I suppose you mean

[tex]\frac{x}{3} = \frac{4}{{x + 4}}[/tex]

Now, multiply both sides with a common denominator to get rid of the denominators, so with for example [itex]3\left( {x + 4} \right)[/itex]
 
  • #3
Hurkyl
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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:

[tex]4/x+4 = \frac{4}{x} + 4[/tex]

but

[tex]4/(x+4) = \frac{4}{x+4}[/tex]


(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?
 
  • #4
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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
 
  • #5
TD
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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:
 
  • #6
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Another Question

so would i then have 3(3) = x2 = 4 ?
 
  • #7
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Never Mind, I think I'll just give up. It is way to hard for me to understand
 
  • #8
TD
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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.

[tex]\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}}[/tex]

Now, try losing the right denominator by multiplying both sides with (x+4) in the same way :smile:
 
  • #9
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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?
 
  • #10
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wait I made a mistake . . .

It would be x(x=4) = 12
Right...
 
  • #11
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I mean x(x+4) = 12

Right
 
  • #12
Hurkyl
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Yes:

x/3 = 4/(x+4)

implies

x(x+4) = 12
 
  • #13
TD
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mlbmaniaco said:
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.
 
  • #14
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We are supposed to solve by factoring, so how do I do that?
 
  • #15
TD
Homework Helper
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So we have

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

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

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