Mastering Rational Expressions: Simple Tips and Tricks for Advanced Math Classes

AI Thread Summary
The discussion focuses on mastering rational expressions in advanced math classes, highlighting the challenges of self-teaching when concepts are difficult to grasp. A user seeks help with solving the equation x/3 = 4/(x+4) and receives guidance on correctly applying mathematical operations, particularly the importance of parentheses. The conversation emphasizes the need to eliminate denominators by multiplying both sides and leads to the formulation of a quadratic equation, x(x+4) = 12. Participants discuss factoring techniques to solve the equation, underscoring the collaborative effort to clarify complex math concepts. Overall, the thread illustrates the learning process and problem-solving strategies for rational expressions.
mlbmaniaco
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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?
 
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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)
 
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?
 
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
 
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:
 
Another Question

so would i then have 3(3) = x2 = 4 ?
 
Never Mind, I think I'll just give up. It is way to hard for me to understand
 
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:
 
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
wait I made a mistake . . .

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

Right
 
  • #12
Yes:

x/3 = 4/(x+4)

implies

x(x+4) = 12
 
  • #13
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
We are supposed to solve by factoring, so how do I do that?
 
  • #15
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'...
 
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