Solving Rational Equations: What is the Missing Step?

  • Thread starter Thread starter Mathnomalous
  • Start date Start date
  • Tags Tags
    Rational
AI Thread Summary
The equation 3/2 + 2/2x-4 = 1/x-2 was analyzed for solutions, leading to the conclusion that x = 2 is not valid since it makes the denominators zero. The attempt to solve the equation involved finding a common denominator and multiplying through, but ultimately resulted in an inconsistency. The discussion highlighted that the original equation has no solutions due to the restrictions on the variable. The key takeaway is recognizing that x = 2 is outside the domain of the function. This exercise serves as an example of how certain rational equations can yield no valid solutions.
Mathnomalous
Messages
83
Reaction score
5

Homework Statement



Solve for x.

Homework Equations



3/2 + 2/2x-4 = 1/x-2

The Attempt at a Solution



LCD = 2(x-2)

3/2 + 2/2x-4 = 1/x-2 Mult. all terms by 2(x-2)

3x - 6 + 2 = 2

3x = 6

x = 2

What am I missing here?
 
Physics news on Phys.org
Why would you think you're missing something?
 
No idea, really. I think I followed all the correct steps. I also re-checked my class notes and this happens to be one of the exercises done in class by my professor. However, he solved the equation in a different and longer manner but with the same result (even though there is a (x-2)(x-2) = x^2 - 4 ?).

It's probably the inconsistent solution I obtain that bothers me.
 
Well, x = 2 makes your denominators zero, so that can't be the right answer.

Suppose you reduce the second term of the equation by canceling the 2.

Now you have 3/2 + 1/(x-2) = 1/(x-2).

That equation doesn't make any sense, right? Subtracting 1/(x-2) from both sides gets you 3/2 = 0.

There aren't any solutions to this equation.
 
Well, that confirms my suspicion. This is the first exercise in this assignment that has no solutions.

Thank you for your time and assistance!
 
Yeah, you really weren't missing anything per se, you just didn't make the conclusion that x = 2 isn't in the domain of the function, as hgfalling illustrated.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top