Unit
Rational Expressions and Functions
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
First thing we want to do whenever we’re solving a rational equation is to get rid of our denominators and the way we do that is by factoring all the denominators we have and looking for our least common denominator.
In this case nothing can factor out so we just have to figure out what our smallest combination is which is just going to be an x times a 3x plus 1. Then what we do is we take that denominator and multiply it by everything, so we end up with x and 3x plus 1 distributed through. To distribute that through for this first term here our 3x plus 1s cancel just leaving us with 2x. For the second term, our xs cancel leaving us with 3x plus 1, and then for our third term our 3x plus 1 cancels leaving us with negative 6x².
We now have basically a quadratic equation. We want to bring everything to one side and then solve it out somehow. So I’m going to bring everything to the left to make my x² term positive, so we end up with 6x², bring that 3x around minus x and then minus 1 and this is all going to be equal to zero and this should be able to factor. So we end up with 2x and 3x. We want the 3 to be negative to balance this out so we have the -1 here and the plus 1 here.
So figuring out what our solutions are then going to be, we end up with 1/2 for this factor and -1/3 for this factor.
The last thing we have to be careful of is checking our domain, making sure we can actually put this numbers into our equations. So we have to go up to our original statement and here we have one statement that’s 1/x so x can’t be zero, and then we have 3x plus 1 which can’t be zero which if we solve that out we actually get -1/3 which was one of our answers.
So this answer can’t be plugged in into this equation because then we’re dividing by zero so we have to get rid of this answer because it can’t get plugged in leaving us with one solution as ½.