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!
Solving a rational equation. The first thing we want to do is get rid of our denominators all together and the easiest way to do that is to multiply by the least common denominator which we have to do, we have to find by factoring everything first.
So looking at this, I see that I have x² minus 4 which can factor to x minus 2 times x plus 2, which so happens to also be my least common denominator. We have plus 2, plus 2, minus see everything we just need with an x minus 2 and an x plus 2.
So if we multiply both sides by our denominator, x minus 2, x plus 2, our fractions will disappear all together. Distributing in this into our first term, x plus 2s cancel leaving us with 3 times x minus 2 and we’re left with minus. Here our x minus 2s cancel leaving us with 1 times x plus 2 and for this one it’s important to remember your parentheses because we are subtracting we’re going to have to distribute this negative sign through. Don’t have to do it quite now but just remember your parentheses so you remember to distribute it.
And last thing we are equal to, our x minus 2, x plus 2s are going to cancel all together just leaving us with equal to 2. Now we just have a linear equation which we have to solve, distribute everything through 3x minus 6, distribute that through minus x minus 2 is equal to 2, combine like terms, 3x minus x is 2x, -6 minus 2 is -8 minus another -2 is -10. And then just solve. Bring the 10 over, 2x is equal to 10 divide by 2x is equal to 5.
The last thing we have to do is check and make sure this actually works with our equation it’s in the domain of what we can put in. Looking at our original statement we can put in 2 or -2, the answer is 5, there is no conflict there, so our answer is valid and we were able to solve this rational equation.