Carl Horowitz

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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Solving a Rational Equation for a parameter - Problem 1

Carl Horowitz
Carl Horowitz

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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Now whenever we're solving for a rational equation, the first thing we always want to do is to get rid of our denominator and we do that by multiplying by our least common denominator.

Typically we're dealing with numbers and just looking for an x or something like that, but if we're solving for a parameter, we're solving for a variable, our steps are going to be exactly the same.

So this one what we're looking for is multiplying by our least common denominator, we're dealing with a, b and c, there's nothing in common so we just multiply by a, b, c.

For our first term our a's cancel leaving us with bc is equal to b's cancel ac plus ab. So now we're trying to solve for a, but you know what this is we actually have 2 a's, so in order to solve for what we need to do is make 1 and we can do that by factoring out the a. Factor out the a, we end up with ac plus b is equal to bc, and c and b are just variables representing numbers, so we can divide just as we would any other number, this is divide by c plus b leaving us with a is equal to bc over c plus b.

So solving for a parameter, multiply by least common denominator and just solve as you would if you were just solving for any variable in any other equation.

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