Unit 3, Lesson 7
Solving Rational Equations
Algebra 2
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What do you notice? What do you wonder?
Jada is working to find values of that make this equation true:
She says, “If I multiply both sides by , I find that the solutions are and , but when I substitute in , the equation does not make any sense.”
Consider the equation . We could solve this equation for by multiplying each expression by to get an equation with no variables in denominators, and then rearranging it into an expression that equals 0. Here is what that looks like:
The last equation, , leads us to believe that the original equation has two solutions: and . Substituting into the original equation, we get , which is true since each side is equal to . But, substituting into the original equation, we get , which isn’t a valid equation since division by 0 is not allowed. This means isn’t a solution, so what happened to make us think that it was?
Let’s consider the simpler equation . This equation has one solution, . But if we multiply each side by the result is a new equation, , which has solutions 5 and 1. The 1 is a solution to the new equation because when , . But if we substitute 1 for into the original equation, we get , which is not a valid equation, so 1 is not a solution to the original equation. Because we multiplied each side of the original equation by an expression that has the value 0 when , the two sides and 0 that were unequal at that specific -value are now equal. For this example, is sometimes called an extraneous solution.
In the original example, is the extraneous solution. While is a solution to the equation we wrote after we multiplied the original equation by on each side, it is not a solution to the original equation since they are not equivalent. It should be noted that even though we multiplied by , , and , only one extraneous solution was added. This shows that multiplying by an expression that can equal 0 does not always cause an extraneous solution. So how do we tell if a solution is extraneous or not? We substitute it into the original equation and make sure the result is a valid equation.