Lesson 4 of 4
Clear denominators on the allowed domain and reject forbidden roots.
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Big ideas
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Every original denominator must be nonzero. These exclusions remain part of the equation even after you clear fractions.
Multiply both sides, including every term on each side, by a common denominator. On the allowed domain this produces an equivalent equation without fractions.
Solve the resulting equation, then substitute each candidate into the original denominators and original equality. A forbidden candidate is an extraneous solution, not a valid answer.
A context can impose more restrictions, such as positive rates or nonnegative time. State units and verify that a proposed value fits the model as well as the algebra.
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The solution is .
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A simplified rule never restores the canceled input.
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Wrong idea: multiplying by a common denominator means excluded values become allowed. Clearing denominators is justified only where the original fractions exist. The final solution set must still satisfy all original restrictions.
Tip
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Why must restrictions be recorded before multiplying by a variable denominator? Multiplication can hide inputs where the original equation is undefined.
Worked example
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Solve .
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Those inputs are excluded throughout.
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What is the first step before clearing rational denominators?
Record where the original denominators are zero.
Solve , with .
.
When multiplying an equation by a common denominator, which terms are multiplied?
Every term on both sides.
What is an extraneous candidate in a rational equation?
A value from transformed algebra that fails the original domain or equality.
Can an excluded candidate become valid after cancellation?
No.
If clearing denominators gives on the allowed domain, what follows?
There are no solutions.
If a rational equation is an identity on its allowed domain, what is its solution set?
All originally allowed inputs, keeping every exclusion.
Why check a rate model's solution for sign and units?
Algebraic candidates may not make sense in the stated situation.
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