Rational Equations and Excluded Values

7.3: Rational Equations and Excluded Values. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 3 of 83 parts Practice Flashcards

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Single rational equation

Objective: Clear a denominator and check its excluded value.

Prerequisite skills: factoring and common denominators. Record forbidden denominator zeros first. Multiply by the denominator only for allowed inputs, solve the resulting equation, and check the candidate in the original equation. Common error: accepting a value at which the original denominator is zero. Restrictions and interpretation: Every original denominator must be nonzero before or after clearing fractions.

Formulas

$a/(x−h)=b ⇒ x=h+a/b for b≠0 and x≠h$

Worked examples

Try itSolve 69/(x − 22) = 3.Show answer

Exclude x=22. Multiply by x−22: 69=3(x−22). Divide by 3 and add 22: x=45. The denominator is 23, not zero, and the original quotient is 3. Therefore the requested answer is 45.

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Single rational equation

Objective: Clear a denominator and check its excluded value.

Formulas

$a/(x−h)=b ⇒ x=h+a/b for b≠0 and x≠h$

Worked examples

Try itSolve 69/(x − 22) = 3.Show answer

Exclude x=22. Multiply by x−22: 69=3(x−22). Divide by 3 and add 22: x=45. The denominator is 23, not zero, and the original quotient is 3. Therefore the requested answer is 45.

Equations with common denominators

Objective: Use a common multiplier without losing restrictions.

When fractions share or relate denominators, multiply every term by a common denominator. Numerator cancellation can simplify the equation, but it does not erase the original input exclusions. Common error: multiplying only one fraction by the common denominator. Restrictions and interpretation: Every original denominator must be nonzero before or after clearing fractions.

Formulas

$a/x + b/x = c ⇒ x=(a+b)/c, x≠0, c≠0$

Worked examples

Try itFor nonzero x, solve 25/x + 27/x = 1.Show answer

Combine the fractions: (25+27)/x=1. Multiply by nonzero x: x=52. Substitution gives (25+27)/52=1. Therefore the requested answer is 52.

Canceled expressions still have restrictions

Objective: Identify a rational equation with an excluded algebraic candidate.

Canceling a factor yields an equivalent expression only on the original domain. If the resulting equation demands a forbidden input, the original equation has no solution. This is a domain failure, not an arithmetic error. Common error: allowing the canceled zero back into the equation. Restrictions and interpretation: Every original denominator must be nonzero before or after clearing fractions. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

(x²−a²)/(x−a)=x+a only for x≠a

Worked examples

Try itHow many real solutions does (x² − 625)/(x − 25) = 50 have?Show answer

The original excludes x=25. Factoring and canceling on that domain gives x+25=50, whose only candidate is x=25. That input is forbidden, so there are no solutions. Therefore the requested answer is No solutions.

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