Lesson library
A.10(D)·Algebra II·Rational FunctionsSupporting

Solving Rational Equations

Solve rational equations and verify the resulting value.

Students solve rational equations by noting excluded values, clearing denominators, solving the resulting equation, and verifying the value in the original equation.

Lesson plan and worked examples by TEKSForge · not a TEA-issued lesson. Videos by independent educators.

Archived course: TEKS code and readiness label from TEA's 2013 Algebra II key. Algebra II is not one of Texas's current required STAAR EOC assessments, and this lesson does not change the Biology and Algebra I Study Plans.

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Excluded values

A rational equation has a variable in a denominator. Values that make a denominator zero are excluded. In 183x−2=6\frac{18}{3x-2}=6, x≠23x\ne\frac{2}{3}.

Clear the denominators

Multiply both sides by the denominator or least common denominator. Distribute carefully: 6(3x−2)=18x−126(3x-2)=18x-12. Then solve the resulting linear or quadratic equation.

Check and record

Substitute the result into the original equation. Reject any excluded value. For a griddable answer, convert fractions to an exact decimal when possible, such as 65=1.2\frac{6}{5}=1.2.

Worked example

Solve 183x−2=6\frac{18}{3x-2}=6.

  1. Excluded value: x≠23x\ne\frac{2}{3}.
  2. Multiply both sides by 3x−23x-2: 18=6(3x−2)18=6(3x-2).
  3. Distribute and solve: 18=18x−1218=18x-12, so 30=18x30=18x and x=53x=\frac{5}{3}.
  4. Check: 3(53)−2=33\left(\frac{5}{3}\right)-2=3, and 183=6\frac{18}{3}=6.

x=53x=\frac{5}{3}

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Texas public-school, Texas educator-created, and TEKS-aligned lessons, hosted by their creators. Each card names its source.

Texas public school video

Algebra 2 Solving Contextual Rational Equations

UT IPSI · Kid2Kid / Texas Education Agency

Solve rational equations, check the solutions, and interpret them in context.

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Try it yourself

TEKSForge practice · unscored

  1. 1. Solve xx−3=4\frac{x}{x-3}=4.

    Show answer and reasoning

    x=4x=4

    Multiply by x−3x-3: x=4x−12x=4x-12, so x=4x=4. The value is not excluded, and 41=4\frac{4}{1}=4.

  2. 2. Solve 2x+12=1\frac{2}{x}+\frac{1}{2}=1.

    Show answer and reasoning

    x=4x=4

    Multiply every term by 2x2x: 4+x=2x4+x=2x, so x=4x=4. Check: 24+12=1\frac{2}{4}+\frac{1}{2}=1.

Released STAAR practice

Texas Education Agency · archived 2013 released EOC items · scored with TEA's answer key; step-by-step explanations written by TEKSForge because this release has no rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Archived released STAAR Algebra II EOC · 2013 · Item 13 · 1 point

What value of pp makes the equation 194p−1=5\frac{19}{4p - 1} = 5 true? Record your answer.

Show STAAR answer and explanation

1.2

Multiply by 4p−14p-1 to get 19=5(4p−1)=20p−519=5(4p-1)=20p-5. Add 5 and divide by 20: p=24/20=1.2p=24/20=1.2.

1: Substituting 1 gives 19/319/3, not 5. Solve 19=20p−519=20p-5 to get p=1.2p=1.2.

Start scored STAAR practice

Learning objectives

By the end of the lesson, students can:

  • Identify values that make a denominator zero.
  • Multiply both sides by a denominator or least common denominator to clear fractions.
  • Verify solutions and record exact or decimal answers accurately.

Instructional sequence

  1. 1

    Engage

    5 min

    Ask what value of x would make 12x=3\frac{12}{x}=3 true and how students know.

  2. 2

    Direct instruction

    10 min

    Model listing excluded values, multiplying both sides by the denominator, and solving.

  3. 3

    Guided practice

    12 min

    Solve one-denominator equations, then equations with two fractions using the least common denominator.

  4. 4

    Graphical check

    12 min

    Graph each side as a function and confirm the intersection matches the algebraic solution.

  5. 5

    Closure

    6 min

    Complete the exit ticket and write the answer as it would be gridded.

Misconceptions to address

““Multiply only the fraction by its denominator.””

Teacher move: To keep the equation balanced, multiply both entire sides by the same expression.

““A value that makes a denominator zero can still be a solution.””

Teacher move: Division by zero is undefined. Excluded values must be rejected even if they appear during solving.