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A.2(A)·Algebra II·Algebraic MethodsSupporting

Transforming and Solving Equations

Use algebraic tools to transform and solve equations, including absolute-value equations.

Students solve absolute-value equations by isolating the absolute value, writing both linear cases, and checking each proposed solution.

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.

Practice the released item Ask the tutorAll Algebra II released practice50 minutes

Learn the concept

Isolate the distance

Undo operations outside the bars first. For a∣u∣=da|u|=d with a≠0a\ne0, divide by a to get ∣u∣=d/a|u|=d/a.

Write both cases

If ∣u∣=c|u|=c and c>0c>0, solve u=cu=c and u=−cu=-c. If c=0c=0, solve just u=0u=0. If c<0c<0, there are no real solutions.

Verify in the original

Substitute each candidate before removing the bars. Fractional coefficients need careful multiplication, and the check catches a sign error in either branch.

Worked example

Solve 4∣2−x3∣=164|2-\frac{x}{3}|=16.

  1. Divide by 4: ∣2−x3∣=4|2-\frac{x}{3}|=4.
  2. Solve 2−x3=42-\frac{x}{3}=4: x=−6x=-6.
  3. Solve 2−x3=−42-\frac{x}{3}=-4: x=18x=18.
  4. Check: 4∣2−(−2)∣=164|2-(-2)|=16 and 4∣2−6∣=164|2-6|=16.

{−6,18}\{-6,18\}

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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 Identifying Parent Functions

UT IPSI · Kid2Kid / Texas Education Agency

Review the absolute-value parent graph and its nonnegative outputs before solving distance equations.

Watch on YouTube

Try it yourself

TEKSForge practice · unscored

  1. 1. Solve 5∣1−x2∣=105|1-\frac{x}{2}|=10.

    Show answer and reasoning

    {−2,6}\{-2,6\}

    The interior equals 2 or −2-2, giving x=−2x=-2 or x=6x=6. Both give an absolute value of 2.

  2. 2. Solve 3∣x+4∣=−93|x+4|=-9.

    Show answer and reasoning

    No real solutions.

    Division gives ∣x+4∣=−3|x+4|=-3, impossible because absolute value is nonnegative.

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 39 · 1 point

What is the solution set for the equation 3∣1−14x∣=93\left|1 - \frac{1}{4}x\right| = 9?

  1. {−12,12}\left\{-\frac{1}{2}, \frac{1}{2}\right\}
  2. {−12,1}\left\{-\frac{1}{2}, 1\right\}
  3. {−8,16}\{-8, 16\}
  4. {−8,−16}\{-8, -16\}
Show STAAR answer and explanation

{−8,16}\{-8, 16\}

Divide by 3: ∣1−14x∣=3\left|1-\frac14x\right|=3. Then 1−14x=31-\frac14x=3 gives x=−8x=-8, and 1−14x=−31-\frac14x=-3 gives x=16x=16.

{−12,12}\left\{-\frac{1}{2}, \frac{1}{2}\right\}: x=12x=\frac12 gives 3∣1−18∣=2183\left|1-\frac18\right|=\frac{21}{8}, not 9.

{−12,1}\left\{-\frac{1}{2}, 1\right\}: x=1x=1 gives 3(34)=943\left(\frac34\right)=\frac94, not 9.

{−8,−16}\{-8, -16\}: x=−16x=-16 gives 3∣1+4∣=153|1+4|=15, not 9.

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Learning objectives

By the end of the lesson, students can:

  • Isolate an absolute-value expression.
  • Solve both cases when the isolated value is positive.
  • Check solutions in the original equation.

Instructional sequence

  1. 1

    Engage

    5 min

    Locate points at a fixed distance from zero.

  2. 2

    Direct instruction

    10 min

    Model isolating absolute value before writing two cases.

  3. 3

    Guided practice

    13 min

    Solve cases containing fractional coefficients.

  4. 4

    Compare and verify

    15 min

    Analyze omitted-case and sign errors by substitution.

  5. 5

    Closure

    7 min

    Complete the exit ticket with checks.

Misconceptions to address

“Absolute-value bars are just parentheses.”

Teacher move: Absolute value gives a nonnegative distance, so negative interior values can produce positive outputs.

“Only the positive case is needed.”

Teacher move: Both an interior value of c and an interior value of negative c have absolute value c.