Lesson library
A.5A·Algebra 1·Linear EquationsReadiness

Solving Linear Equations

Solve linear equations in one variable, including variables on both sides and the distributive property.

Students solve one-variable linear equations by maintaining equivalence, including distribution, combining like terms, and variables on both sides.

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

Learn the concept

Preserve equality

An equation states that two expressions have equal value. Adding, subtracting, multiplying, or dividing both sides by the same permissible value preserves the solution set. Simplify each side first and apply distribution to every term.

Collect and isolate

After distribution, combine like terms. Collect variable terms on one side and constants on the other, then divide by the coefficient. A negative outside parentheses changes every sign: −(x+4)=−x−4-(x+4)=-x-4.

Check the original

Substitute the proposed solution into the original equation, including all parentheses. If variable terms cancel, a true statement means infinitely many solutions and a false statement means no solution.

Visualize the concept

Original TEKSForge instructional diagrams, designed for this TEKS skill and clearly distinguished from released STAAR artwork.

A level balance has three x bags and six negative-one tiles on the left and one x bag and six positive-one tiles on the right, representing 3x minus 6 equals x plus 6.
Balanced linear-equation modelOriginal TEKSForge instructional visual · not a released STAAR figure

Study released STAAR visuals

Use the original graphs, tables, and diagrams from released STAAR questions to practice reading labels, values, and relationships before answering the items below.

Released STAAR Algebra I EOC · 2026 · Item 17 · 2 points

Triangle PQR has side PR labeled x centimeters, side PQ labeled 16 centimeters, and side RQ labeled 3 times the quantity x minus 2 centimeters.
Triangle PQROriginal visual from the released STAAR test.

Worked example

Solve 4(x−2)=2x+104(x-2)=2x+10.

  1. Distribute: 4x−8=2x+104x-8=2x+10.
  2. Subtract 2x2x from both sides: 2x−8=102x-8=10.
  3. Add 8 and divide by 2: 2x=182x=18, so x=9x=9.
  4. Check: 4(9−2)=284(9-2)=28 and 2(9)+10=282(9)+10=28.

x=9x=9

Watch the lesson

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

Texas public school video

Algebra 1 Linear Equations and the Distributive Property

Five Minute Math · Aaron Daffern, Dallas ISD

Distribute, collect variables, and preserve equality when solving one-variable equations.

Watch on YouTube

Try it yourself

TEKSForge practice · unscored

  1. 1. Solve 5x+3=2x+185x+3=2x+18.

    Show answer and reasoning

    x=5x=5

    Subtract 2x2x and 3 to obtain 3x=153x=15. Divide by 3; substitution gives 28 on both sides.

  2. 2. Solve −2(x+3)=8-2(x+3)=8.

    Show answer and reasoning

    x=−7x=-7

    Divide both sides by −2 to obtain x+3=−4x+3=-4, then subtract 3. Check: −2(−7+3)=8-2(-7+3)=8.

Released STAAR practice

Texas Education Agency · released EOC items · explanations adapted from TEA rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Released STAAR Algebra I EOC · 2019 · Item 3 · 1 point

Which value of xx makes the equation 1.25(4x−10)=7.51.25(4x - 10) = 7.5 true?

  1. 3.5
  2. −1
  3. −0.5
  4. 4
Show STAAR answer and explanation

4

Distribute 1.25 to both terms: 5x−12.5=7.55x - 12.5 = 7.5. Add 12.5 to both sides to get 5x=205x = 20, then divide by 5: x=4x = 4.

3.5: The 1.25 was multiplied by 4x4x but not by −10, giving 5x−10=7.55x - 10 = 7.5 and x=3.5x = 3.5. Distribute to every term in the parentheses.

−1: The distribution was correct, but 12.5 was subtracted from the right side instead of added, giving 5x=−55x = -5. Undo −12.5 by adding 12.5 to both sides.

−0.5: The 1.25 wasn't distributed to −10, and then 10 was subtracted instead of added. Distribute to every term, then use inverse operations.

Released STAAR Algebra I EOC · 2019 · Item 42 · 1 point

What is the solution to −(6m+8)=4(17−m)-(6m + 8) = 4(17 - m)?

Show STAAR answer and explanation

-38

Distribute: −6m−8=68−4m-6m - 8 = 68 - 4m. Add 4m4m to both sides: −2m−8=68-2m - 8 = 68. Add 8: −2m=76-2m = 76. Divide by −2: m=−38m = -38.

38: Check the last step: −2m=76-2m = 76 means you divide by a negative, so m=−38m = -38.

-30: Remember the negative sign distributes to the 8 too: −(6m+8)=−6m−8-(6m + 8) = -6m - 8.

Released STAAR Algebra I EOC · 2022 · Item 42 · 1 point

What is the solution to 4(q+56.5)=30q−1124(q + 56.5) = 30q - 112?

Show STAAR answer and explanation

13

Distribute: 4q+226=30q−1124q + 226 = 30q - 112. Subtract 4q4q: 226=26q−112226 = 26q - 112. Add 112: 338=26q338 = 26q. Divide by 26: q=13q = 13.

4.5: The 4 was multiplied by qq but not by 56.5. Distribute to both terms: 4⋅56.5=2264 \cdot 56.5 = 226.

Released STAAR Algebra I EOC · 2022 · Item 51 · 1 point

What is the solution to this equation? 2(40−5y)=10y+5(1−y)2(40 - 5y) = 10y + 5(1 - y)

  1. 7.5
  2. 15
  3. 5
  4. Not here
Show STAAR answer and explanation

5

Distribute: 80−10y=10y+5−5y80 - 10y = 10y + 5 - 5y. Combine like terms: 80−10y=5y+580 - 10y = 5y + 5. Add 10y10y and subtract 5: 75=15y75 = 15y, so y=5y = 5.

15: The signs were lost when distributing, so every term became positive. Keep the subtraction: 2(−5y)=−10y2(-5y) = -10y and 5(−y)=−5y5(-y) = -5y.

7.5: The 2 was distributed only to the 40, giving 80−5y80 - 5y. Multiply the 2 by both terms inside the parentheses.

Not here: A solution is listed. Distribute carefully, combine like terms, and you get y=5y = 5.

Released STAAR Algebra I EOC · 2021 · Item 10 · 1 point

What is the solution to 4(y−3)+19=8(2y+3)+74(y-3)+19=8(2y+3)+7?

  1. −12-\frac{1}{2}
  2. 12\frac{1}{2}
  3. −2
  4. 2
Show STAAR answer and explanation

−2

Distribute: 4y−12+19=16y+24+74y-12+19=16y+24+7, so 4y+7=16y+314y+7=16y+31. Then −24=12y-24=12y, so y=−2y=-2.

2: The solution is negative: −24=12y-24=12y.

−12-\frac{1}{2}: Check the distribution and combine constants before dividing by 12.

12\frac{1}{2}: The variable terms and constants must be collected on opposite sides.

Released STAAR Algebra I EOC · 2021 · Item 54 · 1 point

What is the solution to this equation? −4(2m−7)=3(52−4m)-4(2m-7)=3(52-4m)

  1. 32
  2. 46
  3. −6.4
  4. −40.75
Show STAAR answer and explanation

32

Distribute: −8m+28=156−12m-8m+28=156-12m. Add 12m12m and subtract 28: 4m=1284m=128, so m=32m=32.

46: Check the distribution on the left: −4(2m−7)=−8m+28-4(2m-7)=-8m+28.

−6.4: Collect variable terms on one side and constants on the other before dividing by 4.

−40.75: The constants are 156−28=128156-28=128, and the coefficient is 12m−8m=4m12m-8m=4m.

Released STAAR Algebra I EOC · 2026 · Item 17 · 2 points

Triangle PQR has side PR labeled x centimeters, side PQ labeled 16 centimeters, and side RQ labeled 3 times the quantity x minus 2 centimeters.
Triangle PQROriginal visual from the released STAAR test.

The perimeter of triangle PQRPQR is 5x5x centimeters. What are the lengths of PR‾\overline{PR} and RQ‾\overline{RQ} in centimeters?

Show STAAR answer and explanation

PR=10PR = 10 cm and RQ=24RQ = 24 cm

The perimeter equation is x+16+3(x−2)=5xx + 16 + 3(x - 2) = 5x, which simplifies to 4x+10=5x4x + 10 = 5x, so x=10x = 10. Therefore PR=10PR = 10 cm and RQ=3(10−2)=24RQ = 3(10 - 2) = 24 cm.

Released STAAR Algebra I EOC · 2025 · Item 29 · 1 point

What value of zz makes the equation 43z+6=−4(16z+9)\frac{4}{3}z + 6 = -4\left(\frac{1}{6}z + 9\right) true?

  1. 32\frac{3}{2}
  2. −21-21
  3. 92\frac{9}{2}
  4. −45-45
Show STAAR answer and explanation

−21-21

Distribute −4: 43z+6=−23z−36\frac{4}{3}z + 6 = -\frac{2}{3}z - 36. Add 23z\frac{2}{3}z to both sides: 2z+6=−362z + 6 = -36. Subtract 6: 2z=−422z = -42, so z=−21z = -21.

32\frac{3}{2}: Only the first term in the parentheses was multiplied by −4, giving 2z+6=92z + 6 = 9.

92\frac{9}{2}: Only the first term was multiplied, and by 4 instead of −4.

−45-45: Inverse operations were not used when moving terms across the equal sign, giving 23z=−30\frac{2}{3}z = -30.

Released STAAR Algebra I EOC · 2025 · Item 46 · 1 point

Which value of vv makes this equation true? v−65=v+102\frac{v - 6}{5} = \frac{v + 10}{2}

  1. −623-\frac{62}{3}
  2. −163-\frac{16}{3}
  3. 387\frac{38}{7}
  4. 117\frac{11}{7}
Show STAAR answer and explanation

−623-\frac{62}{3}

Multiply both sides by 10: 2(v−6)=5(v+10)2(v - 6) = 5(v + 10), so 2v−12=5v+502v - 12 = 5v + 50. Subtract 5v5v and add 12: −3v=62-3v = 62, so v=−623v = -\frac{62}{3}.

−163-\frac{16}{3}: Only the first term in each set of parentheses was multiplied, giving 2v−6=5v+102v - 6 = 5v + 10.

387\frac{38}{7}: Terms were moved across the equal sign without inverse operations, giving 7v=387v = 38.

117\frac{11}{7}: The denominators were added to the constants instead of multiplied, and terms were moved without inverse operations.

Released STAAR Algebra I EOC · 2026 · Item 36 · 1 point

What value of vv makes this equation true? 3(v+2)=2(v−9)3(v + 2) = 2(v - 9)

  1. –11
  2. −125-\frac{12}{5}
  3. −75-\frac{7}{5}
  4. –24
Show STAAR answer and explanation

–24

Distribute: 3v+6=2v−183v + 6 = 2v - 18. Subtract 2v2v and 6 from both sides: v=−24v = -24.

–11: Only the first term in each set of parentheses was multiplied, giving 3v+2=2v−93v + 2 = 2v - 9.

−125-\frac{12}{5}: Terms were moved across the equal sign without inverse operations, giving 5v=−125v = -12.

−75-\frac{7}{5}: Only the first terms were multiplied, and terms were moved without inverse operations, giving 5v=−75v = -7.

Start scored STAAR practice

Learning objectives

By the end of the lesson, students can:

  • Simplify each side using the distributive property and like terms.
  • Use inverse operations on both sides to isolate the variable.
  • Check a solution by substitution and explain each equivalence-preserving step.

Instructional sequence

  1. 1

    Engage

    5 min

    Use a balance to ask why the same operation must be applied to both sides of an equation.

  2. 2

    Direct instruction

    12 min

    Model simplify each side → collect variable terms → collect constants → divide by the coefficient. Distribute to every term before moving anything.

  3. 3

    Faded example

    13 min

    Solve 3(x−4)=x+63(x-4)=x+6 together, gradually hiding steps until students supply the operation and resulting equation.

  4. 4

    Error analysis

    12 min

    Diagnose incomplete distribution, sign errors, and dividing only one side; then check corrected solutions by substitution.

  5. 5

    Closure

    8 min

    Complete one mixed equation independently and annotate the reason for each step.

Misconceptions to address

“A term can move to the other side by simply changing its sign.”

Teacher move: Show the underlying operation performed on both sides; the shortcut is a record of that operation.

“The outside factor multiplies only the first term in parentheses.”

Teacher move: Use an area model or arrows to show distribution to every term.