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≈ 50 min · Grade 9 · Mathematics

Solving Equations with Variables on Both Sides

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Checked by VeraTeach before publishing · September 2026

Teacher asked for
solving linear equations with variables on both sides
Teacher's note
“They can solve two-step equations and fall apart when the variable shows up twice, usually by moving a term without changing its sign. Mixed class. About a third took Algebra readiness last year and the rest came straight from Grade 8 math.”
Objective produced
Solve equations with variables on both sides by using inverse operations to keep both sides equivalent and justify each step.

Materials & Prep

  • Project or draw a balance-scale image: left pan has 2 identical sealed bags and 6 counters; right pan has 1 identical sealed bag and 14 counters.
  • Prepare a half-sheet practice set and exit ticket. Students need notebooks and pencils.
  • Post the reminder: “Do the same operation to both sides.”
0–7min

Opening

DisplayShow the balance image without labels or an equation. Ask students to predict: “What could we remove from both pans so that one sealed bag is left on each side? What would you do next to find how many counters are in one bag?” Students first think, then compare reasoning with a partner.

ExpectedRemove one bag from each side, then remove 6 counters from each side, leaving one bag equal to 8 counters. This uses their existing understanding that equal amounts can be removed from equal groups.

ConnectReveal that each bag can be represented by x. The balance is 2x + 6 = x + 14. Explain that a variable on both sides is not a new kind of arithmetic. It is the same balance reasoning, written efficiently.

7–18min

Direct instruction

DefineAn equation stays true only when the same operation is performed on both sides. Our first goal is to collect all variable terms on one side, then solve the resulting two-step equation.

Worked exampleSolve 3x + 5 = x + 17. Think aloud: “There is an x-term on each side. I will subtract x from both sides because that makes the x on the right become zero.” Write:

3x + 5 - x = x + 17 - x 2x + 5 = 17 2x = 12 x = 6 Check: Substitute 6 into the original equation: 3(6) + 5 = 23 and 6 + 17 = 23. The original sides match.

MisconceptionAddress the common claim, “Move the x across and make it negative.” If a student merely adds x to the left in 3x + 5 = x + 17, they get 4x + 5 = 17, but they changed only one side. Test x = 6: the original equation is true, while 4(6) + 5 = 17 is false. Emphasize that terms do not travel. We apply an operation to both sides; the apparent sign change comes from adding the inverse.

ModelSolve 7 - 2x = x + 16. Think aloud: “I could subtract x from both sides, but that leaves -3x. That is still valid, but I will add 2x to both sides so my coefficient is positive.” Write:

7 = 3x + 16 -9 = 3x x = -3 Check in the original: 7 - 2(-3) = 13 and -3 + 16 = 13.

Check for understandingDisplay 4x - 9 = 2x + 7. Students show on fingers or write the first operation only, then explain why it must occur on both sides. Listen for “subtract 2x from both sides,” not “move 2x.”

18–32min

Guided practice

Have studentsSolve each equation with a partner, writing an operation on both sides for every line. Pause after each first step for a quick board check.

  1. 4x - 9 = 2x + 7
  2. 5 + 3x = x - 11
  3. 6 - x = 2x + 15

PromptFor each, ask, “Which variable term will you eliminate first? Why is that choice useful?” Invite partners to compare different valid first operations when they occur.

ScaffoldGive students who need it a three-row organizer labeled “original equation,” “same operation on both sides,” and “simplified equation.” Keep the operation written on both sides before combining like terms. Provide the spoken stem: “I will on both sides because will become zero.”

Look forIn problem 2, students should subtract x from both sides before subtracting 5. In problem 3, students may subtract 2x and obtain -3x = 9, then divide by -3. Validate this path and require a substitution check rather than treating a negative coefficient as an error.

32–44min

Independent work

StudentsSolve and check each equation. Require one sentence of justification for problem 3: “I chose to eliminate first because .”

  1. 5x - 8 = 2x + 13
  2. 18 - 2x = x + 3
  3. 4(x - 2) = 2x + 10
  4. 9x + 4 = 9x - 6

CirculateAsk students who cross out or “move” a term to rewrite that line as an operation on both sides. For problem 4, press students to interpret 4 = -6 after subtracting 9x: no value of x can make a true statement, so there is no solution.

ExtensionStudents create an equation with variables on both sides whose solution is x = -4, exchange with a partner, and verify the partner’s solution by substitution. Students ready for a deeper challenge explain why 3x + 2 = 3x + 2 has infinitely many solutions after the variable terms are eliminated.

44–50min

Closing

Exit ticketSolve 8 + 3x = x - 10. Show the operation performed on both sides, then check the answer in the original equation. On the back, complete: “A term does not move across an equal sign; instead, I .”

ExpectedSubtract x from both sides to get 8 + 2x = -10, subtract 8 to get 2x = -18, so x = -9. The final statement should say that the same inverse operation is performed on both sides. Sort tickets for students who still treat sign changes as term-moving and begin the next lesson with a small-group re-model using the balance image.

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