BetterGrades Algebra · Unit A2 · Lesson

Variables on both sides

Collect variable terms and constants strategically and interpret the resulting coefficient.

Opening situation

Start here

Compare two pricing plans that depend on the same usage variable.

Solve linear equations with variables on both sides by collecting variable terms strategically and preserving signs.

Before this lesson

Prerequisite check

  1. Combine 7x3x7x - 3x.
  2. Solve 4x=204x = 20.
  3. Subtract 2x2x from both sides of a simple equation.
Lesson text

Explanation

When xx appears on both sides, use the addition or subtraction property of equality to collect variable terms on one side. Subtracting the smaller variable term often keeps the remaining coefficient positive, but either direction is valid when signs are handled correctly.

After collecting variable terms, collect constants on the other side and divide by the remaining coefficient. The equals sign should stay aligned conceptually: each new line describes the same solution set, not a collection of terms drifting across a boundary.

Some equations collapse after collection. A true constant statement signals infinitely many solutions; a false constant statement signals no solution. Do not force a numerical value when the variable disappears.

When a variable appears on both sides, the goal is to collect all variable terms on one side and constants on the other. Subtracting the same variable term from both sides preserves equality just as subtracting a number does. In 5x+4=2x+19,5x + 4 = 2x + 19, subtract 2x2x from both sides to obtain 3x+4=193x + 4 = 19. The variable has not ‘moved’; equal variable quantities were removed from both sides.

Either side may be chosen, but collecting the variable where its coefficient becomes positive often reduces sign errors. For 3x7=8x+13,3x - 7 = 8x + 13, subtracting 3x3x gives 7=5x+13-7 = 5x + 13 and then 20=5x-20 = 5x. Subtracting 8x8x instead is also valid but produces 5x7=13-5x - 7 = 13. Both paths must lead to x=4x = -4. Comparing them demonstrates that method choice affects convenience, not the solution set.

Simplify each side before collecting across the equation. Distribution and like terms can reveal that apparent variable terms cancel. For 2(x+3)=2x+6,2(x + 3) = 2x + 6, simplification gives 2x+6=2x+62x + 6 = 2x + 6. Subtracting 2x2x leaves 6=6,6 = 6, a true statement indicating infinitely many solutions. If the remaining statement were false, there would be no solution.

Coefficient comparison can predict the likely classification. In ax +b=+ b = cx ++ d, unequal coefficients a and cc usually produce one solution because aca - c remains nonzero. Equal variable coefficients cause the variable to cancel, leaving a comparison of constants. Equal constants yield an identity; unequal constants yield a contradiction. This prediction helps a solver interpret cancellation rather than panic when xx disappears.

The original-equation check remains important. Substitute a one-solution result into both original sides. For identity or contradiction cases, explain why the simplified statement is always true or always false across the domain. A complete answer reports {x}, ,\varnothing , or the appropriate full domain—not merely the last constant statement.

When variables appear on both sides, collect them on the side that makes the arithmetic convenient. In 7x4=3x+20,7x - 4 = 3x + 20, subtracting 3x3x produces a positive 4x4x; subtracting 7x7x would also be valid but creates 4x-4x. Both paths must lead to the same solution if the balance is maintained. There is no rule that variables must move left—only a preference for clear, efficient work.

After variable terms are collected, the equation may become an ordinary one- or two-step equation, or the variable may disappear. Track the coefficient carefully: subtracting 3x3x from 7x7x leaves 4x,4x, while subtracting 7x7x from 3x3x leaves 4x-4x. A quick check substitutes the candidate into both original expressions rather than into a simplified line, because the original is the claim the solution must satisfy.

Grouping can conceal variable terms on both sides. In 2(x+5)=3(x1),2(x + 5) = 3(x - 1), distribute before collecting: 2x+10=3x32x + 10 = 3x - 3. Subtracting 2x2x gives 10=x3,10 = x - 3, then adding 33 gives x=13x = 13. Substitution confirms both original sides equal 3636. The check is particularly valuable because a single missed distribution sign would create a plausible-looking but incorrect candidate. When fractions are present too, clear denominators only after recording restrictions and distributing the common multiplier across every term. The goal is a simpler equivalent equation, not merely fewer visible fraction bars. Keep variable terms on one side and constants on the other only after both sides have been simplified. This order prevents terms hidden inside grouping from being collected prematurely.

Method

Collect variable quantities with visible balance steps

  1. Distribute and combine like terms separately within each side.
  2. Subtract one variable term from both sides, preferably leaving a convenient coefficient.
  3. Collect constants with an additive inverse and isolate the variable.
  4. If the variable cancels, classify the remaining statement as true or false.

Check: Substitute a unique solution into both original sides, or justify why the original equation is always true or always false.

Reference

Definitions and conditions

collect variable terms
Use balance operations to place variable terms on one side.Subtract or add the same variable expression on both sides.
constant statement
An equation containing no variables after simplification.Its truth determines whether the solution set is all values or empty.
strategic side
The side chosen to hold the remaining variable term.Choosing it can reduce negative coefficients but does not change the solution.
variable cancellation
Removal of equal variable terms from both sides through a balance operation.It signals that classification depends on the remaining constant statement.
contradiction
A statement such as 4=94 = 9 that is false for every input.An equation simplifying to a contradiction has no solution.
Examples

Worked examples

Foundation

Solve7x+2=4x+207x + 2 = 4x + 20

  1. Subtract 4x4x from both sides.
  2. Solve3x+2=203x+2=20
  3. Check x=6x=6 in both original sides.

Answerx=6x = 6

Collecting the smaller variable term keeps a positive coefficient. Subtracting equal variable quantities from both sides preserves the balance and explains the apparent movement.

Representation

Solve52x=3x155 - 2x = 3x - 15

  1. Add 2x2x to both sides.
  2. Add 1515 to both sides.
  3. Solve, then check20=5x20=5x

Answerx=4x = 4

Variable terms and constants are collected with separate balance operations. Choosing the subtraction direction can keep the remaining coefficient positive without changing correctness.

Transfer

Solve3(x+2)=2x+113(x + 2) = 2x + 11

  1. Distribute to get3x+6=2x+113x+6=2x+11
  2. Subtract 2x2x.
  3. Subtract 66 and check.

Answerx=5x = 5

Simplify before collecting variable terms. When the variable cancels, the constant statement—not a guessed value—determines the solution classification.

Practice

20 practice questions

Recall and read the structure

Warm-up

Question 1Retrieval · Foundation

Solve5x+3=2x+185x + 3 = 2x + 18

Need a hint?

Recall the named definition or perform a direct substitution before choosing an operation.

Question 2Retrieval · Foundation

Solve9x4=7x+109x - 4 = 7x + 10

Need a hint?

Recall the named definition or perform a direct substitution before choosing an operation.

Question 3Concept · Standard

Solve4x+12=6x24x + 12 = 6x - 2

Need a hint?

State what must remain true, then connect that condition to the equation.

Question 4Concept · Standard

Solve83x=x128 - 3x = x - 12

Need a hint?

State what must remain true, then connect that condition to the equation.

Build accuracy one step at a time

Core practice

Question 5Procedure · Standard

Solve2x9=55x2x - 9 = 5 - 5x

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 6Procedure · Standard

Solve4x+7=x8-4x + 7 = -x - 8

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 7Procedure · Mixed

Solve3(x+1)=2x+93(x + 1) = 2x + 9

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 8Procedure · Mixed

Solve4(2x1)=5x+174(2x - 1) = 5x + 17

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 9Procedure · Standard

Solve2(x5)+x=5x182(x - 5) + x = 5x - 18

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 10Procedure · Mixed

Solve0.5x+4=0.2x+100.5x + 4 = 0.2x + 10

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 11Representation · Mixed

Which subtraction keeps a positive coefficient in 7x+1=3x+97x+1=3x+9?

Need a hint?

Label the quantities and make the same relationship visible in the new form.

Question 12Representation · Transfer

A student changes 4x+2=7x104x+2=7x-10 to 4x7x=1024x-7x=-10-2. Is the line equivalent?

Need a hint?

Label the quantities and make the same relationship visible in the new form.

Explain, compare, and diagnose

Represent and reason

Question 13Error Analysis · Mixed

Solve ax +b=+ b = cx +d+ d when aca \ne c.

Need a hint?

Locate the first line that no longer preserves the original relationship.

Question 14Transfer · Transfer

Check x=7x = 7 in 4x+12=6x24x+12=6x-2.

Need a hint?

Identify the familiar equation structure before changing any symbols.

Question 15Modeling · Transfer

Solve6x+2=6x+26x + 2 = 6x + 2

Need a hint?

Define the unknown and its units before writing the equation.

Question 16Exit · Standard

Solve and verify5(x1)+2x=3(x+7)25(x-1)+2x = 3(x+7)-2

Need a hint?

Solve, classify the solution set, and verify against the original equation.

Model, transfer, and verify

Finish strong

Question 17Transfer · Transfer

Solve9x14=5x+189x - 14 = 5x + 18

Need a hint?

Identify the familiar equation structure before changing any symbols.

Question 18Modeling · Transfer

Solve 43x=7x+244 - 3x = 7x + 24 using a path that leaves a positive coefficient.

Need a hint?

Define the unknown and its units before writing the equation.

Question 19Error Analysis · Mixed

Classify 6(x2)+5=6x76(x - 2) + 5 = 6x - 7.

Need a hint?

Locate the first line that no longer preserves the original relationship.

Question 20Exit · Standard

A student reaches 0x=120x = 12 and divides by 00 to get x=0x = 0. Repair the classification.

Need a hint?

Solve, classify the solution set, and verify against the original equation.

Common mistakes

Error analysis

Wrong move: A variable term can cross the equals sign by changing sign without explanation.

Why it fails: This hides the equality operation and makes sign errors hard to detect.

Repair: Write the same addition or subtraction of the variable term on both sides.

Open-response checkA2.7

A student reaches 0x=120x = 12 and divides by 00 to get x=0x = 0. Repair the classification.

Write a complete attempt before opening the response guide.

Attempt once to unlock the response guide

Complete a substantive attempt to unlock the protected solution and scoring criteria.

Before continuing

Exit check

  1. Classify 6(x2)+5=6x76(x - 2) + 5 = 6x - 7.
  2. A student reaches 0x=120x = 12 and divides by 00 to get x=0x = 0. Repair the classification.
Summary

What to remember

Collect variable terms with explicit balance operations, then collect constants.

  • If the variable disappears, classify the resulting constant statement instead of inventing a value.

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