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C. Algebra · Algebra: Patterns, Rational Numbers, Equations, Inequalities, Coding and Modelling

Equations with Variables on Both Sides: When Are Two Plans Equal?

C2.3 A1About 50 minutes
Learning goals
  • I can solve equations that have variable terms on both sides, with integers and decimals.
  • I can write an equation to find when two linear situations are equal.
  • I can verify my solution by checking that both sides have the same value.

Words to know

variable term
A term with a variable in it, like 4x or −2.5n.
constant term
A term with no variable, like 11 or −6.
collect like terms
Move all the variable terms to one side and all the constant terms to the other, by doing the same thing to both sides.
verify
Substitute the solution into both sides of the original equation and check that they are equal.

Remember from Grade 7 and earlier in this unit

In the Grade 7 modelling lesson, you found when two ski plans cost the same by solving 45 + 4v = 10v. Earlier in this unit you found where two lines cross, like 120 − 6t = 30 + 4t.

Those equations have the variable on both sides. Today you'll learn a reliable method for solving them, with integers and decimals.

Let's start with two fundraisers

Two Grade 8 classes are raising money for a local food bank.

  • Class A has $120 and raises $35 a day.
  • Class B has $260 and raises $15 a day.

After how many days will they have the same amount? Let d be the number of days:

120 + 35d = 260 + 15d

The idea: get all the d-terms on one side and all the numbers on the other.

  • Subtract 15d from both sides: 120 + 20d = 260.
  • Subtract 120 from both sides: 20d = 140.
  • Divide both sides by 20: d = 7.

Verify: Class A: 120 + 35(7) = 120 + 245 = 365. Class B: 260 + 15(7) = 260 + 105 = 365 ✓. After 7 days, both classes have $365.

024678100100200300400500days (d)money raised ($)Class B (dashed): 260 + 15dClass A (solid): 120 + 35dequal at d = 7: $365 each
Class A starts lower but raises money faster, so its line is steeper. The lines cross at d = 7, where both classes have $365. After that, Class A is ahead.
4x + 2 = x + 11=Remove one x-tile from each side=Remove 2 units from each side: 3x = 9=Share into 3 equal groups: x = 3=
Solving 4x + 2 = x + 11 with algebra tiles. Removing one x-tile from each side keeps the balance. Then removing 2 units from each side leaves 3x = 9, so each x-tile matches 3 units.

Steps for variables on both sides

  1. Simplify each side (combine like terms).
  2. Collect the variable terms on one side by adding or subtracting a variable term on both sides.
  3. Collect the constants on the other side.
  4. Divide by the coefficient.
  5. Verify: both sides of the original equation should give the same number.

A handy choice

You can move the variable terms to either side. Subtracting the smaller variable term usually leaves a positive coefficient, which means fewer sign errors. For 3n − 7 = 5n + 9, subtract 3n from both sides: −7 = 2n + 9.

Worked example

Example 1: Tiles to symbols

Solve 4x + 2 = x + 11.

  1. Subtract x from both sides: 3x + 2 = 11.
  2. Subtract 2 from both sides: 3x = 9.
  3. Divide both sides by 3: x = 3.
  4. Verify: left 4(3) + 2 = 14; right 3 + 11 = 14 ✓.

x = 3

Worked example

Example 2: A negative solution

Solve 3n − 7 = 5n + 9.

  1. Subtract 3n from both sides: −7 = 2n + 9.
  2. Subtract 9 from both sides: −16 = 2n.
  3. Divide both sides by 2: n = −8.
  4. Verify: left 3(−8) − 7 = −31; right 5(−8) + 9 = −31 ✓.

n = −8

Worked example

Example 3: Decimals

Solve 1.5a + 4 = 0.5a − 2.

  1. Subtract 0.5a from both sides: a + 4 = −2.
  2. Subtract 4 from both sides: a = −6.
  3. Verify: left 1.5(−6) + 4 = −9 + 4 = −5; right 0.5(−6) − 2 = −3 − 2 = −5 ✓.

a = −6

Worked example

Example 4: A negative variable term

Solve −2x + 11 = x − 4.

  1. Add 2x to both sides: 11 = 3x − 4.
  2. Add 4 to both sides: 15 = 3x.
  3. Divide both sides by 3: x = 5.
  4. Verify: left −10 + 11 = 1; right 5 − 4 = 1 ✓.

x = 5

Hands-on activity

Mystery envelopes

You need: 4 or 5 envelopes (or small cups turned upside down), about 30 coins or buttons, and a helper.

  1. Your helper secretly puts the same number of coins in every envelope. (Say 3, but don't peek!)
  2. They set up two sides on a table that have the same total. For example: Left: 3 envelopes + 1 coin. Right: 1 envelope + 7 coins.
  3. Write the equation: 3x + 1 = x + 7.
  4. Solve it by removing the same things from both sides: one envelope from each side, then one coin from each side. Write each step.
  5. Open an envelope to check.

What to look for: 3x + 1 = x + 7 → 2x + 1 = 7 → 2x = 6 → x = 3. Each move on the table matches one line of the written solution. Try a few set-ups, then switch roles.

Just for fun

Why did the equation book a meeting with the guidance counsellor?

It had issues on both sides and needed help collecting itself.

Your turn

1

Solve 4x + 3 = x + 18.

2

Solve 7y − 5 = 3y + 11.

3

Solve 2m + 9 = 5m − 6.

4

Solve 6k + 4 = 2k − 8.

5

Solve −3p + 2 = p + 10.

6

Solve 0.8t + 1 = 0.3t + 3.5.

7

Pool pass A costs $20 plus $3 per visit. Pass B costs $5 per visit. After how many visits do they cost the same?

More practice: warm-up

1

Solve 3x = x + 8.

2

Solve 5n = 2n + 12.

3

Solve x + 7 = 2x.

4

Solve 4a − 1 = 3a + 2.

5

Solve 6y = 4y − 10.

More practice: on your own

1

Solve 9x − 4 = 5x + 12 and verify.

2

Solve 2.5n + 6 = n − 3 and verify.

3

Solve −4x − 7 = 2x + 11 and verify.

4

Snow-clearing service A charges $150 for the season plus $10 per snowfall. Service B charges $25 per snowfall. When do they cost the same? Which is cheaper for 14 snowfalls?

5

Hiker A is at 220 m and climbs 15 m each minute. Hiker B is at 400 m and comes down 15 m each minute. When are they at the same height? What height?

Challenge

1

Solve 3(x + 4) = x − 6.

2

An equilateral triangle has sides of 2x + 1. A square has sides of x + 3. Their perimeters are equal. Find x and the perimeter.

3

Solve 0.25x − 3 = 1.5 − 0.5x.

Check your understanding

Quiz

1.

Solve 3x + 5 = x + 13.

2.

Which is a correct first step for 7n − 2 = 4n + 10?

3.

Solve 2y − 9 = 5y + 3.

4.

Plan A costs $30 plus $2 per item. Plan B costs $5 per item. They cost the same at...

5.

Is x = 3 a solution of 4x − 1 = 2x + 5?

6.

Solve 1.5m + 2 = 0.5m − 4.

For parents and teachers

Big idea: C2.3 asks Grade 8 students to solve equations with multiple terms, integers and decimals, in various contexts. Equations with the variable on both sides are the algebra behind "When do two plans cost the same?"

Common difficulties:

  • Moving a term without changing its sign, for example writing 4x + 2 = x + 11 as 5x + 2 = 11. Talk about "subtracting x from both sides" rather than "moving" it.
  • Doing something to one side only.
  • Verifying only one side. Check that both sides give the same number.

How to help: Try the mystery envelope game. Each physical move matches one line of algebra, which makes the balance idea concrete. Then compare two real choices, like two snow-clearing services, and ask your child to find the break-even point. This lesson builds A1 skills: modelling, reasoning and perseverance.