Grade by Grade
C. Algebra · Algebra: Expressions, Equations and Coding

Solving Multi-Step Equations and Verifying Solutions

C1.5 A1About 75 minutes
Learning goals
  • I can solve equations with brackets, with variables on both sides, and with negative numbers.
  • I can explain each step by keeping the equation balanced.
  • I can verify a solution by substituting it into both sides (LS and RS).
  • I can recognize equations with no solution and equations true for every value.

Words to know

equation
A statement that two expressions are equal, such as 3x + 2 = x + 8.
solution (root)
A value of the variable that makes the equation true.
inverse operations
Operations that undo each other: addition and subtraction; multiplication and division.
LS and RS
The left side and the right side of an equation. To verify, calculate both separately and see if they are equal.
identity
An equation that is true for every value of the variable, such as 2(x + 1) = 2x + 2.

Remember from Grade 8

In Grade 8 you solved equations with integers and decimals, including equations with variables on both sides, and checked solutions by substitution. Grade 9 equations have more steps: brackets on one or both sides, negative coefficients, and some surprising special cases.

Let's start with a number puzzle

“I'm thinking of a number. If I triple it and subtract 7, I get the same result as adding 9 to it.”

Let n be the number. The puzzle says 3n − 7 = n + 9.

You could guess and check, but algebra finds the answer directly, every time. By the end of this lesson, puzzles like this will take you less than a minute. (The answer is at the end of Example 1.)

Keep the balance

An equation is like a balanced scale. Whatever you do to one side, you must do to the other side, and the scale stays balanced. You already know the two moves from the history lesson:

  • al-muqābala: remove the same thing from both sides (like 1 x-tile or 2 unit tiles).
  • al-jabr: add the same thing to both sides to “restore” a subtracted term.
=3x + 2 = x + 8start=2x + 2 = 8take 1 x-tile from each side=2x = 6take 2 unit tiles from each side=x = 3split each side into 2 equal groups
Solving 3x + 2 = x + 8 with tiles. Each step does the same thing to both sides. At the end, one x-tile balances 3 unit tiles, so x = 3.
1. Expand brackets2. Collect like terms on each side3. Get variable terms on one side4. Get constants on the other side5. Divide by the coefficient6. Verify: LS = RS?
A reliable order of steps for solving. Not every equation needs every step, but every solution should end with step 6.
Worked example

Example 1: Variables on both sides

Solve 3n − 7 = n + 9 (the number puzzle) and verify.

  1. Subtract n from both sides: 2n − 7 = 9.
  2. Add 7 to both sides: 2n = 16.
  3. Divide both sides by 2: n = 8.
  4. Verify: LS = 3(8) − 7 = 24 − 7 = 17. RS = 8 + 9 = 17. LS = RS ✓.

n = 8. The mystery number is 8.

Worked example

Example 2: Brackets

Solve 4(x − 3) = 2x + 6 and verify.

  1. Expand: 4x − 12 = 2x + 6.
  2. Subtract 2x from both sides: 2x − 12 = 6.
  3. Add 12 to both sides: 2x = 18.
  4. Divide by 2: x = 9.
  5. Verify: LS = 4(9 − 3) = 4(6) = 24. RS = 2(9) + 6 = 24 ✓.

x = 9

Worked example

Example 3: Negatives

Solve 5 − 2(x + 4) = 3x + 7 and verify.

  1. Expand carefully: −2(x + 4) = −2x − 8. So 5 − 2x − 8 = 3x + 7.
  2. Collect on the left: −2x − 3 = 3x + 7.
  3. Add 2x to both sides (to keep x positive): −3 = 5x + 7.
  4. Subtract 7: −10 = 5x.
  5. Divide by 5: x = −2.
  6. Verify: LS = 5 − 2(−2 + 4) = 5 − 2(2) = 1. RS = 3(−2) + 7 = −6 + 7 = 1 ✓.

x = −2

Worked example

Example 4: Special cases

Solve (a) 2(x + 3) = 2x + 5 and (b) 3(x − 1) + 1 = 3x − 2.

  1. (a) Expand: 2x + 6 = 2x + 5. Subtract 2x: 6 = 5, which is false. No value of x can make it true: no solution.
  2. (b) Expand: 3x − 3 + 1 = 3x − 2, so 3x − 2 = 3x − 2. Subtract 3x: −2 = −2, which is always true. Every value of x is a solution. This equation is an identity: the two sides are equivalent expressions.

(a) no solution (b) every real number is a solution

Habits of good solvers

  • Write each new line of the equation under the last one, with the equal signs lined up.
  • Do one step at a time and say what you did (“subtract 2x from both sides”).
  • When you verify, work out LS and RS separately from the original equation, not from a later line (a mistake in an early line would carry through).
  • Choose the side that keeps the x-coefficient positive. It avoids sign slips.
Hands-on activity

Mystery bags

You need: 4 identical paper bags or envelopes, about 30 coins (or buttons, dried beans), a table.

  1. Without your teen watching, a helper puts the same number of coins in each bag and closes them. (Try 4 coins per bag.)
  2. Set up the “equation” on the table: on the left, 3 bags and 2 loose coins; on the right, 1 bag and 10 loose coins. The helper must make the two sides have the same total: with 4 per bag, 3(4) + 2 = 14 and 4 + 10 = 14.
  3. Write the equation: 3x + 2 = x + 10.
  4. Solve by doing the same thing to both sides: remove 1 bag from each side, then 2 loose coins from each side. Share what is left equally among the bags.
  5. Open the bags to verify!
  6. Swap roles and make up your own puzzle.

Model answer: 3x + 2 = x + 10 → 2x + 2 = 10 → 2x = 8 → x = 4.

Just for fun

Why did the equation never fall over?

Because whatever happened to one side, it made sure the same thing happened to the other. Perfect balance.

Your turn

1

Solve 5x − 4 = 2x + 11 and verify.

2

Solve 2(x + 6) = 20.

3

Solve 3(2x − 1) = 4x + 9 and verify.

4

Solve 7 − x = 2x − 8.

5

Solve 4(x + 1) − 3 = 2(x − 2) + 11.

6

Is x = −3 a solution of 2x + 9 = 5 − x? Explain.

7

Write and solve an equation: “Double a number plus 5 equals 4 times the number minus 7.”

More practice: warm-up

1

Solve x + 7 = 12.

2

Solve 4x = −20.

3

Solve 3x − 5 = 16.

4

Solve x/2 + 3 = 9.

5

Solve 2x + 1 = x + 6.

More practice: on your own

1

Solve 6(x − 2) = 3(x + 5) and verify.

2

Solve 10 − 3(x − 2) = 4.

3

Solve 2.5x + 3 = 1.5x − 4.

4

Which equation has no solution and which is true for every x? (a) 4(x + 2) = 4x + 8 (b) 3x + 2 = 3x − 1

5

Solve −2(3 − x) = 5x + 6 and verify.

Challenge

1

Solve 3(2x − 1) − 2(x + 4) = 5(x − 3) and verify.

2

Find the value of k so that the equation 2x + k = 5x − 1 has the solution x = 3.

3

Create an equation with brackets on both sides whose solution is x = −2. Show that it works.

Check your understanding

Quiz

1.

Solve 2x + 5 = 17.

2.

Solve 3(x − 4) = 15.

3.

Solve 5x − 3 = 2x + 9.

4.

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

5.

How many solutions does 2(x + 1) = 2x + 3 have?

6.

Solve 6 − 2(x − 1) = 0.

For parents and teachers

Big idea: C1.5 asks students to create and solve equations for various contexts and verify their solutions. This lesson focuses on solving and verifying; the next two add fractions, formulas and real situations.

Common difficulties:

  • Sign errors when expanding a negative: −2(x + 4) is −2x − 8.
  • Doing a step to only one side.
  • Skipping verification, or verifying with a later line instead of the original equation.
  • Confusion about “no solution”: students may think they made a mistake. If the variables cancel and a false statement remains, there is no solution; if a true statement remains, every number works.

How to help: Do the mystery-bags activity. Then, for written practice, ask your teen to say each step aloud (“I subtracted 2x from both sides”). Insist on the LS/RS check; it catches most errors.