041i - Solving Linear Equations Algebraically

Lesson 41 — Solving Linear Equations Algebraically: Guided Practice

Strand: Algebra | Descriptor: AC9M8A02 | Duration: 45 minutes

Learning Intentions

• To solve linear equations involving negative coefficients and the pronumeral on both sides.
• To solve linear equations containing fractions using algebraic techniques.

Success Criteria

I can:

1. Solve equations with the pronumeral on both sides, including negative coefficients.
2. Solve equations containing a fraction by multiplying both sides by the denominator.
3. Solve equations that combine brackets and fractions.
4. Justify each step of my working using inverse operations.

Warmup

(5 minutes — mini whiteboards, rapid recall)

Solve each equation from memory (Year 7 review):

Answers: 1. ; 2. ; 3. ; 4. .

Teacher note: Questions 3 and 4 preview today’s extensions — a negative-friendly both-sides equation and a fraction. Ask: “What’s different about today’s equations going to be?”

Activities

Activity 1 — Negative Coefficients and Both-sides Equations (10 min)

I do. Solve .

Setting-out rule: always move to collect the variable on the side with the larger (or less negative) coefficient — here the right side started at , so adding to both sides avoids ever writing a negative pronumeral term.

I do — a case with negatives on both sides. Solve .

We do:

(Answers: 1. ; 2. .)

You do:

(Answers: 1. ; 2. ; 3. ; 4. — a deliberate non-whole-number result; discuss that this is still a valid solution.)

Activity 2 — Equations with Fractions (10 min)

I do. Solve .

I do — a fraction covering the whole expression. Solve .

Key distinction to model explicitly: in the first example only the -term is divided by ; in the second, the entire left side is divided by , so multiplying by must be done before anything else.

We do:

(Answers: 1. ; 2. .)

You do:

(Answers: 1. ; 2. ; 3. ; 4. .)

Activity 3 — Applied Inquiry: the Number-puzzle Challenge (14 min)

Pairs. Each puzzle must be translated into an equation before solving.

Puzzle 1. I think of a number, divide it by , then add . The result is . What was my number?

Puzzle 2. I think of a number, double it, subtract the number from , and get twice the original number again. What was my number?

Puzzle 3. A number increased by , then halved, equals the number decreased by . What is the number?

Puzzle 3 rewritten as an equation: .

Socratic scaffolding for Puzzle 3:

Prompt	Purpose
Understand: what two quantities are being set equal?	”The number increased by 6, then halved” and “the number decreased by 2.”
Devise a plan: how do you clear the fraction?	Multiply both sides by .
Carry out the plan	.
Continue	, so .
Look back — check it	and ✓
Looking back — could you have solved it another way?	Yes — treat it as “half of ” without multiplying first, then multiply out fractions of a bracket. Compare efficiency.

Answers: Puzzle 1 — , so . Puzzle 2 — is one valid reading (answers vary depending on translation; accept justified alternative equations), giving . Puzzle 3 — .

Checks for Understanding

(6 minutes — exit ticket, collected)

1. Solve .
2. Solve .
3. Solve .
4. Solve .
5. Reasoning. Explain why it is more efficient to add to both sides of than to subtract from both sides.

Answers: 1. ; 2. ; 3. ; 4. ; 5. Adding removes the negative pronumeral term entirely and leaves a positive coefficient on the other side, avoiding a sign error; subtracting would leave on the left, requiring an extra negative-sign step.

Common Misconceptions

Misconception	How to pre-empt it
Multiplying only the -term by the denominator, not the whole side, e.g. treating as .	Model with a box around the entire numerator before multiplying; use the “whole side” rule.
When collecting pronumerals, subtracting from the side with the smaller coefficient, creating unnecessary negatives.	Explicitly compare both options before starting; choose the side with the larger coefficient.
Losing a sign when moving a negative pronumeral term across the equals sign.	Insist on writing the inverse operation applied to both sides as a separate annotated step.
Believing every equation must give a whole-number answer.	Deliberately include fractional answers (Activity 1, Q4) and normalise them.
Multiplying only one term inside a bracket by the denominator when both a bracket and fraction are present.	Slow down and expand the bracket fully before or after clearing the fraction — show both orders give the same result.

Enrichment — Competition-Style Problems

E1 (Kangaroo style). Solve .

Answer

E2 (AMC Junior style). If has the solution , find .

Answer

, so , giving .

E3 (Challenge). Solve .

Answer

E4 (Challenge). A number, when is subtracted and the result halved, equals the number divided by . Find the number.

Answer

Homework

1. Solve: (a) (b) (c) (d) .
2. Solve: (a) (b) (c) (d) .
3. Solve .
4. A number, tripled and increased by , equals the number subtracted from . Form and solve an equation.
5. Reasoning. Two students solve . One collects on the left, the other on the right. Show both methods reach the same solution, and explain which is less error-prone and why.
6. Challenge. Solve .

Answers: Q1 — (a) (b) (c) (d) . Q2 — (a) (b) (c) (d) . Q3 — , so , giving . Q4 — , so . Q5 — left-collection: gives , ; right-collection: same equation gives , . Both are equally valid; collecting on the side that avoids a negative coefficient (here, the right, since would need subtracting) is generally less error-prone. Q6 — , so , .