Lesson 18 ~30 min Unit 2 · Linear Relationships +85 XP

Simultaneous Equations — Substitution Method

Replace one variable with an expression from the other equation to solve a system of two simultaneous equations. Isolate, substitute, solve and back-substitute.

Today's hook: If $y = 2x + 1$ and $3x + y = 16$, can you replace $y$ in the second equation with $(2x + 1)$? That swap turns two equations into one — and one equation is easy to solve!

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Think First

warm-up

Before you read on — If $y = 2x + 1$, can you replace $y$ in $3x + y = 16$ with $(2x + 1)$? What single equation do you get? Try to solve it before continuing.

Record your answer in your workbook. Then check: $3x + (2x+1) = 16 \Rightarrow 5x = 15 \Rightarrow x = 3$, $y = 7$.

The Big Idea — Substitution

+5 XP

Substitution means replacing one thing with another equal thing. If you know $y = 2x + 1$, then everywhere you see $y$, you can swap in $(2x + 1)$.

The substitution method works in four steps: Isolate one variable in one equation, Substitute that expression into the other equation, Solve the resulting single-variable equation, then Back-substitute to find the other variable. Always verify your solution in both original equations.

Isolate first

Make one variable the subject in one equation.

Substitute

Replace that variable in the other equation.

Verify both

Check your solution satisfies both equations.

Isolate → Substitute → Solve → Back-substitute → Verify

What You'll Master

objectives

Know

Substitution means replacing one variable with an expression from the other equation
The four-step method: isolate, substitute, solve, back-substitute

Understand

Why substitution works: replacing equals with equals always gives an equivalent equation
When substitution is easiest: when one variable is already isolated or has coefficient 1 or −1

Can Do

Solve simultaneous equations using the substitution method
Choose which variable to isolate for the cleanest substitution
Check solutions in both original equations

Words You Need

vocabulary

Substitution methodSolving simultaneous equations by replacing one variable with its equivalent expression from another equation.

IsolateRearrange an equation to make one variable the subject (e.g., $y = \ldots$ or $x = \ldots$).

SubstituteReplace a variable with an equivalent expression from another equation.

Back-substituteAfter finding one variable, substitute its value back to find the other variable.

ReplaceSwap one quantity for another equal quantity — the core idea of substitution.

VerifyCheck that your solution satisfies both original equations.

Worked Example — Easy Substitution

+5 XP

When to use: One equation already has a variable isolated. Solve: $y = x + 3$ and $2x + y = 15$.

In $y = x + 3$, the variable $y$ is already isolated. Replace $y$ in $2x + y = 15$ with $(x + 3)$. This gives one equation with one variable, which we can solve directly. Then back-substitute to find $y$.

Step 1: $y = x + 3$ → Step 2: $2x + (x+3) = 15$ → Step 3: $x = 4$ → Step 4: $y = 7$

Step 1: $y = x + 3$ — already isolated. Use this as the substitution expression.

Step 2: Replace $y$ in $2x + y = 15$: $\quad 2x + (x + 3) = 15$. Use brackets!

Step 3: $3x + 3 = 15 \Rightarrow 3x = 12 \Rightarrow x = 4$.

Step 4: Back-substitute: $y = 4 + 3 = 7$. Verify: Eq1: $7 = 4+3$ ✓; Eq2: $2(4)+7 = 15$ ✓. Solution: $(4, 7)$.

Worked Example — Isolate First

+5 XP

When to use: No variable is isolated yet. Solve: $x + y = 7$ and $2x - y = 5$.

Scan both equations: in $x + y = 7$, the variable $y$ has coefficient 1. Isolate $y$: $y = 7 - x$. Then substitute this expression for $y$ in the second equation. After solving for $x$, back-substitute to find $y$.

$y = 7 - x$ → $2x - (7-x) = 5$ → $3x = 12$ → $x = 4$, $y = 3$

Step 1: From $x + y = 7$, isolate $y$: $y = 7 - x$.

Step 2: Substitute into $2x - y = 5$: $2x - (7 - x) = 5$. Use brackets — the minus sign distributes!

Step 3: Expand: $2x - 7 + x = 5 \Rightarrow 3x = 12 \Rightarrow x = 4$.

Step 4: Back-substitute: $y = 7 - 4 = 3$. Verify: Eq1: $4+3=7$ ✓; Eq2: $2(4)-3=5$ ✓. Solution: $(4, 3)$.

Spot the Trap

heads-up

These are the four most common errors with substitution. Learn to spot them before they cost you marks.

Wrong: Isolating the variable with coefficient 2 or 3 introduces fractions and extra errors. Pick coefficient 1 or −1 whenever possible.

Right: Scan for a variable with coefficient 1 or −1. That gives the cleanest substitution with the fewest fractions.

Wrong: $-2(7 - 2x) = -14 - 4x$. Forgetting to distribute the negative to both terms inside brackets.

Right: $-2(7 - 2x) = -14 + 4x$. The negative multiplies everything inside.

Wrong: Stopping after finding $x$. You must back-substitute to find $y$ — the solution is always a pair $(x, y)$.

Right: Always check in both original equations. A solution must satisfy both — not just one.

Quick-fire Brain Trainer — speed drill on substitution steps. Go!

1 Substitute $y = x + 2$ into $x + y = 8$. What equation do you get?

$x + (x + 2) = 8 \Rightarrow 2x + 2 = 8$
2 $y = 2x$ and $3x + y = 15$. Find $x$.

$3x + 2x = 15 \Rightarrow 5x = 15 \Rightarrow x = 3$
3 If $x = 4$ and $y = x - 1$, what is $y$?

$y = 4 - 1 = 3$
4 Isolate $y$: $2x + y = 10$

$y = 10 - 2x$
5 Expand: $-3(2 - x)$

$-6 + 3x$
6 $y = x + 3$ and $2x + y = 18$. Solve for $x$.

$2x + (x+3) = 18 \Rightarrow 3x = 15 \Rightarrow x = 5$
7 If $x = 2$ and $y = 3x + 4$, find $y$.

$y = 3(2) + 4 = 10$
8 Solve: $y = 3x$ and $x + y = 12$

$x + 3x = 12 \Rightarrow 4x = 12 \Rightarrow x = 3$, $y = 9$. Solution: $(3, 9)$
9 Isolate $x$: $x + 3y = 9$

$x = 9 - 3y$
10 Check: Does $(2, 5)$ satisfy $y = 2x + 1$ and $3x + y = 11$?

Eq1: $5 = 2(2)+1 = 5$ ✓; Eq2: $3(2)+5 = 11$ ✓. Yes!

Complete in your workbook.

Quick Check · 5 questions

Which variable should you isolate? $y = 3x + 2$ and $2x + y = 12$

+10 XP

Substitute $y = x - 2$ into $2x + y = 10$. What equation do you get?

+10 XP

The solution has $x = 3$. If $y = 2x + 1$, what is $y$?

+10 XP

Solve: $y = 2x$ and $x + y = 9$

+10 XP

After substituting, you get $3x = 6$. What is the next step?

+10 XP

Show Your Working · 3 questions

Show Your Working

3 questions

Apply Medium 4 MARKS

Q6. Solve using substitution: $y = x + 3$ and $2x + y = 15$. Show all steps including verification.

Answer in your workbook.

Apply Medium 5 MARKS

Q7. Solve using substitution: $x + y = 7$ and $2x - y = 5$. You need to isolate a variable first.

Answer in your workbook.

Understand Easy 3 MARKS

Q8. Explain the substitution method in your own words. When is it the best method to use?

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — $y = 3x + 2$ already has $y$ isolated. Use it directly for substitution.

2. C — $2x + (x - 2) = 10$. Always use brackets when substituting.

3. D — $y = 2(3) + 1 = 7$. Multiply first, then add.

4. A — $x + 2x = 9 \Rightarrow x = 3$, $y = 6$. Check: $3 + 6 = 9$ ✓.

5. C — $3x = 6 \Rightarrow x = 2$. Then back-substitute to find $y$, verify in both.

Model Answers

Q6 (4 marks): Step 1: $y = x + 3$ already isolated [1]. Step 2: $2x + (x+3) = 15$ [1]. Step 3: $3x + 3 = 15 \Rightarrow x = 4$ [1]. Step 4: $y = 4 + 3 = 7$. Verify: $7 = 4+3$ ✓, $2(4)+7 = 15$ ✓ [1]. Solution: $(4, 7)$.

Q7 (5 marks): Step 1: Isolate $y$ from $x + y = 7$: $y = 7 - x$ [1]. Step 2: $2x - (7 - x) = 5$ [1]. Step 3: $2x - 7 + x = 5 \Rightarrow 3x = 12 \Rightarrow x = 4$ [1]. Step 4: $y = 7 - 4 = 3$ [1]. Verify: $4 + 3 = 7$ ✓, $2(4) - 3 = 5$ ✓ [1]. Solution: $(4, 3)$.

Q8 (3 marks): Substitution works by making one variable the subject of one equation, then replacing (substituting) that variable in the other equation [1]. This creates one equation with one unknown, which can be solved [1]. It is best when a variable is already isolated or has coefficient 1 or −1, to avoid fractions [1].

Stretch Challenges · +25 XP, +10 coins

Push Further

Stretch 1: Solve using substitution: $3x + 2y = 12$ and $y = x - 1$. Show all working.

Reveal solution

Substitute $y = x - 1$ into $3x + 2y = 12$: $3x + 2(x-1) = 12 \Rightarrow 5x - 2 = 12 \Rightarrow x = \frac{14}{5}$. Back-sub: $y = \frac{14}{5} - 1 = \frac{9}{5}$. Check: $3(\frac{14}{5}) + 2(\frac{9}{5}) = \frac{60}{5} = 12$ ✓.

Stretch 2: The sum of two numbers is 15 and their difference is 3. Write two equations and solve using substitution.

Reveal solution

$x + y = 15$ and $x - y = 3$. From eq 2: $x = y + 3$. Sub into eq 1: $(y+3)+y = 15 \Rightarrow 2y = 12 \Rightarrow y = 6$. Back-sub: $x = 9$. The numbers are 9 and 6. Check: $9 + 6 = 15$ ✓, $9 - 6 = 3$ ✓.

Stretch 3: A rectangle has perimeter 28 cm. The length is 4 cm more than the width. Use substitution to find the dimensions.

Reveal solution

$2L + 2W = 28$ and $L = W + 4$. Substitute: $2(W+4) + 2W = 28 \Rightarrow 4W + 8 = 28 \Rightarrow W = 5$ cm. Back-sub: $L = 9$ cm. Check: $2(9) + 2(5) = 28$ ✓, $9 = 5 + 4$ ✓.

Quick Review

The 4 steps

Isolate → Substitute → Solve → Back-substitute

Best variable to isolate

Pick coefficient 1 or −1 — fewest fractions

Use brackets!

Always wrap the substituted expression in (brackets)

Distribute the negative

$-2(7-2x) = -14 + 4x$, not $-14-4x$

Back-substitute

Find the second variable — the answer is a pair $(x,y)$

Verify in BOTH

Check the solution satisfies both original equations

Your Badges

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Substituter

Isolator

Back-Subber

Verifier

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Tick when you've finished Learn, Practice and the Stretch. Earns +85 XP and +25 coins.

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