Lesson 19 ~30 min Unit 2 · Linear Relationships +90 XP

Simultaneous Equations — Elimination Method

Add or subtract equations to make one variable disappear. If coefficients do not match, multiply first. A powerful technique for equations in standard form.

Today's hook: If $x + y = 10$ and $x - y = 4$, what happens if you add the two equations? The $y$ terms cancel: $2x = 14$, so $x = 7$. One variable vanished in a single step — that is the magic of elimination!

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Think First

warm-up

Before you read on — if $x + y = 10$ and $x - y = 4$, what happens when you add the two equations? What happens when you subtract them? Try both before continuing.

Adding: $2x = 14 \Rightarrow x = 7$. Subtracting: $2y = 6 \Rightarrow y = 3$. One variable disappeared each time — that is elimination!

The Big Idea — Elimination

+5 XP

Elimination solves simultaneous equations by adding or subtracting them so one variable cancels to zero. If the coefficients of one variable are opposite in sign (e.g., $+3y$ and $-3y$), add. If they are the same sign, subtract. If they do not match yet, multiply one or both equations first.

The method relies on one principle: if $A = B$ and $C = D$, then $A + C = B + D$ and $A - C = B - D$. Both equations are true, so combining them gives a new true equation. By arranging the combination so one variable becomes zero, we reduce to one equation with one unknown.

Opposite signs

$+3y$ and $-3y$ → ADD the equations.

Same signs

$+2x$ and $+2x$ → SUBTRACT the equations.

Don't match?

Multiply one or both equations first.

Opposite signs → Add · Same signs → Subtract · Don't match → Multiply first

What You'll Master

objectives

Know

Elimination means adding or subtracting equations to remove one variable
If coefficients match, add or subtract directly
If coefficients do not match, multiply one or both equations first

Understand

Why adding equals to equals is mathematically valid
When to add vs subtract: same signs subtract, opposite signs add
Why elimination is often faster than substitution when both equations are in standard form

Can Do

Solve simultaneous equations by elimination (add, subtract, or multiply first)
Decide whether to add or subtract to eliminate a variable
Multiply equations to make coefficients match before eliminating
Check solutions by substituting into both original equations

Words You Need

vocabulary

EliminationA method of solving simultaneous equations by adding or subtracting to remove one variable.

Add equationsCombine two equations by adding their left and right sides; used when coefficients have opposite signs.

Subtract equationsCombine two equations by subtracting one from the other; used when coefficients have the same sign.

Match coefficientsMake the coefficients of one variable equal (or opposite) in both equations so elimination works.

Multiply equationMultiply every term in an equation by the same number to create matching coefficients.

Cancel variableWhen adding or subtracting causes a variable term to become zero, removing it from the equation.

Elimination by Addition and Subtraction

+5 XP

When coefficients already cancel, one operation eliminates a variable directly. Use the same-sign/opposite-sign rule to decide.

By addition: $x + y = 7$ and $x - y = 3$. The $y$ terms are $+y$ and $-y$ (opposite signs). Adding cancels $y$: $2x = 10$, $x = 5$. Back-substitute: $y = 2$.

By subtraction: $2x + 3y = 13$ and $2x - y = 5$. The $x$ terms are both $+2x$ (same sign). Subtracting cancels $x$: $4y = 8$, $y = 2$. Back-substitute: $x = 3.5$.

Same signs → subtract · Opposite signs → add

Addition example: $x + y = 7$ and $x - y = 3$. Add: $2x = 10$, $x = 5$. Sub back: $5 + y = 7$, $y = 2$. Verify: $5+2=7$ ✓, $5-2=3$ ✓.

Subtraction example: $2x + 3y = 13$ and $2x - y = 5$. Subtract: $(3y) - (-y) = 8$, so $4y = 8$, $y = 2$. Sub back: $2x - 2 = 5$, $x = \frac{7}{2}$.

Common error: For same-sign coefficients, adding gives $4x$ instead of $0$ — nothing cancels! Always check the result after your operation.

Multiplying to Match Coefficients

+5 XP

When no variable has matching coefficients yet, multiply one (or both) equations to create a match.

Multiply one equation: $3x + 2y = 12$ and $x - y = 1$. The $y$ terms are $+2y$ and $-y$. Multiply the second by 2 to get $-2y$. Then add the equations to cancel $y$.

Multiply both equations: $2x + 3y = 17$ and $3x + 2y = 13$. Neither pair of coefficients matches. Multiply (1) by 2 and (2) by 3 to make both $y$ coefficients equal to 6, then subtract.

Multiply (2) by 2 → $y$ coefficients become $+2y$ and $-2y$ → Add to cancel

One multiply: $3x+2y=12$ and $x-y=1$. Multiply (2) by 2: $2x-2y=2$. Now add: $5x=14$, $x=\frac{14}{5}$. Back-sub: $y=\frac{9}{5}$. Check: $\frac{42}{5}+\frac{18}{5}=12$ ✓, $\frac{14}{5}-\frac{9}{5}=1$ ✓.

Both multiply: $2x+3y=17$ and $3x+2y=13$. Multiply (1) by 2: $4x+6y=34$. Multiply (2) by 3: $9x+6y=39$. Subtract: $5x=5$, $x=1$. Back-sub: $y=5$. Check: $2+15=17$ ✓, $3+10=13$ ✓.

Critical rule: When multiplying an equation, every term on both sides must be multiplied. $3 \times (x+2y=5)$ gives $3x+6y=15$ — not $3x+2y=5$!

Elimination vs Substitution + Brain Trainer

strategy

Which method is best? Here is a quick guide to decide:

Situation	Best Method	Why
One variable has coefficient 1 or −1	Substitution	Easy to rearrange and substitute
Coefficients already match	Elimination (add/subtract)	One step to cancel a variable
All coefficients are different	Elimination (multiply first)	Multiply to match, then eliminate
Both equations in $ax+by=c$ form	Elimination	Line them up and operate directly

Brain Trainer — elimination speed drill. Go!

1 $x + y = 9$, $x - y = 3$ — solve by elimination

Add: $2x = 12$, $x = 6$. Back-sub: $y = 3$. Solution: $(6, 3)$
2 $2x + y = 11$, $2x - y = 5$ — add or subtract?

Add (opposite $y$ signs): $4x = 16$, $x = 4$. Back-sub: $y = 3$. Solution: $(4, 3)$
3 $3x + 2y = 14$, $3x - y = 8$ — same $x$ coefficients

Subtract: $3y = 6$, $y = 2$. Back-sub: $3x = 10$, $x = \frac{10}{3}$
4 $x + 2y = 8$, $x + y = 5$ — subtract

Subtract: $y = 3$. Back-sub: $x = 2$. Solution: $(2, 3)$
5 $5x + 3y = 22$, $2x + 3y = 13$ — same $3y$ coefficient

Subtract: $3x = 9$, $x = 3$. Back-sub: $y = \frac{7}{3}$
6 $4x - 3y = 6$, $2x + 3y = 12$ — opposite $3y$ signs

Add: $6x = 18$, $x = 3$. Back-sub: $y = 2$. Solution: $(3, 2)$
7 $3x + 2y = 17$, $2x + 3y = 13$ — multiply both

×3 and ×2: $9x+6y=51$ and $4x+6y=26$. Subtract: $5x=25$, $x=5$, $y=1$
8 $x + 4y = 14$, $3x - 2y = 0$ — multiply (2) by 2

×2 on (2): $6x-4y=0$. Add: $7x=14$, $x=2$. Back-sub: $y=3$
9 Identify: $2x + 3y = 11$ and $4x - y = 3$. Best operation?

Multiply (1) by 2 to get $4x+6y=22$. Subtract (2) from this: $7y=19$, $y=\frac{19}{7}$
10 Which equation pair would you solve faster with substitution vs elimination? $y = 3x$ and $x + y = 8$

Substitution (y already isolated): $x + 3x = 8$, $x = 2$, $y = 6$. Elimination also works but substitution is faster here.

Complete in your workbook.

Quick Check · 5 questions

Add or subtract to eliminate from $2x + y = 8$ and $3x - y = 7$?

+10 XP

To eliminate $x$ from $2x + 3y = 11$ and $4x - y = 9$, which operation?

+10 XP

Solve: $x + y = 6$ and $x - y = 2$

+10 XP

After eliminating $y$, you find $5x = 15$, so $x = 3$. If $x + y = 7$, what is $y$?

+10 XP

Which method is best for $2x + 3y = 12$ and $5x - 3y = 9$?

+10 XP

Show Your Working · 3 questions

Show Your Working

3 questions

Apply Easy 4 MARKS

Q6. Solve by elimination: $x + y = 8$ and $2x - y = 7$. Show all steps including verification.

Answer in your workbook.

Apply Medium 5 MARKS

Q7. Solve by elimination: $3x + 2y = 14$ and $x + 2y = 10$. (Hint: the $2y$ terms are the same — try subtracting.)

Answer in your workbook.

Apply Hard 6 MARKS

Q8. Solve by elimination: $2x + 3y = 19$ and $3x - 2y = 9$. (Hint: multiply both equations to make $y$ coefficients 6 and −6.)

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — Add. The $y$ terms $+y$ and $-y$ have opposite signs; adding cancels them.

2. D — Multiply (1) by 2 to get $4x+6y=22$, then subtract. $x$ terms cancel.

3. C — $x=4$, $y=2$. Adding gives $2x=8$, $x=4$; back-sub $y=2$.

4. B — $y=4$. Substitute $x=3$ into $x+y=7$: $3+y=7 \Rightarrow y=4$.

5. A — Elimination by addition. $+3y$ and $-3y$ are opposites; adding cancels $y$ immediately.

Model Answers

Q6 (4 marks): $y$ terms $+y$ and $-y$ → add [1]. $3x = 15$, $x = 5$ [1]. Back-sub: $y = 3$ [1]. Verify: $5+3=8$ ✓, $2(5)-3=7$ ✓ [1]. Solution: $(5, 3)$.

Q7 (5 marks): $2y$ terms same sign → subtract [1]. $(3x+2y)-(x+2y)=4 \Rightarrow 2x=4 \Rightarrow x=2$ [2]. Back-sub: $2+2y=10 \Rightarrow y=4$ [1]. Verify: $3(2)+2(4)=14$ ✓, $2+2(4)=10$ ✓ [1]. Solution: $(2, 4)$.

Q8 (6 marks): Multiply (1)×2: $4x+6y=38$ [1]. Multiply (2)×3: $9x-6y=27$ [1]. Add: $13x=65 \Rightarrow x=5$ [2]. Back-sub: $2(5)+3y=19 \Rightarrow y=3$ [1]. Verify: $10+9=19$ ✓, $15-6=9$ ✓ [1]. Solution: $(5, 3)$.

Stretch Challenges · +25 XP, +10 coins

Elimination Challenge Set

Stretch 1: Solve: $4x + 5y = 23$ and $3x - 2y = 4$. (Hint: multiply to make $y$ coefficients 10 and −10.)

Reveal solution

(1)×2: $8x+10y=46$; (2)×5: $15x-10y=20$. Add: $23x=66$, $x=\frac{66}{23}$. Back-sub: $y=\frac{61}{23}$.

Stretch 2: Solve: $5x + 2y = 16$ and $3x + 4y = 18$. (Hint: multiply (1) by 2 to match $y$ coefficients, then subtract.)

Reveal solution

(1)×2: $10x+4y=32$. Subtract (2): $7x=14$, $x=2$. Back-sub: $3(2)+4y=18 \Rightarrow y=3$. Check: $5(2)+2(3)=16$ ✓, $3(2)+4(3)=18$ ✓.

Stretch 3: A coffee and a muffin cost $8 together. Two coffees and three muffins cost $19. Use elimination to find the price of each.

Reveal solution

$c + m = 8$ and $2c + 3m = 19$. Multiply (1) by 2: $2c+2m=16$. Subtract from (2): $m=3$. Back-sub: $c=5$. Coffee = $5, muffin = $3. Check: $5+3=8$ ✓, $2(5)+3(3)=19$ ✓.

Stretch 4: A rectangular paddock satisfies $3x + 2y = 36$ and $2x - y = 6$. Find $x$ and $y$.

Reveal solution

Multiply (2) by 2: $4x-2y=12$. Add to (1): $7x=48$, $x=\frac{48}{7}$. Back-sub: $y=\frac{54}{7}$.

Quick Review

The principle

If $A = B$ and $C = D$, then $A \pm C = B \pm D$

Opposite signs

$+3y$ and $-3y$ → ADD to cancel

Same signs

$+2x$ and $+2x$ → SUBTRACT to cancel

Don't match?

Multiply one or both equations first

Multiply all terms

Every term on both sides must be multiplied

Always verify

Check solution in both original equations

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