Before we begin: If $2x + 3y = 13$ and $2x - y = 5$, notice that both equations have $2x$. What would happen if you subtracted one equation from the other? Try to work out what $y$ would be.
Come back to this after you have worked through the lesson.
Elimination is powerful when both equations are in standard form ($ax + by = c$) and neither variable has a coefficient of 1. By making the coefficients of one variable match, you can add or subtract the equations to eliminate that variable entirely. This leaves a single equation with one unknown that you can solve directly.
- Identify when coefficients already match for elimination.
- Multiply one or both equations to create matching coefficients.
- Correctly add or subtract equations to eliminate a variable.
- Choose the most efficient method (substitution or elimination) for a given pair of equations.
Wrong: Adding equations when coefficients are equal. In $3x + 2y = 12$ and $3x + 5y = 21$, adding gives $6x + 7y = 33$, which does not eliminate anything.
Right: When coefficients are equal, subtract the equations: $(3x + 5y) - (3x + 2y) = 21 - 12$ gives $3y = 9$.
Wrong: Forgetting to multiply every term when scaling an equation. Multiplying $2x + 3y = 7$ by 2 to get $4x + 6y = 7$.
Right: Multiply every term: $2 \times (2x + 3y = 7)$ becomes $4x + 6y = 14$.
The elimination method has six steps: line up both equations with $x$ terms, $y$ terms, and constants aligned. Match coefficients for one variable by multiplying one or both equations. Add or subtract the equations to eliminate that variable. Solve the resulting equation. Substitute back to find the other variable. Check both values in both original equations.
Same sign → subtract. Opposite signs → add. Then solve and check.
The decision to add or subtract depends on the signs of the matching coefficients. If the coefficients have the same sign (both positive or both negative), subtract the equations. If they have opposite signs, add the equations. The goal is always to make the target variable disappear.
+5x and +5x → subtract. +4y and -4y → add. Gone.
When no coefficients already match, multiply one or both equations by numbers that create matching coefficients. Use the LCM (lowest common multiple) of the coefficients to decide what to multiply by. Remember to multiply every term on both sides of the equation.
2x and 3x → LCM is 6. Eq 1 x3, Eq 2 x2. Then subtract.
Both methods solve the same problems, but one is usually more efficient depending on the form of the equations. Use substitution when a variable is already isolated or has coefficient 1. Use elimination when both equations are in standard form with no easy isolation.
Isolated variable → substitution. Standard form → elimination. Both work.
Identify: The $x$ coefficients are equal (both 5). Same sign → subtract.
Subtract Equation 2 from Equation 1: $(5x + 3y) - (5x - 2y) = 23 - 8$
$5y = 15$ → $y = 3$
Substitute back: $5x + 3(3) = 23$ → $5x = 14$ → $x = \dfrac{14}{5} = 2.8$
Check: Equation 2: $5(2.8) - 2(3) = 14 - 6 = 8$ ✓
Identify: The $y$ coefficients are opposites ($+4$ and $-4$). Opposite signs → add.
Add the equations: $(3x + 4y) + (2x - 4y) = 18 + 2$
$5x = 20$ → $x = 4$
Substitute back: $3(4) + 4y = 18$ → $12 + 4y = 18$ → $4y = 6$ → $y = 1.5$
Check: Equation 2: $2(4) - 4(1.5) = 8 - 6 = 2$ ✓
Strategy: Eliminate $x$. LCM of 2 and 3 is 6. Multiply Eq 1 by 3, Eq 2 by 2.
Multiply: $\begin{cases} 6x + 9y = 42 \\ 6x + 10y = 44 \end{cases}$
Subtract: $(6x + 10y) - (6x + 9y) = 44 - 42$ → $y = 2$
Back-substitute: $2x + 3(2) = 14$ → $2x = 8$ → $x = 4$
Check: Eq 2: $3(4) + 5(2) = 12 + 10 = 22$ ✓
Brain Trainer
4 quick-fire drills. Beat the clock.
5 MCQs and 3 short-answer questions. Target: 80% accuracy.
Your answer:
The $y$ coefficients are equal (both 2). Same sign → subtract Equation 2 from Equation 1.
$(5x + 2y) - (3x + 2y) = 24 - 16$
$2x = 8$ → $x = 4$
Substitute back: $3(4) + 2y = 16$ → $12 + 2y = 16$ → $2y = 4$ → $y = 2$
Check Eq 1: $5(4) + 2(2) = 20 + 4 = 24$ ✓. Check Eq 2: $3(4) + 2(2) = 12 + 4 = 16$ ✓.
Answer: $x = 4,\; y = 2$
Your answer:
Eliminate $x$. LCM of 2 and 3 is 6.
Equation 1 $\times$ 3: $6x + 9y = 51$
Equation 2 $\times$ 2: $6x + 10y = 56$
Subtract: $(6x + 10y) - (6x + 9y) = 56 - 51$ → $y = 5$
Back-substitute: $2x + 3(5) = 17$ → $2x = 2$ → $x = 1$
Check: $3(1) + 5(5) = 3 + 25 = 28$ ✓.
Answer: $x = 1,\; y = 5$
Your answer:
(a) Let $l$ = length (m), $w$ = width (m).
Perimeter: $2l + 2w = 240$
Length-width: $l = w + 20$ or $l - w = 20$
(b) Simplify perimeter: divide by 2 → $l + w = 120$
Now solve: $\begin{cases} l + w = 120 \\ l - w = 20 \end{cases}$
Add: $2l = 140$ → $l = 70$ m
$w = 120 - 70 =$ $50$ m
(c) Area $= 70 \times 50 =$ $3500$ m²
Verify perimeter: $2(70) + 2(50) = 140 + 100 = 240$ m ✓
Answer: Length 70 m, width 50 m, area 3500 m²
Consolidate and reflect before moving on.
Solve by elimination: $$\begin{cases} 3x + 4y = 26 \\ 5x - 2y = 18 \end{cases}$$. You will need to multiply one equation before eliminating. Show every step clearly. (Hint: which variable needs the smaller multiplier?)
Elimination: line up equations, match coefficients, add or subtract to eliminate a variable, solve, back-substitute, and verify both equations.
Adding equations when coefficients are equal (instead of subtracting). Same sign always means subtract. Opposite signs always means add.
Substitution and elimination solve the same problems. Substitution wins when a variable is isolated; elimination wins when both equations are in standard form.
Solve: $4x + 3y = 23$ and $4x - y = 11$ by elimination. Time yourself, can you do it in under 60 seconds?
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