Solve pairs of equations by substituting one into the other, then verify your solution in both equations.
40 min 8 cards 130 XP
Think First
warm-up
Apples cost $2 each and bananas cost $1 each. If you buy 3 apples and 2 bananas for $8, is there only one possible combination? What if you know you bought 5 pieces of fruit in total for $8 — can you work out exactly how many of each?
Record your answer in your workbook.
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The Big Idea
+5 XP to read
A simultaneous equation is a pair of equations with two unknowns. The solution is the pair of values that satisfies both equations at the same time. The substitution method works by expressing one variable from one equation, then substituting that expression into the other equation to get a single equation with one unknown.
Two equations, two unknowns. The solution is where both are true.
one point, both true
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Lesson Objectives
what you will learn
Recognise when a problem requires simultaneous equations
Solve simultaneous equations using the substitution method
Express one variable in terms of the other
Back-substitute to find the second variable
Verify solutions by checking both original equations
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Key Vocabulary
terms to know
Simultaneous equationsTwo or more equations that must be true at the same time, sharing the same variables.
SubstitutionReplacing a variable with an equivalent expression from another equation.
Back-substitutionPlugging a found value back into an equation to find the remaining variable.
Unique solutionOne pair of values that satisfies both equations — represented by two lines crossing.
Break-even pointThe point where two cost models give the same total price.
VerifyTo check a solution by substituting into the original equations.
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Spot the Trap
heads-up
Wrong: Substituting back into the same equation you started with. If you rearranged Equation 1 to get $y = \dots$, then substitute into Equation 1, you get $0 = 0$.
Right: Always substitute into the other equation — the one you did not rearrange.
Wrong: Forgetting to find the second variable after finding the first.
Right: After finding one variable, always back-substitute to find the other. A solution needs both values.
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The Substitution Method
+5 XP
The substitution method has four steps: isolate one variable in one equation, substitute that expression into the other equation, solve for the remaining variable, then back-substitute to find the first variable. Always verify by checking both original equations.
$y = 3x - 2$ into $2x + 5y = 24$ → $x = 2, y = 4$.
substitute into the other
Pick the easy one
Choose the equation where a variable has coefficient 1 or is already isolated.
Substitute into the OTHER
Never substitute back into the same equation you rearranged. Use the other one.
Verify both equations
A solution only counts if it satisfies BOTH original equations. Check both.
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When to Use Substitution
+5 XP
Substitution is easiest when one equation already has a variable isolated (like $y = 3x + 2$) or when one coefficient is 1 (making isolation trivial). If both equations are in standard form with no coefficient of 1, elimination might be faster. Choose the method that minimises fractions and arithmetic errors.
If x or y has no number in front of it, isolating that variable is trivial.
Avoid fractions if possible
If isolating gives fractions, consider elimination instead.
Both methods work
Substitution and elimination always give the same answer. Choose the cleaner path.
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Setting Up from Words
+5 XP
Word problems with two unknowns naturally produce simultaneous equations. One sentence usually gives the total quantity, another gives the total value. Define two variables, write two equations, then solve.
80 coffees, $400 total. Regular $4.50, Large $5.50.
quantity and value
Two unknowns, two facts
If the problem mentions two things you do not know, you need two equations.
Label clearly
Let r = regular coffees, l = large coffees. Write this explicitly.
Check with context
40 regular and 40 large coffees at $4.50 and $5.50: 40(4.50) + 40(5.50) = 180 + 220 = 400. Correct.
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Comparing Plans
+5 XP
When comparing two pricing plans, the break-even point is where both plans cost the same. Set the cost equations equal and solve. Below the break-even, the plan with the lower base cost wins. Above it, the plan with the lower per-unit rate wins.
Plan A: $30 + $0.20n. Plan B: $15 + $0.50n. Break-even at n = 50.
set costs equal
Set costs equal
To find break-even, write both cost formulas and set them equal to each other.
Compare above and below
Test a value on each side of break-even to see which plan is cheaper.
Justify with numbers
When asked which plan is better, calculate the actual cost for the given usage.
Worked Example 1 — Substitution
Q.Solve the simultaneous equations: $y = 2x + 1$ and $3x + 4y = 18$.
Solution
1
$y$ is already isolated in the first equation.
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Substitute $y = 2x + 1$ into the second equation: $3x + 4(2x + 1) = 18$
Q.A shop sells pens at $1.20 each and pencils at $0.80 each. A school buys 150 items in total and spends $152. How many pens and how many pencils did they buy?
Solution
1
Let $p$ = number of pens, $c$ = number of pencils.
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Total items: $p + c = 150$
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Total cost: $1.20p + 0.80c = 152$
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From the first equation: $c = 150 - p$. Substitute into the second.
Q.Phone Plan A costs $30 per month plus $0.20 per call. Plan B costs $15 per month plus $0.50 per call. Find the break-even point and determine which plan is cheaper for 80 calls per month.
Solution
1
Plan A cost: $C_A = 30 + 0.20n$ where $n$ = number of calls
Not quite. Substitute the expression for $y$ into the second equation, expand, and solve for $x$ first. Then back-substitute.
2.Which equation pair is best solved by substitution?
Correct! When one variable is already isolated (like $y = 5x + 2$), substitution is the natural choice.
Think about which pair has a variable already isolated or with coefficient 1. That is when substitution shines.
3.A cafe sells regular coffees for $4.50 and large for $5.50. They sell 80 coffees for $400 total. How many large coffees?
Correct! Let $r$ = regular, $l$ = large. $r + l = 80$ and $4.5r + 5.5l = 400$. Substitute $r = 80 - l$ to get $l = 40$.
Set up two equations: one for total count, one for total value. Use substitution with $r = 80 - l$.
4.What is the most common error when using substitution?
Correct! If you rearrange Equation 1 to get $y = \dots$, then substitute that into Equation 1, you get $0 = 0$ — a useless identity. Always substitute into the OTHER equation.
Think about what happens if you rearrange one equation and then substitute that expression back into itself.
5.Plan A: $25 + $0.40n. Plan B: $10 + $0.70n. When is Plan A cheaper than Plan B?
Correct! Break-even: $25 + 0.40n = 10 + 0.70n$ gives $15 = 0.30n$, so $n = 50$. Plan A has a higher base cost but lower per-call rate, so it becomes cheaper when you make many calls. Plan A is cheaper for more than 50 calls.
Set the costs equal to find break-even. Then test values on each side to see which plan wins.
6.Solve by substitution: $y = 4x - 5$ and $3x + 2y = 23$
7.A school buys 150 pens and pencils in total. Pens cost $1.20 each and pencils cost $0.80 each. The total cost is $152. Set up simultaneous equations and solve to find how many pens and how many pencils were bought.
Your answer:
Variables defined clearly
Two correct equations set up
Correct substitution method used
80 pens, 70 pencils
Model Answer
Let $p$ = pens, $c$ = pencils.
$p + c = 150$ (total items)
$1.20p + 0.80c = 152$ (total cost)
From (1): $c = 150 - p$
Substitute into (2): $1.20p + 0.80(150 - p) = 152$
8.Phone Plan A costs $30 per month plus $0.20 per call. Plan B costs $15 per month plus $0.50 per call. Find the break-even number of calls and determine which plan is cheaper for 80 calls per month. Justify your answer with calculations.
Your answer:
Two cost equations written
Break-even found correctly (n = 50)
Costs calculated at 80 calls
Correct plan identified with justification
Model Answer
Plan A: $C_A = 30 + 0.20n$
Plan B: $C_B = 15 + 0.50n$
At break-even: $30 + 0.20n = 15 + 0.50n$
$15 = 0.30n$ → $n = 50$ calls
At 80 calls:
$C_A = 30 + 0.20(80) = 30 + 16 = \$46$
$C_B = 15 + 0.50(80) = 15 + 40 = \$55$
Since $80 > 50$, we are past break-even. Plan A has the lower per-call rate, so Plan A is cheaper at 80 calls ($46 vs $55).
Answer: Break-even at 50 calls. Plan A is cheaper for 80 calls.
Review
Consolidate and reflect before moving on.
Stretch Challenge
A rectangle has perimeter 34 cm. If the length were 3 cm longer and the width 2 cm shorter, the perimeter would still be 34 cm. Find the original dimensions of the rectangle. (Hint: write two equations in $l$ and $w$, then solve.)
Key Idea
Substitution: isolate one variable, substitute into the other equation, solve, then back-substitute. Always verify both original equations.
Common Trap
Substituting back into the same equation you rearranged. This gives $0 = 0$ — always substitute into the OTHER equation.
Connection
Word problems with two unknowns produce simultaneous equations. One equation for total quantity, one for total value.
Interactive: Substitution Step Builder — drag and drop the steps of substitution into the correct order.
Substitution Solver
Word Problem Warrior
Plan Comparator
Daily Challenge
Solve: $2x + y = 11$ and $y = x + 2$ by substitution. Time yourself — can you do it in under 60 seconds?
Printable Worksheets
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