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?
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.
- 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
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.
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 $2x + 5y = 24$ → $x = 2, y = 4$.
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.
$y = 3x + 2$ → substitute. $2x + 3y = 12$ → substitution works.
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 $\$5.50$.
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.
$y$ is already isolated in the first equation.
Substitute $y = 2x + 1$ into the second equation: $3x + 4(2x + 1) = 18$
Expand: $3x + 8x + 4 = 18$ → $11x + 4 = 18$ → $11x = 14$ → $x = \dfrac{14}{11}$ wait, let us recheck.
Let us use a cleaner example: Solve $y = 3x - 2$ and $2x + 5y = 24$.
Substitute $y = 3x - 2$ into the second equation: $2x + 5(3x - 2) = 24$
Expand: $2x + 15x - 10 = 24$ → $17x - 10 = 24$ → $17x = 34$ → $x = 2$
Back-substitute: $y = 3(2) - 2 = 6 - 2 =$ $4$
Check: $2(2) + 5(4) = 4 + 20 = 24$ ✓ and $4 = 3(2) - 2$ ✓
Let $p$ = number of pens, $c$ = number of pencils.
Total items: $p + c = 150$
Total cost: $1.20p + 0.80c = 152$
From the first equation: $c = 150 - p$. Substitute into the second.
$1.20p + 0.80(150 - p) = 152$
$1.20p + 120 - 0.80p = 152$ → $0.40p = 32$ → $p = 80$
$c = 150 - 80 =$ $70$
Check: $80 + 70 = 150$ ✓ and $80(1.20) + 70(0.80) = 96 + 56 = 152$ ✓
Plan A cost: $C_A = 30 + 0.20n$ where $n$ = number of calls
Plan B cost: $C_B = 15 + 0.50n$
At break-even, $C_A = C_B$: $30 + 0.20n = 15 + 0.50n$
$30 - 15 = 0.50n - 0.20n$ → $15 = 0.30n$ → $n = 50$ calls
For 80 calls: $C_A = 30 + 0.20(80) = 30 + 16 = \$46$
$C_B = 15 + 0.50(80) = 15 + 40 = \$55$
Since $80 > 50$, the plan with the lower per-call rate (Plan A) is cheaper.
Brain Trainer
4 quick-fire drills. Beat the clock.
5 MCQs and 3 short-answer questions. Target: 80% accuracy.
Your answer:
Substitute $y = 4x - 5$ into $3x + 2y = 23$:
$3x + 2(4x - 5) = 23$
$3x + 8x - 10 = 23$
$11x = 33$ → $x = 3$
$y = 4(3) - 5 = 12 - 5 =$ $7$
Verify: $3(3) + 2(7) = 9 + 14 = 23$ ✓ and $7 = 4(3) - 5$ ✓
Answer: $x = 3,\; y = 7$
Your 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): $1.20p + 0.80(150 - p) = 152$
$1.20p + 120 - 0.80p = 152$
$0.40p = 32$ → $p = 80$
$c = 150 - 80 =$ $70$
Check: $80 + 70 = 150$ ✓ and $80(1.20) + 70(0.80) = 96 + 56 = 152$ ✓
Answer: 80 pens and 70 pencils
Your 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.
Consolidate and reflect before moving on.
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.)
Substitution: isolate one variable, substitute into the other equation, solve, then back-substitute. Always verify both original equations.
Substituting back into the same equation you rearranged. This gives $0 = 0$, always substitute into the OTHER equation.
Word problems with two unknowns produce simultaneous equations. One equation for total quantity, one for total value.
Solve: $2x + y = 11$ and $y = x + 2$ by substitution. Time yourself, can you do it in under 60 seconds?
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