Simultaneous equations - Year 10 Mathematics

What you will learn

understand that simultaneous equations are a pair of equations with common unknowns,
solve simultaneous equations graphically by finding the point of intersection,
solve simultaneous equations algebraically by substitution,
solve simultaneous equations algebraically by elimination,
apply simultaneous equations to real-world problems including break-even analysis.

Worked example 0 Real-world example: break-even analysis

A bakery has fixed costs of $200 per day and variable costs of $1.50 per loaf. Each loaf sells for $4.00. How many loaves must be sold to break even?

Let $n$ = number of loaves. Cost: $C = 200 + 1.5n$ . Revenue: $R = 4n$ .
Break-even means $C = R$ : $200 + 1.5n = 4n$ .
$200 = 2.5n$ , so $n = 80$ .
The bakery must sell $80$ loaves per day to break even.

Key idea: break-even is the intersection of the cost and revenue lines.

1. Graphical method

Two linear equations represent two straight lines. Their point of intersection is the solution — the pair $(x, y)$ that satisfies both equations simultaneously.

Three possible outcomes:

Outcome	Lines	Solutions
One solution	Lines intersect at one point	Unique pair $(x, y)$
No solution	Lines are parallel (same gradient, different intercept)	None
Infinitely many	Lines are identical	Every point on the line

Worked example 1 Solving graphically

Solve $y = 2x + 1$ and $y = -x + 7$ by reading the graph above.

The lines cross at the point $(2, 5)$ .
Check: $y = 2(2) + 1 = 5$ and $y = -(2) + 7 = 5$ . Both equations satisfied.
Solution: $x = 2$ , $y = 5$ .

2. Substitution method

When to use: when one equation already has a variable expressed as the subject (or can easily be rearranged).

Formula reference

\text{Step 1: Express one variable in terms of the other.}

\text{Step 2: Substitute into the other equation.}

\text{Step 3: Solve the resulting single-variable equation.}

\text{Step 4: Back-substitute to find the second variable.}

Worked example 2 Substitution method

Solve $y = 3x - 1$ and $2x + y = 9$ .

The first equation gives $y = 3x - 1$ .
Substitute into the second: $2x + (3x - 1) = 9$ .
$5x - 1 = 9$ , so $5x = 10$ , $x = 2$ .
Back-substitute: $y = 3(2) - 1 = 5$ .
Solution: $x = 2$ , $y = 5$ .
Check in second equation: $2(2) + 5 = 9$ . Correct.

Worked example 3 Substitution when rearrangement is needed

Solve $3x - 2y = 7$ and $x + y = 1$ .

From the second equation: $x = 1 - y$ .
Substitute into the first: $3(1 - y) - 2y = 7$ .
$3 - 3y - 2y = 7$ , so $-5y = 4$ , $y = -\dfrac{4}{5}$ .
$x = 1 - \left(-\dfrac{4}{5}\right) = \dfrac{9}{5}$ .
Solution: $x = \dfrac{9}{5}$ , $y = -\dfrac{4}{5}$ .

3. Elimination method

When to use: when both equations are in the form $ax + by = c$ and you can make the coefficients of one variable match (or be opposites) by multiplying.

Worked example 4 Elimination by addition

Solve $3x + 2y = 16$ and $x - 2y = 0$ .

The $y$ -coefficients are $+2$ and $-2$ — they already cancel on addition.
Add the equations: $(3x + 2y) + (x - 2y) = 16 + 0$ , so $4x = 16$ , $x = 4$ .
Substitute into equation 2: $4 - 2y = 0$ , so $y = 2$ .
Solution: $x = 4$ , $y = 2$ .

Worked example 5 Elimination requiring multiplication

Solve $2x + 3y = 12$ and $5x + 2y = 11$ .

To eliminate $y$ $y$ , multiply equation 1 by $2$ $2$ and equation 2 by $3$ $3$ :
- $4x + 6y = 24$
- $15x + 6y = 33$
Subtract: $(15x + 6y) - (4x + 6y) = 33 - 24$ , so $11x = 9$ , $x = \dfrac{9}{11}$ .
Substitute into equation 1: $2\!\left(\dfrac{9}{11}\right) + 3y = 12$ , so $\dfrac{18}{11} + 3y = 12$ , $3y = \dfrac{114}{11}$ , $y = \dfrac{38}{11}$ .
Solution: $x = \dfrac{9}{11}$ , $y = \dfrac{38}{11}$ .

Worked example 6 Recognising no solution

Solve $2x + 4y = 10$ and $x + 2y = 8$ .

Multiply equation 2 by $2$ : $2x + 4y = 16$ .
But equation 1 says $2x + 4y = 10$ .
$10 \neq 16$ , so there is no solution — the lines are parallel.

4. Real-world applications

Worked example 7 Mixture problem

A chemist mixes solution A (30% acid) with solution B (70% acid) to make 200 mL of 45% acid. How much of each is needed?

Let $a$ = mL of A and $b$ = mL of B.
Total volume: $a + b = 200$ .
Acid content: $0.30a + 0.70b = 0.45 \times 200 = 90$ .
From equation 1: $a = 200 - b$ . Substitute: $0.30(200 - b) + 0.70b = 90$ .
$60 - 0.30b + 0.70b = 90$ , so $0.40b = 30$ , $b = 75$ .
$a = 200 - 75 = 125$ .
The chemist needs $125$ mL of A and $75$ mL of B.

Worked example 8 Break-even with two products

A company makes widgets and gadgets. Each widget costs $5 to produce and sells for $12. Each gadget costs $8 to produce and sells for $15. Fixed costs are $1400 per week. The company makes $w$ widgets and $g$ gadgets. If it makes twice as many widgets as gadgets and wants to break even, find $w$ and $g$ .

Profit per widget: $12 - 5 = 7$ . Profit per gadget: $15 - 8 = 7$ .
Break-even: $7w + 7g = 1400$ , i.e. $w + g = 200$ .
Constraint: $w = 2g$ .
Substitute: $2g + g = 200$ , so $3g = 200$ , $g = \dfrac{200}{3} \approx 66.7$ .
Since production must be whole numbers: $g = 67$ , $w = 134$ (with slight profit), or $g = 66$ , $w = 132$ (slight loss). The exact break-even requires $g = 66\tfrac{2}{3}$ .

Key idea: mathematical solutions sometimes need practical adjustment.

Practice

Fluency

Tier 1: basic solving

Solve by substitution: $y = x + 3$ and $y = 2x - 1$ .
Solve by substitution: $y = 4x$ and $3x + y = 14$ .
Solve by elimination: $x + y = 10$ and $x - y = 4$ .
Solve by elimination: $3x + y = 11$ and $x + y = 5$ .
Solve: $2x + y = 7$ and $x - y = 2$ .
Solve: $y = 2x + 3$ and $3x - y = -1$ .
Solve: $4x + 3y = 18$ and $2x + 3y = 12$ .
Solve: $x + 2y = 8$ and $3x + 2y = 12$ .
Write the equations for: “Two numbers add to 20 and differ by 6.” Solve them.
Write the equations for: “A pen costs $2 more than a pencil. Three pens and two pencils cost $21.” Solve them.

Reasoning

Tier 2: multi-step and applications

Solve $\dfrac{x}{2} + \dfrac{y}{3} = 4$ and $x - y = 3$ .
Solve $5x + 2y = 1$ and $3x - 4y = 11$ .
Determine whether $2x - 6y = 4$ and $x - 3y = 7$ have no solution, one solution, or infinitely many solutions.
A fruit shop sells apples at $3 per kg and bananas at $2 per kg. A customer buys 5 kg of fruit for $12. How many kg of each?
Two cars leave the same point. Car A travels north at $80$ km/h and car B travels north at $60$ km/h but left $1$ hour earlier. When and where does car A overtake car B?
The perimeter of a rectangle is $36$ cm and the length is $4$ cm more than the width. Find the dimensions.
Solve $2x + 3y = 1$ and $4x + 6y = 2$ . How many solutions are there? Explain.
A test has $30$ questions. Correct answers score $4$ marks; wrong answers lose $1$ mark. A student scores $75$ . How many correct answers?

Reasoning

Tier 3: extended problems

Three friends buy cinema tickets and snacks. Use the information below to set up and solve simultaneous equations: 2 tickets and 1 snack cost $35; 1 ticket and 2 snacks cost $25. Find the cost of a ticket and a snack.
A boat travels $30$ km upstream in $3$ hours and $30$ km downstream in $2$ hours. Find the speed of the boat in still water and the speed of the current.
A company’s cost function is $C = 500 + 8n$ and revenue function is $R = 20n - 0.05n^2$ . Find the break-even point(s).
The line $ax + by = 1$ passes through $(2, 1)$ and $(4, -1)$ . Find $a$ and $b$ , then write the equation in the form $y = mx + c$ .

Challenge

Reasoning

Harder reasoning

A two-digit number has digits that add to $9$ . If the digits are reversed, the new number is $27$ less than the original. Find the number. (Hint: let the tens digit be $t$ and units digit be $u$ , so the number is $10t + u$ .)
Solve the system $\dfrac{2}{x} + \dfrac{3}{y} = 7$ and $\dfrac{5}{x} - \dfrac{1}{y} = 9$ by letting $u = \dfrac{1}{x}$ and $v = \dfrac{1}{y}$ .
Find the equation of the line passing through the intersection of $2x + y = 5$ and $x - y = 1$ that also passes through $(0, 4)$ .
A shop sells two sizes of coffee. On Monday, 40 small and 25 large coffees earned $285. On Tuesday, 30 small and 35 large coffees earned $295. Find the price of each size, then find the day’s revenue if 50 small and 50 large are sold.

Interactive

Try it yourself: two lines, one solution

Work through 3 examples. Drag the blue or green points to reposition each line.

Example 1 (easy). Drag either line so the two lines cross at (2, 1). The red dot shows their current intersection.

Line A slope: 0.67
Line B slope: -0.67
Intersection: (0.00, 1.00)

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 10 core - answers

Fluency

Tier 1: basic solving

$x + 3 = 2x - 1$ , $x = 4$ , $y = 7$ .
$3x + 4x = 14$ , $7x = 14$ , $x = 2$ , $y = 8$ .
Add: $2x = 14$ , $x = 7$ , $y = 3$ .
Subtract: $2x = 6$ , $x = 3$ , $y = 2$ .
Add: $3x = 9$ , $x = 3$ , $y = 1$ .
Substitute: $3x - (2x + 3) = -1$ , $x - 3 = -1$ , $x = 2$ , $y = 7$ .
Subtract: $2x = 6$ , $x = 3$ , $y = 2$ .
Subtract: $2x = 4$ , $x = 2$ , $y = 3$ .
$x + y = 20$ and $x - y = 6$ . Add: $2x = 26$ , $x = 13$ , $y = 7$ .
Let pencil $= p$ , pen $= p + 2$ . $3(p + 2) + 2p = 21$ , $5p + 6 = 21$ , $p = 3$ . Pencil $3, pen $5.

Reasoning

Tier 2: multi-step and applications

From eq 2: $x = y + 3$ . Substitute: $\dfrac{y+3}{2} + \dfrac{y}{3} = 4$ . Multiply by $6$ : $3(y+3) + 2y = 24$ , $5y + 9 = 24$ , $y = 3$ , $x = 6$ .
Multiply eq 1 by $2$ : $10x + 4y = 2$ . Add to eq 2: $13x = 13$ , $x = 1$ , $y = -2$ .
Multiply eq 2 by $2$ : $2x - 6y = 14$ . But eq 1 is $2x - 6y = 4$ . Since $14 \neq 4$ , there is no solution (parallel lines).
$a + b = 5$ and $3a + 2b = 12$ . From eq 1: $b = 5 - a$ . Substitute: $3a + 2(5-a) = 12$ , $a + 10 = 12$ , $a = 2$ kg apples, $b = 3$ kg bananas.
Car B position: $60t$ (where $t$ is hours after B left). Car A position: $80(t - 1)$ (A left 1 hour later). Overtake: $80(t-1) = 60t$ , $80t - 80 = 60t$ , $20t = 80$ , $t = 4$ hours after B left. Position: $60 \times 4 = 240$ km from start.
$2l + 2w = 36$ and $l = w + 4$ . Substitute: $2(w+4) + 2w = 36$ , $4w + 8 = 36$ , $w = 7$ cm, $l = 11$ cm.
Eq 2 is exactly $2 \times$ eq 1: $4x + 6y = 2(2x + 3y) = 2(1) = 2$ . The equations are identical, so there are infinitely many solutions.
Let $c$ = correct, $w$ = wrong. $c + w = 30$ and $4c - w = 75$ . Add: $5c = 105$ , $c = 21$ correct answers.

Reasoning

Tier 3: extended problems

$2t + s = 35$ and $t + 2s = 25$ . Multiply eq 2 by $2$ : $2t + 4s = 50$ . Subtract eq 1: $3s = 15$ , $s = 5$ . $t = \dfrac{35 - 5}{2} = 15$ . Ticket $15, snack $5.
Let boat speed $= b$ , current $= c$ . Upstream: $\dfrac{30}{b - c} = 3$ , so $b - c = 10$ . Downstream: $\dfrac{30}{b + c} = 2$ , so $b + c = 15$ . Add: $2b = 25$ , $b = 12.5$ km/h, $c = 2.5$ km/h.
$500 + 8n = 20n - 0.05n^2$ . $0.05n^2 - 12n + 500 = 0$ . $n^2 - 240n + 10\,000 = 0$ . $n = \dfrac{240 \pm \sqrt{57\,600 - 40\,000}}{2} = \dfrac{240 \pm \sqrt{17\,600}}{2} = \dfrac{240 \pm 40\sqrt{11}}{2} = 120 \pm 20\sqrt{11}$ . So $n \approx 120 - 66.3 = 53.7$ or $n \approx 120 + 66.3 = 186.3$ . Break-even at approximately $54$ and $186$ units.
Through $(2,1)$ : $2a + b = 1$ . Through $(4,-1)$ : $4a - b = 1$ . Add: $6a = 2$ , $a = \dfrac{1}{3}$ , $b = 1 - \dfrac{2}{3} = \dfrac{1}{3}$ . Equation: $\dfrac{x}{3} + \dfrac{y}{3} = 1$ , so $x + y = 3$ , $y = -x + 3$ .

Reasoning

Challenge

Number $= 10t + u$ . $t + u = 9$ and $10t + u - (10u + t) = 27$ , so $9t - 9u = 27$ , $t - u = 3$ . Add: $2t = 12$ , $t = 6$ , $u = 3$ . The number is $63$ .
Let $u = \dfrac{1}{x}$ , $v = \dfrac{1}{y}$ . $2u + 3v = 7$ and $5u - v = 9$ . From eq 2: $v = 5u - 9$ . Substitute: $2u + 3(5u - 9) = 7$ , $17u - 27 = 7$ , $u = 2$ , $v = 1$ . So $x = \dfrac{1}{2}$ , $y = 1$ .
Solve $2x + y = 5$ and $x - y = 1$ : add, $3x = 6$ , $x = 2$ , $y = 1$ . The intersection is $(2, 1)$ . Line through $(2, 1)$ and $(0, 4)$ : gradient $= \dfrac{4 - 1}{0 - 2} = -\dfrac{3}{2}$ . Equation: $y = -\dfrac{3}{2}x + 4$ .
$40s + 25l = 285$ and $30s + 35l = 295$ . Multiply eq 1 by $7$ and eq 2 by $5$ : $280s + 175l = 1995$ and $150s + 175l = 1475$ . Subtract: $130s = 520$ , $s = 4$ . $40(4) + 25l = 285$ , $25l = 125$ , $l = 5$ . Small $4, large $5. Revenue for 50 of each: $50(4) + 50(5) = 200 + 250 = 450$ .

Prefer paper? Print the answer key as a separate booklet: open print view ->