Linear Relationships in Context
The final lesson of Unit 2. Bring together gradient, equations, sketching, and simultaneous equations to solve real-world problems. Every concept connects back to $y = mx + c$.
Printable Worksheets
Print or save as PDF — or build a custom worksheet from any module's questions.
Before you read on — A rectangle has perimeter 24 cm and the length is twice the width. Set up two equations and find the dimensions. Think about which concept to use.
Every concept in Unit 2 builds toward solving real-world problems. The thread connecting them all is $y = mx + c$.
Unit 2 told a story: we started with the Cartesian plane and coordinate pairs, built up to gradient and the equation of a line, learned to sketch lines by three methods, and finally solved simultaneous equations by graphing, substitution, and elimination. This lesson shows how it all connects in context.
Know
- All key concepts from Unit 2 and how they connect
- How to choose the right method for a given problem
Understand
- How gradient, equation, sketching, and simultaneous equations are all aspects of the same big idea: $y = mx + c$
- When to use substitution vs elimination vs graphing
Can Do
- Solve mixed problems involving any concept from the unit
- Set up and solve word problems using simultaneous equations
- Choose the appropriate method for each problem type
You can find the gradient from any of three sources, then use it to build the equation $y = mx + c$.
| Given | Method | Formula |
|---|---|---|
| Two points $(x_1, y_1)$ and $(x_2, y_2)$ | Gradient formula | $m = \frac{y_2 - y_1}{x_2 - x_1}$ |
| Graph of the line | Read rise and run | $m = \frac{\text{rise}}{\text{run}}$ |
| Equation in $y = mx + c$ form | Read the coefficient | $m$ is the number before $x$ |
From two points: $(1, 3)$ and $(5, 11)$. $m = \frac{11-3}{5-1} = \frac{8}{4} = 2$.
Find the equation through $(2,5)$ and $(4,11)$: $m = \frac{11-5}{4-2} = 3$. Sub $(2,5)$: $5 = 3(2) + c \Rightarrow c = -1$. Equation: $y = 3x - 1$. Check: $3(4)-1 = 11$ ✓.
Common error: Writing $m = \frac{x_2 - x_1}{y_2 - y_1}$ (run over rise). Gradient is rise over run, i.e., change in $y$ over change in $x$.
| Method | When to Use | Key Steps |
|---|---|---|
| Gradient-intercept | Equation is $y = mx + c$ | Plot $(0, c)$, use $m$ to find second point |
| Intercepts | Equation in form $ax + by = d$ | Find $x$-int ($y=0$) and $y$-int ($x=0$), join |
| Table of values | Any equation; good for checking | Choose $x$ values, calculate $y$, plot points |
Intercept method example: Sketch $2x + 3y = 12$.
$y$-intercept ($x=0$): $3y = 12$, $y = 4$ → $(0, 4)$.
$x$-intercept ($y=0$): $2x = 12$, $x = 6$ → $(6, 0)$.
Plot both intercepts and join with a straight line.
Substitution (best when one equation has a variable isolated): $y = 2x + 3$ and $3x + 2y = 23$. Substitute: $3x + 2(2x+3) = 23 \Rightarrow 7x = 17 \Rightarrow x = \frac{17}{7}$.
Elimination (best when coefficients match): $2x + 3y = 13$ and $5x - 3y = 1$. $y$ terms $+3$ and $-3$ (opposite). Add: $7x = 14$, $x = 2$. Back-sub: $y = 3$. Check: $4+9=13$ ✓, $10-9=1$ ✓.
| Situation | Best Method |
|---|---|
| One variable has coefficient 1 or $-1$, or already isolated | Substitution |
| Both equations in $ax+by=c$ form, coefficients match | Elimination (direct) |
| Both in $ax+by=c$ form, coefficients don't match | Elimination (multiply first) |
For word problems with two unknowns: Read → Define → Set up → Solve → Check.
Cinema tickets example: 2 adults + 3 children = $72. 1 adult + 4 children = $58. Let $a$ = adult price, $c$ = child price. Equations: $2a+3c=72$ and $a+4c=58$. From eq (2): $a=58-4c$. Sub: $2(58-4c)+3c=72 \Rightarrow 116-5c=72 \Rightarrow c=8.80$. Then $a=22.80$. Check: $2(22.80)+3(8.80)=72$ ✓.
Linear model example: A car rental costs $60 per day plus $0.30 per km. Model: $C = 0.30k + 60$. For 200 km: $C = 0.30(200) + 60 = 60 + 60 = \$120$.
Brain Trainer — mixed Unit 2 revision. Go!
-
1 Find the gradient through $(2, 3)$ and $(6, 11)$.
$m = \frac{11-3}{6-2} = \frac{8}{4} = 2$ -
2 What is the $y$-intercept of $y = -4x + 7$?
$c = 7$, so the $y$-intercept is $(0, 7)$ -
3 Solve $y = 3x$ and $y = x + 4$ simultaneously.
$3x = x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$, $y = 6$. Solution: $(2, 6)$ -
4 Find the equation of the line with $m = 3$ through $(1, 5)$.
$5 = 3(1) + c \Rightarrow c = 2$. Equation: $y = 3x + 2$ -
5 A line has intercepts $(4, 0)$ and $(0, -8)$. Find its gradient.
$m = \frac{-8-0}{0-4} = \frac{-8}{-4} = 2$ -
6 Solve by elimination: $x + y = 10$ and $x - y = 4$.
Add: $2x = 14$, $x = 7$. Then $y = 3$. Solution: $(7, 3)$ -
7 What is the gradient of $2x + 5y = 10$? (Rearrange first.)
$y = -\frac{2}{5}x + 2$, so $m = -\frac{2}{5}$ -
8 The sum of two numbers is 18 and their difference is 4. Find the numbers.
$x+y=18$, $x-y=4$. Add: $2x=22$, $x=11$, $y=7$. Numbers: 11 and 7 -
9 A taxi costs $5 flagfall plus $2 per km. Write a linear equation for cost $C$.
$C = 2k + 5$ where $k$ = km travelled -
10 Does $(3, -1)$ lie on the line $y = 2x - 7$? Show working.
When $x=3$: $y = 2(3) - 7 = -1$. Yes, $(3,-1)$ lies on the line ✓
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. A line passes through $(2, 5)$ and $(4, 11)$. Find its equation and state the gradient and $y$-intercept.
Q7. Solve simultaneously by elimination: $2x + y = 9$ and $3x - y = 11$. Check your solution.
Q8. Two numbers add to 15 and differ by 3. Write two equations, solve using simultaneous equations, and verify your answer.
Quick Check
1. B — $m = \frac{10-2}{5-1} = \frac{8}{4} = 2$.
2. B — $y = -2x + 5$. Gradient $-2$ as coefficient of $x$, intercept $+5$.
3. D — $(3, 6)$. Set $2x = x+3 \Rightarrow x=3$, $y=6$.
4. A — $m = -2$. Points $(3,0)$ and $(0,6)$: $m = \frac{6}{-3} = -2$.
5. B — $\$120$. $C = 60 + 0.30(200) = 120$.
Model Answers
Q6 (4 marks): $m = \frac{11-5}{4-2} = 3$ [1]. Use $(2,5)$: $5 = 3(2)+c \Rightarrow c=-1$ [1]. Equation: $y=3x-1$ [1]. Check: $3(4)-1=11$ ✓ [1].
Q7 (5 marks): $y$ terms $+1$ and $-1$ → add [1]. $5x=20 \Rightarrow x=4$ [2]. Back-sub: $8+y=9 \Rightarrow y=1$ [1]. Verify: $2(4)+1=9$ ✓, $3(4)-1=11$ ✓ [1]. Solution: $(4,1)$.
Q8 (6 marks): Define: $x+y=15$ and $x-y=3$ [2]. Add: $2x=18 \Rightarrow x=9$ [2]. Back-sub: $y=6$ [1]. Check: $9+6=15$ ✓, $9-6=3$ ✓ [1]. Numbers: 9 and 6.
Unit 2 Challenge Set
Stretch 1: A line passes through $(-1, 7)$ and has gradient $-3$. Find its equation, then find where it crosses the $x$-axis.
Reveal solution
$7 = -3(-1) + c \Rightarrow c = 4$. Equation: $y = -3x + 4$. For $x$-intercept: $0 = -3x + 4 \Rightarrow x = \frac{4}{3}$. The $x$-intercept is $\left(\frac{4}{3}, 0\right)$.
Stretch 2: The sum of two numbers is 20. Three times the first plus twice the second is 48. Find the numbers using simultaneous equations.
Reveal solution
$x + y = 20$ and $3x + 2y = 48$. From (1): $y = 20-x$. Sub: $3x + 2(20-x) = 48 \Rightarrow x = 8$. $y = 12$. Check: $8+12=20$ ✓, $3(8)+2(12)=48$ ✓.
Stretch 3: A plumber charges a call-out fee plus an hourly rate. A 2-hour job costs $150 and a 5-hour job costs $330. Find the call-out fee and hourly rate using a linear model.
Reveal solution
$150 = 2m + f$ and $330 = 5m + f$. Subtract: $180 = 3m \Rightarrow m = 60$. Back-sub: $f = 30$. Call-out fee: $30, hourly rate: $60/hour. Check: $5(60)+30=330$ ✓.
Stretch 4: Sketch $y = 2x + 1$ and $y = -x + 4$ on the same axes and find their intersection algebraically.
Reveal solution
Set equal: $2x+1 = -x+4 \Rightarrow 3x = 3 \Rightarrow x=1$. Then $y=3$. Intersection: $(1, 3)$. Check: $y = -(1)+4 = 3$ ✓.
Gradient formula
$m = \frac{y_2-y_1}{x_2-x_1} = \frac{\text{rise}}{\text{run}}$
Equation of a line
$y = mx + c$: $m$ = gradient, $c$ = $y$-intercept
From two points
Find $m$, then substitute a point to find $c$
Three sketch methods
Gradient-intercept, intercepts, table of values
Substitution
Isolate → Substitute → Solve → Back-sub → Verify
Elimination
Match coefficients → Add/subtract → Solve → Verify
Word problems
Read → Define → Set up → Solve → Check
Linear model
$y = mx + c$: $m$ = rate of change, $c$ = starting value
Unit 2 Badges
0 of 8Mark Unit 2 as complete!
You have finished all 20 lessons of Unit 2: Linear Relationships. Tick to earn +100 XP and +50 coins — a milestone achievement!