Linear Functions & Simultaneous Equations
Understand linear functions and their graphs, solve simultaneous equations using substitution and elimination, and get an introduction to quadratic expressions.
What You'll Learn
- Find the gradient and y-intercept of a linear function
- Plot linear graphs from equations and tables of values
- Find the equation of a line from its graph or two points
- Solve simultaneous equations by substitution and elimination
- Understand parallel and perpendicular line relationships
- Recognise and expand quadratic expressions (introduction)
IB Assessment Focus
Criterion A: Apply algebraic methods to unfamiliar problems with multiple steps.
Criterion B: Discover the relationship between gradient, y-intercept, and the graph. Justify your solutions.
Criterion C: Use correct algebraic notation and communicate reasoning clearly.
Criterion D: Apply linear functions to real-world contexts (distance-time, cost models).
Key Vocabulary
| Term | Definition |
|---|---|
| Function | A rule that assigns each input exactly one output. Written f(x) or y = ... |
| Linear function | A function whose graph is a straight line: y = mx + c |
| Gradient (m) | The steepness of a line; rise ÷ run; rate of change |
| y-intercept (c) | Where the line crosses the y-axis (when x = 0) |
| x-intercept | Where the line crosses the x-axis (when y = 0) |
| Simultaneous equations | Two equations with two unknowns; solved for values satisfying both |
| Quadratic | An expression where the highest power of x is 2 |
| Arithmetic sequence | A sequence with a constant difference between consecutive terms |
Linear Functions: y = mx + c
Every linear function can be written in the form y = mx + c, where m is the gradient and c is the y-intercept.
Understanding Gradient (m)
The gradient measures how steep a line is and the direction it slopes. It is the rate of change — how much y changes for each unit increase in x.
| Gradient | Meaning | Graph |
|---|---|---|
| m > 0 | Line slopes upward left to right | Positive slope ↗ |
| m < 0 | Line slopes downward left to right | Negative slope ↘ |
| m = 0 | Horizontal line (no slope) | Flat — |
| Large |m| | Steep line | Nearly vertical |
| Small |m| | Gentle line | Nearly horizontal |
Understanding the y-intercept (c)
The y-intercept is the value of y when x = 0. It tells you where the line crosses the y-axis. In y = 3x + 5, the y-intercept is 5, meaning the line passes through (0, 5).
Finding the Equation of a Line
Simply substitute into y = mx + c.
If m = 3 and c = −2, the equation is y = 3x − 2.
- Start with y = mx + c and substitute the known gradient.
- Substitute the coordinates of the given point to find c.
- Write the final equation.
Example: Gradient = 2, passes through (3, 7).
y = 2x + c → 7 = 2(3) + c → 7 = 6 + c → c = 1.
Equation: y = 2x + 1
- Calculate the gradient: m = (y&sub2; − y&sub1;) / (x&sub2; − x&sub1;).
- Substitute m and one point into y = mx + c to find c.
- Write the final equation.
Example: Points (1, 4) and (3, 10).
m = (10 − 4) / (3 − 1) = 6/2 = 3.
Using (1, 4): 4 = 3(1) + c → c = 1.
Equation: y = 3x + 1
Parallel and Perpendicular Lines
| Relationship | Gradient Rule | Example |
|---|---|---|
| Parallel lines | Same gradient (m&sub1; = m&sub2;) | y = 3x + 1 and y = 3x − 5 are parallel |
| Perpendicular lines | Product of gradients = −1 (m&sub1; × m&sub2; = −1) | y = 2x + 3 and y = −½x + 1 are perpendicular |
Arithmetic Sequences as Linear Functions
An arithmetic sequence has a constant difference (d) between terms. The nth term formula is a linear function:
a = 5, d = 3. Tn = 5 + (n − 1)(3) = 5 + 3n − 3 = 3n + 2
Check: T&sub1; = 3(1) + 2 = 5 ✓, T&sub2; = 3(2) + 2 = 8 ✓
Graphing Linear Functions
Every linear equation can be drawn as a straight line. You need at least two points to plot a line.
Method 1: Table of Values
| x | y = 2x − 1 | (x, y) |
|---|---|---|
| −2 | 2(−2) − 1 = −5 | (−2, −5) |
| 0 | 2(0) − 1 = −1 | (0, −1) |
| 1 | 2(1) − 1 = 1 | (1, 1) |
| 3 | 2(3) − 1 = 5 | (3, 5) |
Plot these points and draw a straight line through them. The y-intercept is −1 and the gradient is 2.
Method 2: Gradient-Intercept Method
- Plot the y-intercept (c) on the y-axis.
- From that point, use the gradient: rise ÷ run. If m = 3, go up 3, right 1.
- If m is negative (e.g., m = −2), go down 2, right 1.
- If m is a fraction (e.g., m = 2/3), go up 2, right 3.
- Plot at least 2 points and draw the line.
Method 3: Intercepts Method
- Find the y-intercept: set x = 0 and solve for y.
- Find the x-intercept: set y = 0 and solve for x.
- Plot both intercepts and draw the line through them.
Example: 3x + 2y = 12
y-intercept: x = 0 → 2y = 12 → y = 6 → (0, 6)
x-intercept: y = 0 → 3x = 12 → x = 4 → (4, 0)
Plot (0, 6) and (4, 0) and draw the line.
Special Lines
| Equation | Type | Description |
|---|---|---|
| y = c (e.g., y = 3) | Horizontal line | All points have y-coordinate = 3; gradient = 0 |
| x = c (e.g., x = −2) | Vertical line | All points have x-coordinate = −2; gradient is undefined |
| y = x | Identity line | Passes through origin at 45°; gradient = 1 |
| y = −x | Reflection line | Passes through origin at −45°; gradient = −1 |
Simultaneous Equations
Simultaneous equations are two (or more) equations with two unknowns. You need to find the values of x and y that satisfy BOTH equations at the same time.
Method 1: Elimination
The goal is to make the coefficient of one variable the same in both equations, then add or subtract to eliminate it.
- The coefficient of y is already the same (2) in both equations.
- Subtract equation 1 from equation 2: (5x + 2y) − (3x + 2y) = 22 − 16 → 2x = 6 → x = 3
- Substitute x = 3 into equation 1: 3(3) + 2y = 16 → 9 + 2y = 16 → 2y = 7 → y = 3.5
- Verify: Eq.1: 3(3) + 2(3.5) = 9 + 7 = 16 ✓. Eq.2: 5(3) + 2(3.5) = 15 + 7 = 22 ✓
- To eliminate y: multiply equation 2 by 3 → 12x + 3y = 42
- Subtract equation 1: (12x + 3y) − (2x + 3y) = 42 − 12 → 10x = 30 → x = 3
- Substitute into eq. 2: 4(3) + y = 14 → y = 14 − 12 = 2
- Verify: Eq.1: 2(3) + 3(2) = 6 + 6 = 12 ✓. Eq.2: 4(3) + 2 = 14 ✓
Method 2: Substitution
Rearrange one equation to express one variable in terms of the other, then substitute into the second equation.
Graphical Interpretation
Solving simultaneous equations is the same as finding where two lines intersect (cross). The solution (x, y) is the point of intersection.
Introduction to Quadratics
A quadratic expression has x² as the highest power. At Grade 8, you need to recognise, expand, and factorise simple quadratic expressions.
Expanding Double Brackets (FOIL)
= x² + 5x + 3x + 15 = x² + 8x + 15
= x² + 2x − 4x − 8 = x² − 2x − 8
= 2x² − 6x + x − 3 = 2x² − 5x − 3
Special Products
| Pattern | Formula | Example |
|---|---|---|
| Perfect square | (a + b)² = a² + 2ab + b² | (x + 4)² = x² + 8x + 16 |
| Perfect square | (a − b)² = a² − 2ab + b² | (x − 3)² = x² − 6x + 9 |
| Difference of squares | (a + b)(a − b) = a² − b² | (x + 5)(x − 5) = x² − 25 |
Factorising Quadratics
Factorising is the reverse of expanding: turning x² + bx + c into (x + p)(x + q).
- Find two numbers p and q that multiply to give c and add to give b.
- Write as (x + p)(x + q).
- Check by expanding back.
Need two numbers that multiply to 12 and add to 7. Try: 3 × 4 = 12, 3 + 4 = 7 ✓
Answer: (x + 3)(x + 4)
Check: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓
Need two numbers that multiply to 6 and add to −5. Try: (−2) × (−3) = 6, (−2) + (−3) = −5 ✓
Answer: (x − 2)(x − 3)
This is a difference of two squares: x² − 4² = (x + 4)(x − 4)
Worked Examples
Multi-step problems showing Grade 8 level reasoning with full justification.
Step 2: Use y = mx + c with m = 2 and point (2, 5):
5 = 2(2) + c → 5 = 4 + c → c = 1
Equation: y = 2x + 1
Verify: At (6, 13): y = 2(6) + 1 = 13 ✓
(2x + y) + (x − y) = 7 + 2
3x = 9 → x = 3
Substitute into x − y = 2: 3 − y = 2 → y = 1
Verify: Eq.1: 2(3) + 1 = 7 ✓. Eq.2: 3 − 1 = 2 ✓
Solution: (x, y) = (3, 1)
This is a linear function with gradient 2 (cost per km) and y-intercept 5 (base fee).
Cost for 12 km: C = 2(12) + 5 = 24 + 5 = $29
Distance for $31: 31 = 2d + 5 → 26 = 2d → d = 13 km
Verify: 2(13) + 5 = 26 + 5 = 31 ✓
Step 2: Use y = mx + c with m = −1/3 and point (6, 2):
2 = (−1/3)(6) + c → 2 = −2 + c → c = 4
Equation: y = −(1/3)x + 4
Tn = a + (n − 1)d = 7 + (n − 1)(4) = 7 + 4n − 4 = 4n + 3
T50 = 4(50) + 3 = 200 + 3 = 203
Check: T&sub1; = 4(1) + 3 = 7 ✓, T&sub2; = 4(2) + 3 = 11 ✓
Try: 5 × (−3) = −15 and 5 + (−3) = 2 ✓
x² + 2x − 15 = (x + 5)(x − 3)
Solve: (x + 5)(x − 3) = 0
Either x + 5 = 0 → x = −5, or x − 3 = 0 → x = 3
Solutions: x = −5 or x = 3
Practice Q&A
Attempt each question before revealing the answer. Show full working and justify each step.
The y-intercept is 5: the line crosses the y-axis at (0, 5). This is the value of y when x = 0.
Substitute into eq. 2: 4 + 4y = 10 → 4y = 6 → y = 1.5.
Verify: Eq.1: 3(4) + 4(1.5) = 12 + 6 = 18 ✓. Eq.2: 4 + 4(1.5) = 4 + 6 = 10 ✓.
Solution: (4, 1.5)
y = 4x + c. Substitute (1, 6): 6 = 4(1) + c → c = 2.
Equation: y = 4x + 2
(x + 2)(x − 2) = x² − 4 (difference of squares)
Subtraction: (x² + 12x + 36) − (x² − 4) = x² + 12x + 36 − x² + 4 = 12x + 40
x² − 9x + 20 = (x − 4)(x − 5)
Solve: x − 4 = 0 or x − 5 = 0 → x = 4 or x = 5
Check: 4² − 9(4) + 20 = 16 − 36 + 20 = 0 ✓
T&sub3;: a + 2d = 14 ... (1)
T&sub7;: a + 6d = 26 ... (2)
Subtract (1) from (2): 4d = 12 → d = 3
Substitute into (1): a + 6 = 14 → a = 8
Sequence: 8, 11, 14, 17, 20, 23, 26, ... ✓
Set equal: 0.05t + 20 = 30 → 0.05t = 10 → t = 200 texts.
At 200 texts, both cost $30. After 200 texts, Plan A is more expensive.
This is a simultaneous equations problem solved graphically: the intersection is at t = 200.
Flashcard Review
Tap each card to reveal the answer.
= rise / run
m&sub1; × m&sub2; = −1
2. Substitution (express one variable in terms of the other)
First, Outer, Inner, Last
(Difference of two squares)
(Perfect square expansion)
where a = first term, d = common difference