GraphsStudy Notes

Worked Examples

Part of Linear Graphs Problems · GCSE GCSE Mathematics revision

This study notes covers Worked Examples within Linear Graphs Problems for GCSE Mathematics. Revise Linear Graphs Problems in Graphs for GCSE Mathematics with 16 exam-style questions and 11 flashcards. Use this page as part of a wider topic revision path rather than treating it as an isolated fact. It is section 8 of 10 in this topic. Use this study notes to connect the idea to the wider topic before moving on to questions and flashcards.

Topic position

Section 8 of 10

Practice

16 questions

Recall

11 flashcards

✏️ Worked Examples

Example 1: Interpreting a Real-Life Linear Graph

Question: A mobile phone plan costs £15 per month plus 5p per minute of calls. Write an equation for the total monthly cost C (in £) in terms of minutes m. How much does it cost to make 200 minutes of calls?

Show Solution

Step 1: Identify m and c from the context

Fixed monthly fee (y-intercept): c = 15

Cost per minute (gradient): m = 0.05 (= 5p = £0.05)

Step 2: Write the equation — C = 0.05m + 15

Step 3: Substitute m = 200 — C = 0.05 × 200 + 15 = 10 + 15 = £25

Answer: C = 0.05m + 15. For 200 minutes: £25

Example 2: Solving Simultaneous Equations Graphically

Question: By drawing both lines, solve simultaneously: y = 2x + 1 and y = -x + 7.

Show Solution

Step 1: Plot y = 2x + 1 — three points: (0, 1), (1, 3), (2, 5)

Step 2: Plot y = -x + 7 — three points: (0, 7), (2, 5), (4, 3)

Step 3: Find the intersection — both lines pass through (2, 5).

Check: Line 1: 2(2) + 1 = 5 ✓ Line 2: -(2) + 7 = 5 ✓

Answer: x = 2, y = 5

Example 3: Finding the Equation from Two Points in Context

Question: A car is travelling at constant speed. After 1 hour it is 80 km from home; after 3 hours it is 200 km from home. Write an equation for distance d (km) in terms of time t (hours). How far will it be after 5 hours?

Show Solution

Step 1: Calculate the gradient (speed) — m = (200 - 80) ÷ (3 - 1) = 120 ÷ 2 = 60 km/h

Step 2: Find c (starting distance) — using (1, 80): 80 = 60(1) + c → c = 20

Step 3: Write the equation — d = 60t + 20

Step 4: Substitute t = 5 — d = 60(5) + 20 = 300 + 20 = 320 km

Answer: d = 60t + 20. After 5 hours: 320 km from home

Keep building this topic

Read this section alongside the surrounding pages in Linear Graphs Problems. That gives you the full topic sequence instead of a single isolated revision point.

Exam Tips for Linear Graphs Problems Common Misconceptions Lines in the Real World The Equation y = mx + c — Quick Reference

Practice Questions for Linear Graphs Problems

A taxi company charges a fixed fee plus an amount per mile. On a cost graph (£ against miles), what does the y-intercept represent?

A. The cost per mile
B. The total cost of the journey
C. The fixed charge before any miles are travelled
D. The gradient of the line

1 markfoundation

A plumber charges according to the formula C = 40t + 30, where C is the total cost in pounds and t is the time in hours. Explain what the values 40 and 30 represent in this context.

2 marksstandard

Quick Recall Flashcards

Steps to find the equation of a line from a graph

1. Read the y-intercept (c) where line crosses y-axis 2. Choose two clear points on the line 3. Calculate gradient m = (y2 - y1)/(x2 - x1) 4. Write y = mx + c Example: crosses (0, 1), gradient 2 → y = 2x + 1

What does each letter in y = mx + c represent?

y = mx + c m = gradient (steepness of the line) c = y-intercept (where the line crosses the y-axis) Example: y = 3x + 2 has gradient 3 and crosses y-axis at (0, 2).

16 questions on Linear Graphs Problems — practise free

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Exam Tips for Linear Graphs Problems

Common Misconceptions