GraphsStudy Notes

Worked Examples

Part of Linear Graphs ProblemsGCSE Mathematics

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. This topic appears less often, but it can still be a useful differentiator on mixed-topic papers. 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.

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).

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