Friday, May 8, 2026

📐 Maths Handout

New Study Guide

Study Guide · Quick Reference Cheatsheet

32. Plotting Points and Finding Gradient

This section covers plotting points on a Cartesian plane and determining the gradient of a line passing through them.

a. Plotting Points

To plot points on a graph, we use the Cartesian coordinate system, which consists of a horizontal x-axis and a vertical y-axis. The point where they intersect is the origin (0,0).

Given points: A(0,2), B(2,4), C(4,6), D(6,8)

To plot these points:

Point A (0,2): Start at the origin. Move 0 units along the x-axis (stay at x=0). Move 2 units up along the y-axis. Mark point A.
Point B (2,4): Start at the origin. Move 2 units to the right along the x-axis. From there, move 4 units up parallel to the y-axis. Mark point B.
Point C (4,6): Start at the origin. Move 4 units to the right along the x-axis. From there, move 6 units up parallel to the y-axis. Mark point C.
Point D (6,8): Start at the origin. Move 6 units to the right along the x-axis. From there, move 8 units up parallel to the y-axis. Mark point D.

b. Drawing the Line

Once the points are plotted, use a ruler to draw a straight line that passes through all four points. You will observe that these points are collinear, meaning they lie on the same straight line.

c. Finding the Gradient

The gradient (or slope) of a straight line measures its steepness and direction. It is calculated as the ratio of the vertical change (rise) to the horizontal change (run) between any two distinct points on the line.

The formula for the gradient m between two points (x₁, y₁) and (x₂, y₂) is:

m = (y₂ − y₁) / (x₂ − x₁)

Let’s use points A(0,2) and B(2,4):

m = (4 − 2) / (2 − 0) = 22 = 1

Let’s verify with points C(4,6) and D(6,8):

m = (8 − 6) / (6 − 4) = 22 = 1

The gradient of the line is 1.

33. Perimeter and Length of a Rectangle

Given: Perimeter P = 36 cm, width w = 8 cm.

Formula for the perimeter of a rectangle: P = 2 × (length + width)

Let l represent the length.

36 = 2(l + 8)

18 = l + 8

l = 10

Length = 10 cm

34. Finding the Mean of a Set of Numbers

Numbers: 10, 15, 20, 25, 30

Sum = 10 + 15 + 20 + 25 + 30 = 100

Count = 5

Mean = 100 ÷ 5 = 20

Mean = 20

35. Rounding to the Nearest Hundred Thousand

Number: 568,349

Hundred thousands digit = 5.
Next digit (ten thousands) = 6 (≥5), so round up.
5 becomes 6, all following digits become 0.

600,000

36. Area of a Water Trough (Requires Diagram)

⚠️

Diagram missing

The water trough diagram is not provided. To calculate the painted outer surface area, we need the shape, dimensions, and which faces were painted. Please provide the diagram or a detailed description.

37. Laptop Hire Purchase Calculation

Cash price = sh.40,000. Hire Purchase (HP) price = 20% more.

20% of 40,000 = 0.20 × 40,000 = 8,000

HP price = 40,000 + 8,000 = 48,000

Deposit = 20,000 → Balance = 48,000 − 20,000 = 28,000

14 equal monthly instalments: 28,000 ÷ 14 = 2,000

(a) Monthly instalment = sh.2,000

Extra paid = HP price − Cash price = 48,000 − 40,000 = 8,000

(b) Extra paid = sh.8,000

38. Area of a Shaded Segment of a Circle

Radius r = 13 cm, central angle θ = 60°.

Step 1: Area of sector

A_sector = (θ / 360°) × π × r²

A_sector = (60 / 360) × π × 169 = 16 × π × 169 = 169π6 cm²

Step 2: Area of triangle

A_triangle = ½ × a × b × sin C

a = 13, b = 13, C = 60°, sin 60° = √3/2 ≈ 0.8660

A_triangle = ½ × 13 × 13 × (√3/2) = (169√3)/4 cm²

Step 3: Shaded segment

A_segment = A_sector − A_triangle = 169π6 − 169√34 cm²

Numerical: π ≈ 3.1416, √3 ≈ 1.73205

169π/6 ≈ 88.488, 169√3/4 ≈ 73.179

Difference ≈ 15.309 cm²

Area ≈ 15.31 cm² (rounded to 2 decimal places)

39. Expression for Total Number of Bananas

Let x = bananas Kamene has.

Morries: x + 2
Kamene + Morries = x + (x+2) = 2x + 2
Mary = (2x+2) − 5 = 2x − 3

Total = x + (x+2) + (2x−3) = 4x − 1

Total bananas = 4x − 1

40. Working with Fractions

Work out: 45 ÷ 1015

1015 = 515

45 ÷ 515 = 45 × 551 = 4×55×51 = 451

451

Quick Reference Cheatsheet

Mathematics Quick Reference Sheet (Grade 9 CBC)

Algebra

Linear Equations: ax + b = c → x = (c − b)/a
Quadratic Formula: x = [−b ± √(b² − 4ac)] / (2a)
Expressions: Combine like terms, distributive property a(b + c) = ab + ac
Inequalities: Reverse inequality sign when multiplying/dividing by negative.

Geometry

Area: Rectangle A = lw, Square A = s², Triangle A = ½ × base × height, Circle A = πr², Sector A = (θ/360°) × πr²
Perimeter/Circumference: Rectangle P = 2(l + w), Square P = 4s, Circle C = 2πr or πd
Pythagoras: a² + b² = c² (right triangle)
Circle Segment: Area = Sector area − Triangle area

Mensuration (3D)

Volume: Cuboid V = lwh, Cube V = s³, Cylinder V = πr²h, Prism V = Base area × height
Surface Area: Sum of all faces.

Number System

Fractions: Common denominator for addition/subtraction; multiply numerator×numerator, denominator×denominator; divide by multiplying by reciprocal.
Decimals: Rounding – look at the digit to the right.
Percentages: Convert to fraction or decimal; percentage increase/decrease.
Mean: Sum ÷ count.

Coordinate Geometry

Points: (x, y)
Gradient: m = (y₂ − y₁) / (x₂ − x₁)
Equation of a line: y = mx + c (c = y‑intercept)

✅ All fractions now use pure CSS – no LaTeX `\frac` commands – works in every browser, even without MathJax.

Thursday, May 7, 2026

📐 CBC Mathematics Handout

📈 Question 32: Gradient of a Line

Points: A(0,2), B(2,4), C(4,6), D(6,8)

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Choose A(0,2) and D(6,8):
$$m = \frac{8 - 2}{6 - 0} = \frac{6}{6} = 1$$

Gradient = 1

Gradient = 1 means the line rises 1 unit for every 1 unit to the right.

      Y
      ↑
      8|                             ● D(6,8)
      7|
      6|                          ● C(4,6)
      5|
      4|                       ● B(2,4)
      3|
      2|                    ● A(0,2)
      1|
      0└────────────────────────────────→ X
       0 1 2 3 4 5 6 7 8 9

📏 Question 33: Rectangle Perimeter

Perimeter = 36 cm, width = 8 cm. Find length.

$$P = 2(L + W)$$

$$36 = 2(L + 8)$$
$$L + 8 = 18$$
$$L = 10\ \text{cm}$$

Length = 10 cm

Perimeter is the total distance around the rectangle.

📊 Question 34: Mean of Numbers

Numbers: 10, 15, 20, 25, 30

$$\text{Mean} = \frac{\text{Sum}}{\text{Count}}$$

Sum = 10 + 15 + 20 + 25 + 30 = 100
Count = 5
Mean = 100 ÷ 5 = 20

Mean = 20

The mean is the average, or “fair share”.

🔢 Question 35: Rounding to Nearest Hundred Thousand

Round 568 349 to the nearest hundred thousand.

The hundred‑thousands digit is 5 (since 568 349 = 5 68 349).
Look at the ten‑thousands digit: 6 (≥5), so round up.
5 becomes 6, all following digits become 0.

600 000

Rounding to nearest 100 000 means the last five digits become zero.

💧 Question 36: Painted Trough Area

⚠️ Diagram missing from the exam paper. The trough is typically a trapezoidal prism or rectangular open tank.
Without the dimensions, the exact area cannot be calculated. However, the method is to add the areas of all outer faces.
If you provide the diagram or measurements, I can compute the exact painted area.

       /‾‾‾‾‾‾‾‾‾‾\      (Example trapezoidal trough)
      /           \
     /             \
    | ‾‾‾‾‾‾‾‾‾‾‾‾‾‾ |
    |________________|

💻 Question 37: Hire Purchase

Cash price = sh. 40 000. Hire purchase price = 20% more. Deposit = sh. 20 000. Balance in 14 equal monthly instalments.

$$\text{HP} = \text{Cash} + 20\% \times \text{Cash}$$

HP = 40 000 × 1.20 = 48 000 shillings.
Balance after deposit = 48 000 – 20 000 = 28 000 shillings.
Monthly instalment = 28 000 ÷ 14 = 2 000 shillings per month.
Extra paid on HP = 48 000 – 40 000 = 8 000 shillings.

(a) 2 000 shillings per month (b) 8 000 shillings extra

Hire purchase usually adds a percentage or flat fee; here it is 20% of cash price.

🟠 Question 38: Shaded Area (Ring)

⚠️ Diagram missing. A common shaded area problem involves two concentric circles.
For example, if the larger circle radius = 15 cm and smaller radius = 10 cm, the shaded ring area = π(R² – r²).
With those numbers: π(225 – 100) = 125π ≈ 392.7 cm².

        ╭─────────────────────╮
       /  ┌─────────────┐    /    Outer circle radius R
      │  │    ●        │    │    Inner circle radius r
      │  │   Shaded    │    │    Shaded = R²π – r²π
       \  └─────────────┘    /
        ╰─────────────────────╯

🍌 Question 39: Banana Algebra

Kamene has x bananas. Morris has 2 more than Kamene. Mary has 5 less than the total of Kamene and Morris.

Kamene: $x$
Morris: $x + 2$
Kamene + Morris = $x + (x+2) = 2x + 2$
Mary = $(2x + 2) - 5 = 2x - 3$
Total = Kamene + Morris + Mary = $x + (x+2) + (2x - 3) = 4x - 1$

Total bananas = $4x - 1$

Always define variables clearly before combining.

🧮 Question 40: Fraction Division

Work out: $\displaystyle \frac{4}{5} \div 10\frac{1}{5}$

Convert mixed number to improper fraction: $10\frac{1}{5} = \frac{51}{5}$
Division: $\frac{4}{5} \div \frac{51}{5} = \frac{4}{5} \times \frac{5}{51} = \frac{4}{51}$

$\displaystyle \frac{4}{51}$

Dividing by a fraction is the same as multiplying by its reciprocal.

📌 All missing diagrams are due to the original exam paper not providing them. With the correct diagrams or measurements, the area problems can be solved precisely.

🧑‍🏫 Master Quadratic Equations, Logarithms & Systems — Teacher-Approved Guide

📐 MATHEMATICS SOLUTIONS | STEP-BY-STEP

1. Make x the subject: ax² + bx + c = 0

📐

Quadratic Formula Derivation

Starting from the standard form:

ax² + bx + c = 0

Step 1: Move c to the other side

ax² + bx = -c

Step 2: Divide both sides by a

x² + (b/a)x = -c/a

Step 3: Complete the square

x² + (b/a)x + (b/(2a))² = (b/(2a))² - c/a

Step 4: Simplify

(x + b/(2a))² = (b² - 4ac)/(4a²)

Step 5: Take square root

x + b/(2a) = ± √(b² - 4ac)/(2a)

Step 6: Solve for x

x = -b ± √(b² - 4ac) 2a

x = (-b ± √(b² - 4ac)) / 2a

✅

Final Answer

x = -b ± √(b² - 4ac)2a

2. Simplify: (x²√x)(√x∛x²) / (x³y³)^(1/2)

Step 1: Convert all radicals to powers

√x = x¹ᐟ²    ∛x² = x²ᐟ³

Step 2: Numerator:

x² · x¹ᐟ² = x52
x¹ᐟ² · x²ᐟ³ = x76
Numerator = x52 · x76 = x226 = x113

Step 3: Denominator:

(x³y³)¹ᐟ² = x³ᐟ² · y³ᐟ²

Step 4: Division:

x¹¹ᐟ³ x³ᐟ² · y³ᐟ² = x136 · y⁻³ᐟ²

x¹³ᐟ⁶ / y³ᐟ²

3. Solve: log(x+1) + log(x+1)² = 2 log(x+2)

Step 1: Use logarithm power rule: log A² = 2 log A

log(x+1) + 2 log(x+1) = 3 log(x+1)

Step 2: Rewrite equation

3 log(x+1) = 2 log(x+2)

Step 3: Divide by 3

log(x+1) = 23 log(x+2) = log(x+2)²ᐟ³

Step 4: Remove logs (assuming base 10)

x + 1 = (x + 2)²ᐟ³

Step 5: Cube both sides

(x + 1)³ = (x + 2)²

x³ + 3x² + 3x + 1 = x² + 4x + 4
x³ + 2x² - x - 3 = 0

Step 6: Domain: x > -1, x > -2 → x > -1

Step 7: Approximate solution by testing values

Test x = 1.5 → 3.375 + 4.5 - 1.5 - 3 = 3.375 (too high)
Test x = 1.2 → 1.728 + 2.88 - 1.2 - 3 = 0.408 (close)
Test x = 1.15 → 1.520875 + 2.645 - 1.15 - 3 = 0.015875 ≈ 0

x ≈ 1.15

4. Make r the subject: x/y = (1 + r²)/(1 - r²)

Step 1: Cross multiply

x(1 - r²) = y(1 + r²)

x - xr² = y + yr²

Step 2: Collect r terms

x - y = xr² + yr² = r²(x + y)

Step 3: Isolate r²

r² = x - yx + y

Step 4: Square root (assuming r positive)

r = √[(x - y)/(x + y)]

5. Solve the system of equations

2l₁ + 3l₂ - 4l₃ = 26 ... (1)

l₁ - 5l₂ - 3l₃ = -87 ... (2)

-7l₁ + 2l₂ + 6l₃ = 16 ... (3)

Step 1: Express l₁ from equation (2)

l₁ = 5l₂ + 3l₃ - 87

Step 2: Substitute into (1) and (3)

Equation (1):

2(5l₂ + 3l₃ - 87) + 3l₂ - 4l₃ = 26

10l₂ + 6l₃ - 174 + 3l₂ - 4l₃ = 26

13l₂ + 2l₃ = 200 ... (4)

Equation (3):

-7(5l₂ + 3l₃ - 87) + 2l₂ + 6l₃ = 16

-35l₂ - 21l₃ + 609 + 2l₂ + 6l₃ = 16

-33l₂ - 15l₃ = -593

33l₂ + 15l₃ = 593 ... (5)

Step 3: Solve (4) and (5)

From (4): l₂ = (200 - 2l₃)/13

Substitute into (5):

33·(200 - 2l₃)/13 + 15l₃ = 593

33(200 - 2l₃) + 195l₃ = 7709

6600 - 66l₃ + 195l₃ = 7709

129l₃ = 1109

l₃ ≈ 8.60

Step 4: Find l₂ and l₁

l₂ = (200 - 2×8.60)/13 ≈ 14.06

l₁ = 5(14.06) + 3(8.60) - 87 ≈ 9.09

l₁ ≈ 9.09, l₂ ≈ 14.06, l₃ ≈ 8.60

📊 Final Answers Summary

🎯

Boxed Answers

1. x = (-b ± √(b² - 4ac)) / 2a
2. x¹³ᐟ⁶ / y³ᐟ²
3. x ≈ 1.15
4. r = √[(x - y)/(x + y)]
5. l₁ ≈ 9.09, l₂ ≈ 14.06, l₃ ≈ 8.60

📌 Summary: All five problems have been solved step by step. The quadratic formula was derived, the algebraic expression was simplified using exponent rules, the logarithmic equation was solved numerically, the formula was rearranged for r, and the linear system was solved using substitution.

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📐 MATHEMATICS SOLUTIONS | STEP-BY-STEP

1. Make x the subject: ax² + bx + c = 0

📐

Quadratic Formula Derivation

Starting from the standard form:

ax² + bx + c = 0

Step 1: Move c to the other side

ax² + bx = -c

Step 2: Divide both sides by a

x² + (b/a)x = -c/a

Step 3: Complete the square

x² + (b/a)x + (b/(2a))² = (b/(2a))² - c/a

Step 4: Simplify

(x + b/(2a))² = (b² - 4ac)/(4a²)

Step 5: Take square root

x + b/(2a) = ± √(b² - 4ac)/(2a)

Step 6: Solve for x

x = -b ± √(b² - 4ac) 2a

x = (-b ± √(b² - 4ac)) / 2a

2. Simplify: \frac{(x²√x)(√x∛x²)}{(x³y³)^{1/2}}

Step 1: Convert all radicals to powers

√x = x¹ᐟ²    ∛x² = x²ᐟ³

Step 2: Numerator:

x² · x¹ᐟ² = x52
x¹ᐟ² · x²ᐟ³ = x76
Numerator = x52 · x76 = x226 = x113

Step 3: Denominator:

(x³y³)¹ᐟ² = x³ᐟ² · y³ᐟ²

Step 4: Division:

x¹¹ᐟ³ x³ᐟ² · y³ᐟ² = x136 · y⁻³ᐟ²

x¹³ᐟ⁶ / y³ᐟ²

3. Solve: log(x+1) + log(x+1)² = 2 log(x+2)

Step 1: Use logarithm power rule: log A² = 2 log A

log(x+1) + 2 log(x+1) = 3 log(x+1)

Step 2: Rewrite equation

3 log(x+1) = 2 log(x+2)

Step 3: Divide by 3

log(x+1) = 23 log(x+2) = log(x+2)²ᐟ³

Step 4: Remove logs (assuming base 10)

x + 1 = (x + 2)²ᐟ³

Step 5: Cube both sides

(x + 1)³ = (x + 2)²

x³ + 3x² + 3x + 1 = x² + 4x + 4
x³ + 2x² - x - 3 = 0

Step 6: Domain: x > -1, x > -2 → x > -1

Step 7: Approximate solution by testing values

Test x = 1.5 → 3.375 + 4.5 - 1.5 - 3 = 3.375 (too high)
Test x = 1.2 → 1.728 + 2.88 - 1.2 - 3 = 0.408 (close)
Test x = 1.15 → 1.520875 + 2.645 - 1.15 - 3 = 0.015875 ≈ 0

x ≈ 1.15

4. Make r the subject: x/y = (1 + r²)/(1 - r²)

Step 1: Cross multiply

x(1 - r²) = y(1 + r²)

x - xr² = y + yr²

Step 2: Collect r terms

x - y = xr² + yr² = r²(x + y)

Step 3: Isolate r²

r² = x - yx + y

Step 4: Square root (assuming r positive)

r = √[(x - y)/(x + y)]

5. Solve the system of equations

2l₁ + 3l₂ - 4l₃ = 26 ... (1)

l₁ - 5l₂ - 3l₃ = -87 ... (2)

-7l₁ + 2l₂ + 6l₃ = 16 ... (3)

Step 1: Express l₁ from equation (2)

l₁ = 5l₂ + 3l₃ - 87

Step 2: Substitute into (1) and (3)

Equation (1):

2(5l₂ + 3l₃ - 87) + 3l₂ - 4l₃ = 26

10l₂ + 6l₃ - 174 + 3l₂ - 4l₃ = 26

13l₂ + 2l₃ = 200 ... (4)

Equation (3):

-7(5l₂ + 3l₃ - 87) + 2l₂ + 6l₃ = 16

-35l₂ - 21l₃ + 609 + 2l₂ + 6l₃ = 16

-33l₂ - 15l₃ = -593

33l₂ + 15l₃ = 593 ... (5)

Step 3: Solve (4) and (5)

From (4): l₂ = (200 - 2l₃)/13

Substitute into (5):

33·(200 - 2l₃)/13 + 15l₃ = 593

33(200 - 2l₃) + 195l₃ = 7709

6600 - 66l₃ + 195l₃ = 7709

129l₃ = 1109

l₃ ≈ 8.60

Step 4: Find l₂ and l₁

l₂ = (200 - 2×8.60)/13 ≈ 14.06

l₁ = 5(14.06) + 3(8.60) - 87 ≈ 9.09

l₁ ≈ 9.09, l₂ ≈ 14.06, l₃ ≈ 8.60

📊 Final Answers Summary

🎯

Boxed Answers

1. x = (-b ± √(b² - 4ac)) / 2a
2. x¹³ᐟ⁶ / y³ᐟ²
3. x ≈ 1.15
4. r = √[(x - y)/(x + y)]
5. l₁ ≈ 9.09, l₂ ≈ 14.06, l₃ ≈ 8.60

📌 Summary: All five problems have been solved step by step. The quadratic formula was derived, the algebraic expression was simplified using exponent rules, the logarithmic equation was solved numerically, the formula was rearranged for r, and the linear system was solved using substitution.

📐 The Gradient Of A Straight Line

📝 Name: ____________________

📅 Date: ____________________

🏫 Class: Grade 9 _____

CBC Curriculum - Kenya Mathematics - Grade 8

📐 GRADE 9 MATHS PROMPT (CBC Kenya Curriculum)

MATHS ADVENTURE: Find the Hidden Number!

🎯

THE CHALLENGE

A straight line has a secret code written like this:

y = 1 3 x + 7

Look carefully at the fraction! The number 1 is on top of the number 3 with a horizontal line between them. This horizontal line is called a VINCULUM.

There is a hidden number inside this code. It is called the GRADIENT (or slope).

Your mission: Find the gradient of this line! 🕵️

🧠 WHAT IS A GRADIENT?

Think of a gradient as the steepness of a line.

If you walk...	The gradient is...
Up a very steep hill	Large (like 5 or 10)
Up a gentle slope	Small (like one-third)
On flat ground	Zero (0)
Downhill	Negative (like -2)

🔢 LET'S UNDERSTAND THE FRACTION WITH VINCULUM

The fraction in our equation is written with a horizontal vinculum:

1 3

This means: ONE divided by THREE or ONE part out of THREE equal parts.

Imagine a chocolate bar cut into 3 equal pieces:

┌─────┬─────┬─────┐
│  1  │  2  │  3  │
└─────┴─────┴─────┘

Taking 1 piece means you have one divided by three (1 ÷ 3) of the chocolate.

📍 FINDING POINTS ON THE LINE

Let's find some points that lie on this line. We use the equation: y = 13 x + 7

Point 1: Choose x = 0 → y = 13 × 0 + 7 = 0 + 7 = 7 → Point A = (0, 7)

Point 2: Choose x = 3 → y = 13 × 3 + 7 = 1 + 7 = 8 → Point B = (3, 8)

Point 3: Choose x = 6 → y = 13 × 6 + 7 = 2 + 7 = 9 → Point C = (6, 9)

Point 4: Choose x = 9 → y = 13 × 9 + 7 = 3 + 7 = 10 → Point D = (9, 10)

📈 THE LINE DRAWN ON A GRAPH

      Y
      ↑
     10|                    ● D (9,10)
       |
      9|                 ● C (6,9)
       |
      8|              ● B (3,8)
       |
      7|           ● A (0,7)
       |
      6|         /
       |       /
      5|     /
       |   /
      4|   /
       | /
      3|
       |
      2|
       |
      1|
       |
      0└────────────────────────────────→ X
       0 1 2 3 4 5 6 7 8 9 10 11 12

Can you see the straight line going up? The line crosses the y-axis at 7 (this is called the y-intercept).

🚶‍♂️ UNDERSTANDING THE STEPS (RISE AND RUN)

From Point	To Point	Move RIGHT (Run)	Move UP (Rise)	RISE ÷ RUN
A (0,7)	B (3,8)	3 steps	1 step	1 ÷ 3
B (3,8)	C (6,9)	3 steps	1 step	1 ÷ 3
C (6,9)	D (9,10)	3 steps	1 step	1 ÷ 3

Look at the pattern! Every time you move 3 steps to the right, you move 1 step up.

        ___
       /   ↑ 1 step up
      /
     /
    /
   └──→ 3 steps right

The GRADIENT is calculated as:

        RISE (how much you go UP)
GRADIENT = —————————————————————
         RUN (how much you go RIGHT)

For this line: GRADIENT = 1 ÷ 3

1 3

🏠

REAL-LIFE EXAMPLE: A Gentle Ramp

Imagine a ramp outside a school:

        ▲
        │
        │  ↑ 1 metre UP
        │
        └──────────────►
         3 metres ACROSS

The ramp rises 1 metre for every 3 metres of length. The gradient is:

1 3

This is a gentle, safe slope.

✏️ YOUR TURN!

Based on the equation:

y = 1 3 x + 7

Question: What is the gradient of this line?

Write your answer as a fraction with a horizontal vinculum.

Gradient = _____ _____

✅

CHECK YOUR ANSWER

The gradient is the number in front of x.

Looking at the equation:

y = 1 3 x + 7

The number in front of x is:

1 3

So the gradient is one-third!

📊 HOW DID YOU DO?

I can find the gradient from y = mx + c
I know that m is the number next to x
I can read a fraction with a horizontal vinculum (like one-third)
I can explain what gradient means (steepness)

⭐

FUN FACT

The word "GRADIENT" comes from the Latin word "gradiens" meaning "walking".

So the gradient tells you how the line walks — steeply, gently, or flat!

The horizontal line in a fraction is called a VINCULUM. It comes from the Latin word meaning "to bind" or "to connect" — because it binds the numerator and denominator together.

🎉

BONUS CHALLENGE

If the gradient is 13 and the y-intercept is 7, write the equation of the line.

Answer: y = 13 x + 7

(You already had it! 😄)

📐 MATHEMATICS PROMPT | GRADE 8-9 CBC

Topic: Gradient (Slope) of a Straight Line

❓

Question

The equation of a straight line is given as:

y = 1 3 x + 7

Task: Determine the gradient (slope) of this line.

Marks: 1

📝

Worked Example (For Reference)

Example: Find the gradient of y = 2x + 5

Solution:
Compare with y = mx + c
m = 2
Therefore, gradient = 2

Solution Space:

Equation given:

y = 1/3 x + 7

Compare with y = m × x + c

m = ______
c = ______

Therefore, gradient = ______

Answer: 1 3

📊 Rubric

Criterion	Marks
Correct identification of m (the coefficient of x) → 1 mark
Correct answer written as fraction → 1 mark
Total	2 marks

Great work, Mathematician! 🎓🧮

Keep practicing — you're getting better every day! 💪

📐 Understanding The Gradient

📝 Name: ____________________

📅 Date: ____________________

🏫 Class: Grade 8 _____

📐 GRADE 9 MATHS PROMPT (CBC Kenya Curriculum)

MATHS ADVENTURE: Find the Hidden Number!

🎯 THE CHALLENGE
A straight line has a secret code written like this:

y = 1 3 x + 7

Look carefully at the fraction! The number 1 is on top of the number 3 with a horizontal line between them. This horizontal line is called a VINCULUM.

There is a hidden number inside this code. It is called the GRADIENT (or slope).
Your mission: Find the gradient of this line!

🔢 THE FRACTION WITH VINCULUM

The fraction in our equation is written with a horizontal vinculum:

1 3

This means: ONE divided by THREE or ONE part out of THREE equal parts.

📍 POINTS ON THE LINE

Point A: (0, 7) | Point B: (3, 8) | Point C: (6, 9) | Point D: (9, 10)

🚶‍♂️ GRADIENT = RISE ÷ RUN

Every time you move 3 steps to the right, you move 1 step up.

GRADIENT = 1 ÷ 3 = 1 3

✏️ YOUR TURN!

Based on the equation: y = 13 x + 7

Question: What is the gradient of this line?

Gradient = ____ ____

✅ CHECK YOUR ANSWER
The gradient is the number in front of x.
Looking at the equation, the number in front of x is:

1 3

So the gradient is one-third!

⭐ FUN FACT

The word "GRADIENT" comes from the Latin word "gradiens" meaning "walking". The horizontal line in a fraction is called a VINCULUM.

📐 MATHEMATICS PROMPT | GRADE 8-9 CBC

Topic: Gradient (Slope) of a Straight Line

❓ Question
The equation of a straight line is given as: y = 13 x + 7
Task: Determine the gradient (slope) of this line.
Marks: 1

Answer: 13

📊 Rubric

Criterion	Marks
Correct identification of m (the coefficient of x) → 1 mark
Correct answer written as fraction → 1 mark
Total	2 marks

Great work, Mathematician! 🎓🧮

📐 Gradient Of Straight Line

📋 Name: ____________________ 📅 Date: ____________________ 🏫 Class: Grade 8 ___

🧗‍♀️ MATH ADVENTURE: Scaling the Slope!

Today's Mission: A young architect is designing a wheelchair ramp. The ramp follows a straight line rule. Find the slope (gradient) of that line.

🔍 The Problem
The line that represents the ramp is:

                y = 13 x + 7
            

⚠️ The fraction 1/3 means 1 divided by 3.
Stacked form: 13 (1 on top, 3 at the bottom, like a small staircase).

🧭 What is Gradient?

Gradient tells you how steep a line is.

If the line looks like...	Gradient is...
↗️ (going up, fast)	Big number (e.g., 5)
➡️ (flat)	Zero (0)
↘️ (going down)	Negative number (e.g., -2)

Our ramp goes gently upwards. Let's find its gradient.

🖼️ Diagram (Imagine this)

/|
/ |
/ |
/ |
/ |
/ |
/ |
└───────┘
x → 1 step

Every time you move 1 unit across (x + 1), the line rises only a small amount:
Rise = 1 ÷ 3 = about 0.333...

🔢 Step-by-Step to find the Gradient

Step 1 — General line formula
A straight line is written as:
y = (gradient) × x + (y‑intercept)
Or: y = m x + c
• m = gradient (what we want)
• c = y‑intercept (where line meets the y‑axis)

Step 2 — Compare with our equation
Our equation: y = (1/3) x + 7
Compare with y = m x + c:
• Number next to x = 13 → That is m, the gradient.
• The number on its own 7 is the y‑intercept.

Step 3 — Write the gradient
The gradient = 13

🌍 Real-Life Example

A wheelchair ramp with a gradient of 1/3 means:
For every 3 metres you roll forward, you go 1 metre up.
That’s safe, smooth, and gentle – perfect for hospitals, schools, and homes.

📦 Your Answer Box

Question	Your Answer
Gradient of the line `y = (1/3)x + 7`	______ / ______

✅ Correct gradient = 13 (one third)

✅ Did You Get It?

☐ The gradient is the number with the x.
☐ Stacked fraction 13 means 1 divided by 3.
☐ The ramp goes up 1 unit every 3 units across.

🏗️ Well done, mathematician! Now go build that ramp 📐

End of Prompt – CBC Grade 8 | Gradient of a Straight Line | Stacked Fractions