What is a Fractional Equation?
A fractional equation is an equation that contains fractions with variables in the denominator.
For example:
\frac{x + 2}{3} = \frac{2x – 1}{5}
or
The goal is to find the value(s) of the variable that make the equation true.
Steps for Solving Fractional Equations
- Find the denominators — list all the denominators in the equation.
- Find the LCM (Lowest Common Multiple) of all the denominators.
- Multiply through by the LCM to eliminate fractions.
- Simplify the resulting equation (it should now be free of denominators).
- Solve the equation as a normal algebraic equation.
- Check for restrictions — values that make any denominator zero are not allowed (they make the equation undefined).
Example 1
Solve:
\frac{x + 1}{2} = \frac{3x – 5}{4}
Step 1: Denominators are 2 and 4.
LCM = 4
Step 2: Multiply through by 4:
4 \times \frac{x + 1}{2} = 4 \times \frac{3x – 5}{4}
What is a Quadratic Equation?
A quadratic equation is an equation in which the highest power of the variable is 2.
It has the general form:
ax^2 + bx + c = 0
where
- (a, b, c) are real numbers, and
- (a \neq 0).
Examples
- (x^2 – 5x + 6 = 0)
- (2x^2 + 3x – 5 = 0)
- (3x^2 = 12x – 9)
Methods of Solving Quadratic Equations
There are 4 main methods:
Factorization Method
Example:
x^2 – 5x + 6 = 0 [/katex]
Find two numbers that multiply to 6 and add to -5 → (-2) and (-3)
x^2 – 2x – 3x + 6 = 0
x(x – 2) – 3(x – 2) = 0
(x – 2)(x – 3) = 0
x = 2 or x = 3
Completing the Square
Example:
x^2 + 6x + 5 = 0
Move the constant to the right:
x^2 + 6x = -5
Take half of the coefficient of (x), square it:
\left(\frac{6}{2}\right)^2 = 9
Add 9 to both sides:
[katext]
x^2 + 6x + 9 = -5 + 9
[/katex]
(x + 3)^2 = 4
x + 3 = \pm 2
x = -3 \pm 2
x = -1 or x = -5
Using the Quadratic Formula
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}Example:
2x^2 + 3x – 2 = 0
Here, (a = 2), (b = 3), (c = -2)
x = \frac{-3 \pm \sqrt{3^2 – 4(2)(-2)}}{2(2)}
x = \frac{-3 \pm \sqrt{9 + 16}}{4}
x = \frac{-3 \pm 5}{4}
<strong>x = (\frac{1}{2})</strong> or <strong>x = -2
Overview
Coordinate geometry, also called analytic geometry, is the study of geometric figures using a coordinate system. It helps us represent points, lines, and shapes on the Cartesian plane using algebraic equations.
Module 1: The Cartesian Plane
Lesson Content
The Cartesian plane consists of two perpendicular lines called axes:
– The x-axis (horizontal)
– The y-axis (vertical)
These axes intersect at a point called the origin (0, 0). Each point on the plane is represented as an ordered pair (x, y).
Example
Plot the points A(2, 3), B(–1, 2), and C(0, –3) on the Cartesian plane.
– A is 2 units right and 3 units up.
– B is 1 unit left and 2 units up.
– C is on the y-axis, 3 units below the origin.
Quick Quiz (MCQs)
- The point (0, 0) is called the:
A. Midpoint
B. Origin
C. Centre
D. Pole
Answer: B - The x-axis and y-axis divide the plane into how many quadrants?
A. Two
B. Three
C. Four
D. Five
Answer: C
Assignment
Identify the quadrant in which the following points lie:
(a) (3, 2)
(b) (–4, 1)
(c) (–2, –3)
(d) (5, –1)
Module 2: Distance Between Two Points
Lesson Content
The distance (d) between two points A(x₁, y₁) and B(x₂, y₂) is given by:
d = √((x₂ – x₁)² + (y₂ – y₁)²)
Example
Find the distance between P(2, 3) and Q(7, 11):
d = √((7 – 2)² + (11 – 3)²) = √(25 + 64) = √89
Quick Quiz (MCQs)
- The distance between (1, 2) and (4, 6) is:
A. 4
B. 5
C. √20
D. √25
Answer: B
Assignment
Find the distance between the following pairs of points:
(a) (2, 5) and (6, 9)
(b) (–1, –3) and (4, 2)
Module 3: Midpoint of a Line Segment
Lesson Content
The midpoint (M) of a line joining A(x₁, y₁) and B(x₂, y₂) is:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Example
Find the midpoint of A(2, 3) and B(6, 11):
M = ((2 + 6)/2, (3 + 11)/2) = (4, 7)
Quick Quiz (MCQs)
- The midpoint of (2, 4) and (6, 8) is:
A. (8, 12)
B. (2, 8)
C. (4, 6)
D. (3, 5)
Answer: C
Assignment
Find the midpoint between (–3, 5) and (7, –9).
Module 4: Gradient (Slope) of a Line
Lesson Content
The gradient (m) of a line passing through points A(x₁, y₁) and B(x₂, y₂) is:
m = (y₂ – y₁)/(x₂ – x₁)
If the line rises as it moves right, the slope is positive.
If it falls as it moves right, the slope is negative.
Example
Find the gradient of a line through (1, 2) and (4, 8):
m = (8 – 2)/(4 – 1) = 6/3 = 2
Quick Quiz (MCQs)
- The gradient of the line joining (2, 3) and (5, 9) is:
A. 1
B. 2
C. 3
D. 4
Answer: B
Assignment
Find the slope of the line passing through (–2, –1) and (3, 4).
Module 5: Equation of a Straight Line
Lesson Content
The general form of a straight-line equation is:
y = mx + c
where:
– m = gradient
– c = y-intercept
If the line passes through (x₁, y₁) with slope m, then:
y – y₁ = m(x – x₁)
Example
Find the equation of a line passing through (2, 3) with gradient 2:
y – 3 = 2(x – 2) ⇒ y = 2x – 1
Quick Quiz (MCQs)
- The line passing through (0, 4) with slope 3 is:
A. y = 3x + 4
B. y = 3x – 4
C. y = 4x + 3
D. y = 4x – 3
Answer: A
Assignment
Find the equation of a line passing through (1, 2) and (3, 6).
Module 6: Parallel and Perpendicular Lines
Lesson Content
Two lines are parallel if they have equal gradients.
Two lines are perpendicular if the product of their gradients = –1.
Example
If line 1 has slope 2, then:
– Any parallel line has slope 2.
– Any perpendicular line has slope –½.
Quick Quiz (MCQs)
- Which of the following lines is perpendicular to y = 2x + 3?
A. y = 2x – 4
B. y = –½x + 1
C. y = ½x – 1
D. y = –2x + 4
Answer: B
Assignment
Find the equation of a line perpendicular to y = 3x + 2 that passes through (0, 4).