Simultaneous Equations

Grade 12 Math - Simultaneous Equations

Lesson Objectives

Solve systems of two linear equations using substitution and elimination methods.
Solve systems of one linear and one quadratic equation algebraically.
Apply simultaneous equations to real-life problems including geometry and finance.

Lesson Introduction

Simultaneous equations involve finding the values of variables that satisfy two or more equations at the same time. These systems can be:

Linear-Linear: both equations are linear.
Linear-Quadratic: one equation is linear, the other is quadratic.

This lesson explores strategies for solving both types, including advanced word problems.

Core Lesson Content

Worked Example

Solving Two Linear Equations

Common methods:

Substitution Method: Solve one equation for one variable and substitute into the other.
Elimination Method: Multiply equations to eliminate one variable and solve for the other.

Example 1 (Substitution): Solve

x + y = 5

and

2x - y = 4

From the first:

y = 5 - x

Substitute into second:

2x - (5 - x) = 4 \Rightarrow 2x - 5 + x = 4 \Rightarrow 3x = 9 \Rightarrow x = 3

Then

y = 5 - 3 = 2

Solution:

x = 3, y = 2

Example 2 (Elimination): Solve

3x + 2y = 12

and

2x - 2y = 4

Add equations:

5x = 16 \Rightarrow x = \frac{16}{5}

Substitute into first:

3 \cdot \frac{16}{5} + 2y = 12 \Rightarrow \frac{48}{5} + 2y = 12

2y = 12 - \frac{48}{5} = \frac{60 - 48}{5} = \frac{12}{5} \Rightarrow y = \frac{6}{5}

Solution:

x = \frac{16}{5}, y = \frac{6}{5}

Solving Linear-Quadratic Systems

These occur when one equation is linear and the other is quadratic. Use substitution.

Example 3: Solve

y = 2x + 1

and

y = x^2 + x - 2

Substitute the first into the second:

2x + 1 = x^2 + x - 2

0 = x^2 - x - 3

Use quadratic formula:

x = \frac{1 \pm \sqrt{(-1)^2 + 4 \cdot 1 \cdot 3}}{2} = \frac{1 \pm \sqrt{13}}{2}

Find

y

from

y = 2x + 1

Solutions:

\left( \frac{1 \pm \sqrt{13}}{2},\ 2\left(\frac{1 \pm \sqrt{13}}{2}\right) + 1 \right)

Example 4: Solve

x + y = 7

and

x^2 + y^2 = 25

From the first:

y = 7 - x

Substitute into second:

x^2 + (7 - x)^2 = 25

x^2 + 49 - 14x + x^2 = 25 \Rightarrow 2x^2 - 14x + 24 = 0

Solve:

x = \frac{14 \pm \sqrt{(-14)^2 - 4(2)(24)}}{2 \cdot 2} = \frac{14 \pm \sqrt{196 - 192}}{4} = \frac{14 \pm 2}{4}

x = 4, x = 3

, then

y = 3, y = 4

Solutions:

(4, 3) \text{ and } (3, 4)

Word Problem Involving Linear-Quadratic System

Example 5: The sum of a number and its square is 42. The number and another number add to 9. Find the numbers.
Let

x

be the first number,

y

be the second.

x^2 + x = 42

and

x + y = 9 \Rightarrow y = 9 - x

Solve

x^2 + x - 42 = 0 \Rightarrow (x - 6)(x + 7) = 0 \Rightarrow x = 6 \text{ or } -7

Then

y = 3 \text{ or } 16

Solutions:

(6, 3) \text{ and } (-7, 16)

Exercises

Solve the system: $3x - y = 7$ , $x + y = 5$
[WAEC] Solve: $x + y = 4$ , $x^2 + y^2 = 10$ [Past Question]
[NECO] Solve: $x + 2y = 9$ , $x^2 + y^2 = 41$ [Past Question]
Solve: $y = 3x - 2$ , $x^2 + y^2 = 13$
The sum of a number and twice another is 10. Their squares add to 58.
[WAEC] Find the points of intersection of $y = x^2$ and $y = 2x + 3$ [Past Question]
A rectangle has length $x$ and width $y$ . If $x + y = 10$ and $xy = 21$ , find $x$ and $y$ .
Solve: $y = x^2 + 1$ , $x^2 + y^2 = 10$
The sum and product of two numbers are 7 and 10 respectively. Find the numbers.
A line intersects a parabola at two points. If $y = x + 1$ and $y = x^2 + 2x$ , find the points of intersection.

Conclusion/Recap

Solving simultaneous equations—both linear and linear-quadratic—requires solid algebraic skills and logical reasoning. The substitution and elimination methods allow us to solve even complex systems, while real-life problems require modeling and careful interpretation. Practice is essential to mastering these techniques.