Introduction to Simultaneous Equations

Simultaneous equations involve solving two or more equations together to find common solutions.

In this chapter, we deal with:

• One linear equation: a straight-line relationship, e.g. y = mx + c

• One non-linear equation: often a quadratic, circle or curved function, e.g. y = x² − 3

The solutions represent points where the graphs intersect.

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Key Concepts

Linear Equation

General form:

y = mx + c

Where:

• m = slope/gradient

• c = y-intercept

Example

y = 2x + 1

A line with:

• gradient = 2

• y-intercept = 1

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Non-Linear Equation

Common types:

• Quadratic: y = ax² + bx + c

• Circle: x² + y² = r²

• Cubic: y = x³

• Root graph: y = √x

Example

y = x² − 3

A parabola opening upward with vertex at (0, −3)

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Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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Solution of Simultaneous Equations

Solutions are intersection points of graphs.

Possible cases:

• One point → graphs touch once

• Two points → graphs intersect twice

• No point → graphs do not intersect

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Methods of Solving

Substitution Method

Steps

1. Make one equation the subject

2. Substitute into second equation

3. Solve resulting equation

4. Substitute x-values back to find y-values

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Elimination Method

• Align terms carefully

• Add/subtract equations

• Eliminate one variable

Usually used when x² and y² appear together.

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Graphical Method

1. Draw both graphs

2. Locate intersection points

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Example 1 — Quadratic and Linear

Given:

• y = 2x + 1

• y = x² − 3

Step 1 — Substitute

2x + 1 = x² − 3

Rearrange:

x² − 2x − 4 = 0

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Step 2 — Solve Quadratic

Using quadratic formula:

x = (−b ± √(b² − 4ac))/2a

Where:

• a = 1

• b = −2

• c = −4

Substitute:

x = [−(−2) ± √((−2)² − 4(1)(−4))]/2(1)

x = [2 ± √(4 + 16)]/2

x = [2 ± √20]/2

x = 1 ± √5

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Step 3 — Find y-values

Using y = 2x + 1

For x₁ = 1 + √5:

y₁ = 2(1 + √5) + 1

For x₂ = 1 − √5:

y₂ = 2(1 − √5) + 1

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Example 2 — Circle and Line

Given:

• x² + y² = 25

• y = 2x + 3

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Step 1 — Substitute

x² + (2x + 3)² = 25

Expand:

x² + 4x² + 12x + 9 = 25

Combine terms:

5x² + 12x − 16 = 0

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Step 2 — Solve Quadratic

Solve for x using factorisation or quadratic formula.

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Step 3 — Find y-values

Substitute x-values into:

y = 2x + 3

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Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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Nature of Solutions

Using discriminant:

Δ = b² − 4ac

Cases

• Δ > 0 → two distinct solutions

• Δ = 0 → one repeated solution

• Δ < 0 → no real solutions

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Graphical Interpretation

Graphs help visualize:

• intersections

• tangency

• no intersection

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Introduction to Quadratic Functions

General form:

y = ax² + bx + c

Where:

• a, b, c are constants

• a ≠ 0

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Shape of Graph

• a > 0 → opens upward

• a < 0 → opens downward

Graph is called a parabola.

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Maximum and Minimum Values

• a > 0 → minimum point

• a < 0 → maximum point

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Key Features

Vertex

Vertex = (h, k)

Formula for h

h = −b/2a

Formula for k

k = f(h)

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Axis of Symmetry

x = h

or

x = −b/2a

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Intercepts

y-intercept

At x = 0:

y = c

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x-intercepts

Solve:

ax² + bx + c = 0

Using:

x = (−b ± √(b² − 4ac))/2a

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Example 1 — Upward Opening Parabola

Given:

y = 2x² − 4x + 3

Identify coefficients

• a = 2

• b = −4

• c = 3

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Find h

h = −(−4)/2(2)

= 1

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Find k

k = 2(1)² − 4(1) + 3

= 1

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Vertex

(1, 1)

Minimum value = 1

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Example 2 — Downward Opening Parabola

Given:

y = −x² + 4x + 1

Coefficients

• a = −1

• b = 4

• c = 1

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Find h

h = −4/2(−1)

= 2

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Find k

k = −(2)² + 4(2) + 1

= 5

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Vertex

(2, 5)

Maximum value = 5

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Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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Applications of Quadratics

Optimization

Used for:

• maximizing profit

• minimizing cost

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Physics

Projectile motion follows parabolic paths.

Maximum height = vertex.

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Engineering

Used in:

• bridge design

• arches

• satellite dishes

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Quadratic Functions and the Line y = x

Line y = x:

• diagonal line

• passes through origin

• gradient = 1

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Intersection with Quadratic

Given:

y = x

and

y = f(x)

Set equal:

x = f(x)

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Example

f(x) = x² − 4x + 3

Set equal:

x = x² − 4x + 3

Rearrange:

x² − 5x + 3 = 0

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Solve

x = [5 ± √(25 − 12)]/2

x = (5 ± √13)/2

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Nature of Intersections

Using discriminant:

Δ = (b − 1)² − 4ac

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Cases

• Δ > 0 → two intersections

• Δ = 0 → tangent

• Δ < 0 → no intersection

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Quadratic Inequalities

General forms:

• ax² + bx + c > 0

• ax² + bx + c ≥ 0

• ax² + bx + c < 0

• ax² + bx + c ≤ 0

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Solving Quadratic Inequalities

Step 1

Write in standard form.

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Step 2

Solve corresponding quadratic equation.

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Step 3

Use roots to divide number line.

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Step 4

Test intervals.

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Step 5

Write final solution.

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Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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Example

Solve:

x² − 3x − 4 > 0

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Step 1 — Factorise

(x − 4)(x + 1) > 0

Roots:

• x = 4

• x = −1

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Step 2 — Test Intervals

Intervals:

• (−∞, −1)

• (−1, 4)

• (4, ∞)

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Step 3 — Final Solution

x ∈ (−∞, −1) ∪ (4, ∞)

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Quadratic Equations

General form:

ax² + bx + c = 0

Where:

a ≠ 0

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Methods of Solving

Factorisation

Example:

x² − 5x + 6 = 0

(x − 2)(x − 3) = 0

Roots:

• x = 2

• x = 3

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Completing the Square

Example:

x² + 6x + 5 = 0

Rewrite:

(x + 3)² − 4 = 0

(x + 3)² = 4

x = −3 ± 2

Roots:

• x = −1

• x = −5

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Quadratic Formula

x = (−b ± √(b² − 4ac))/2a

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Discriminant

Δ = b² − 4ac

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Nature of Roots

• Δ > 0 → two real roots

• Δ = 0 → repeated root

• Δ < 0 → complex roots

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Complex Roots Example

x² + 2x + 5 = 0

Δ = 2² − 4(1)(5)

= −16

Since Δ < 0:

x = −1 ± 2i

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Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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Intersection of a Line and Curve

Intersection points satisfy BOTH equations simultaneously.

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General Steps

1. Write equations clearly

2. Substitute line into curve

3. Form polynomial

4. Solve for x

5. Find y-values

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Example

Given:

• y = 2x + 1

• y = x² − 3x + 2

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Step 1

2x + 1 = x² − 3x + 2

Rearrange:

x² − 5x + 1 = 0

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Step 2 — Solve

x = [5 ± √21]/2

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Step 3 — Find y-values

Using y = 2x + 1:

For x₁ = (5 + √21)/2:

y₁ = 5 + √21

For x₂ = (5 − √21)/2:

y₂ = 5 − √21

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Tangent Case

If Δ = 0:

• line touches curve once

• one repeated solution

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Applications

Physics

Collision points and equilibrium points.

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Engineering

Stress intersections and structural modelling.

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Optimization

Finding critical points.

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Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.