What is a linear equation?

A linear equation has the variable mostly in the first power only.

General form:

\(ax + b = 0\)

where \(a \ne 0\).

Examples:

\(2x + 5 = 13\)
\(3x - 7 = 2x + 4\)
\(\dfrac{x}{2} + 3 = 9\)

The graph of a linear equation is a straight line.

\(y = mx + b\)

Main rule

Whatever you do to one side, you must do to the other side.

You can:

  • add/subtract both sides
  • multiply/divide both sides
  • combine like terms
  • move variables to one side and constants to the other

Moving a term to the other side

When a term moves across the equal sign, its sign changes. A positive term becomes negative, and a negative term becomes positive.

\(x + 7 = 12\)

Move \(+7\) to the right side. It becomes \(-7\).

\(x = 12 - 7\)
\(x = 5\)

Another quick example:

\(x - 4 = 9 \Rightarrow x = 9 + 4 \Rightarrow x = 13\)

Linear Equation Example 1

Solve:

\(2x + 5 = 17\)

Step 1: Subtract 5 from both sides.

\(2x + 5 - 5 = 17 - 5\)
\(2x = 12\)

Step 2: Divide both sides by 2.

\(x = 6\)

Answer:

\(x = 6\)

Check:

\(2(6) + 5 = 12 + 5 = 17\)

Correct.

Linear Equation Example 2

Solve:

\(3x - 4 = x + 10\)

Step 1: Move \(x\) to the left side.

\(3x - x - 4 = 10\)
\(2x - 4 = 10\)

Step 2: Add 4 to both sides.

\(2x = 14\)

Step 3: Divide by 2.

\(x = 7\)

Answer:

\(x = 7\)

Check:

Left side:

\(3(7) - 4 = 21 - 4 = 17\)

Right side:

\(7 + 10 = 17\)

Correct.

Linear Equation Example 3

Solve:

\(\dfrac{x}{3} + 5 = 11\)

Step 1: Subtract 5.

\(\dfrac{x}{3} = 6\)

Step 2: Multiply both sides by 3.

\(x = 18\)

Answer:

\(x = 18\)

Check:

\(\dfrac{18}{3} + 5 = 6 + 5 = 11\)

Correct.

Chapter 1

Linear Equations


1
Definition & Standard Form

A linear equation with one variable has the form:

ax + b = c

where a, b, c are constants and x is the unknown variable.

📌 Move all known values to one side and all unknowns to the other.
Example
In 2x + 5 = 3y − 5, treating y as known:

2x = 3y − 5 − 5 → 2x = 3y − 10
x = (3y − 10) / 2
2
Transposition Rules

When a term crosses the equals sign, its operation flips:

+Addition → subtract when moving
Subtraction → add when moving
×Multiplication → divide when moving
÷Division → multiply when moving
x + 7 = 12
x = 12 − 7 → x = 5
x − 4 = 9
x = 9 + 4 → x = 13
5x = 20
x = 20 ÷ 5 → x = 4
x / 3 = 7
x = 7 × 3 → x = 21

Multi-step: (2x + 5) / 3 = 7

  • 1Multiply both sides by 3: 2x + 5 = 21
  • 2Subtract 5: 2x = 16
  • 3Divide by 2: x = 8
⚠️
Special cases:
· Simplifies to x = x → infinitely many solutions
· False equality (e.g. 8 = 3) → no solution

What is a quadratic equation?

A quadratic equation has \(x^2\) as the highest power.

General form:

\(ax^2 + bx + c = 0\)

where \(a \ne 0\).

Examples:

\(x^2 - 5x + 6 = 0\)
\(2x^2 + 3x - 2 = 0\)
\(x^2 = 16\)

The graph of a quadratic equation is a parabola.

The solutions of a quadratic equation are the \(x\)-values where the parabola crosses the \(x\)-axis.

Main methods for solving quadratic equations

There are three common methods:

  1. Factoring
  2. Square root method
  3. Quadratic formula
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Quadratic Equation Example 1 — Factoring

Solve:

\(x^2 - 5x + 6 = 0\)

We need two numbers that:

  • multiply to 6
  • add to \(-5\)

Those numbers are \(-2\) and \(-3\).

So:

\(x^2 - 5x + 6 = (x - 2)(x - 3)\)

Now solve:

\((x - 2)(x - 3) = 0\)

If two factors multiply to zero, at least one of them must be zero.

So:

\(x - 2 = 0\)
\(x = 2\)

or

\(x - 3 = 0\)
\(x = 3\)

Answer:

\(x = 2,\ x = 3\)

Check with \(x = 2\):

\(2^2 - 5(2) + 6 = 4 - 10 + 6 = 0\)

Check with \(x = 3\):

\(3^2 - 5(3) + 6 = 9 - 15 + 6 = 0\)

Correct.

Quadratic Equation Example 2 — Square Root Method

Solve:

\(x^2 = 49\)

Take the square root of both sides.

Important: when solving \(x^2 = a\), we take both positive and negative roots.

\(x = \pm 7\)

Answer:

\(x = -7,\ 7\)

Why both?

Because:

\(7^2 = 49\)

and

\((-7)^2 = 49\)

Quadratic Equation Example 3 — Quadratic Formula

Solve:

\(2x^2 + 3x - 2 = 0\)

Here:

\(a = 2,\quad b = 3,\quad c = -2\)

Use the quadratic formula:

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substitute:

\(x = \dfrac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)}\)

Simplify inside the square root:

\(x = \dfrac{-3 \pm \sqrt{9 + 16}}{4}\)
\(x = \dfrac{-3 \pm \sqrt{25}}{4}\)
\(x = \dfrac{-3 \pm 5}{4}\)

Now split into two answers.

First:

\(x = \dfrac{-3 + 5}{4} = \dfrac{2}{4} = \dfrac{1}{2}\)

Second:

\(x = \dfrac{-3 - 5}{4} = \dfrac{-8}{4} = -2\)

Answer:

\(x = \dfrac{1}{2},\ -2\)
Chapter 2
Chapter 2

Quadratic Equations


1
Definition & Discriminant

A quadratic equation has the form:

ax² + bx + c = 0  (a ≠ 0)
·a = 1: x₁ + x₂ = −b
·a = 1: x₁ · x₂ = c
·a ≠ 1: x₁ + x₂ = −b/a
·a ≠ 1: x₁ · x₂ = c/a

Discriminant: D = b² − 4ac

D > 0
Two distinct real solutions
D = 0
One repeated real solution
D < 0
No real solutions
2
Solving Methods

Method 1 — Factoring

Find two numbers: multiply to c, add to b (when a = 1).

Example — x² − 5x + 6 = 0
p · q = 6, p + q = −5 → −2 and −3
(x − 2)(x − 3) = 0 → x = 2 or x = 3

Method 2 — Quadratic Formula

x = (−b ± √(b² − 4ac)) / 2a
Example — 2x² − 3x − 5 = 0
a = 2, b = −3, c = −5
D = 9 + 40 = 49
x = (3 ± 7) / 4 → x = 2.5 or x = −1

What is a linear inequality?

A linear inequality is like a linear equation, but with inequality signs.

General forms:

\(ax + b \gt 0\)
\(ax + b \lt 0\)
\(ax + b \ge 0\)
\(ax + b \le 0\)

Example:

\(2x + 5 \lt 13\)

Very important rule

When you multiply or divide both sides by a negative number, you must flip the inequality sign.

For example:

\(-2x \lt 6\)

Divide by \(-2\), and flip \(\lt\) to \(\gt\):

\(x \gt -3\)

This is one of the most important inequality rules.

Change the sign and values. The blue shadow shows all \(x\)-values that make the inequality true.

\(mx + b\) < \(0\)

Linear Inequality Example 1

Solve:

\(2x + 5 \lt 17\)

Step 1: Subtract 5.

\(2x \lt 12\)

Step 2: Divide by 2.

\(x \lt 6\)

Answer:

\(x \lt 6\)

This means every number less than 6 works.

Examples that work:

\(x = 5,\quad x = 0,\quad x = -100\)

Check with \(x = 5\):

\(2(5) + 5 = 15 \lt 17\)

Correct.

Linear Inequality Example 2 — Negative Coefficient

Solve:

\(-3x + 4 \ge 16\)

Step 1: Subtract 4.

\(-3x \ge 12\)

Step 2: Divide by \(-3\).

Because we divide by a negative number, flip the sign:

\(x \le -4\)

Answer:

\(x \le -4\)

Check with \(x = -5\):

\(-3(-5) + 4 = 15 + 4 = 19\)
\(19 \ge 16\)

Correct.

Check with \(x = 0\):

\(-3(0) + 4 = 4\)
\(4 \ge 16\)

False.

So \(x \le -4\) makes sense.

Linear Inequality Example 3 — Variables on Both Sides

Solve:

\(5x - 7 \lt 2x + 8\)

Step 1: Move \(2x\) to the left.

\(5x - 2x - 7 \lt 8\)
\(3x - 7 \lt 8\)

Step 2: Add 7.

\(3x \lt 15\)

Step 3: Divide by 3.

\(x \lt 5\)

Answer:

\(x \lt 5\)

Check with \(x = 4\):

Left side:

\(5(4) - 7 = 20 - 7 = 13\)

Right side:

\(2(4) + 8 = 8 + 8 = 16\)
\(13 \lt 16\)

Correct.

Chapter 3
Chapter 3

Linear Inequalities


1
Definition & Cases

Same as linear equations but uses >, <, ≥, ≤ instead of =.

🔁 Golden rule: When multiplying or dividing by a negative number, flip the inequality sign.
Case 1 — Simple   2x + 3 > 7
Subtract 3 → 2x > 4 → Divide by 2 → x > 2
Case 2 — Negative coefficient   −3x + 5 ≤ 2
Subtract 5 → −3x ≤ −3 → Divide by −3 (flip!) → x ≥ 1
Case 3 — Always true   2x + 5 > 2x + 3
Subtract 2x → 5 > 3 ✓ → All real numbers ℝ
Case 4 — No solution   3x + 4 < 3x − 2
Subtract 3x → 4 < −2 ✗ → No solution ∅

What is a quadratic inequality?

A quadratic inequality has \(x^2\) and an inequality sign.

Examples:

\(x^2 - 5x + 6 \gt 0\)
\(x^2 - 4 \le 0\)
\(2x^2 + 3x - 2 \ge 0\)

A quadratic inequality asks:

For which values of \(x\) is the quadratic expression positive, negative, greater than zero, or less than zero?

The graph is a parabola.

Change the sign and values. The blue shadow shows the solution intervals.

\(ax^2 + bx + c\) > \(0\)

The main strategy:

  1. Move everything to one side.
  2. Set the expression equal to zero.
  3. Solve the quadratic equation.
  4. Use the roots to divide the number line into intervals.
  5. Test each interval.
  6. Choose the interval that satisfies the inequality.

Key idea: roots split the number line

Suppose:

\((x - 2)(x - 3) \gt 0\)

The roots are:

\(x = 2,\quad x = 3\)

These divide the number line into three intervals:

\((-\infty, 2),\quad (2, 3),\quad (3, \infty)\)

Then we test each interval.

Quadratic Inequality Example 1

Solve:

\(x^2 - 5x + 6 \gt 0\)

Step 1: Factor.

\(x^2 - 5x + 6 = (x - 2)(x - 3)\)

So:

\((x - 2)(x - 3) \gt 0\)

Step 2: Find roots.

\(x - 2 = 0 \Rightarrow x = 2\)
\(x - 3 = 0 \Rightarrow x = 3\)

Step 3: Split number line.

\((-\infty, 2),\quad (2, 3),\quad (3, \infty)\)

Step 4: Test each interval.

Interval 1: \(x \lt 2\)

Choose \(x = 0\).

\((0 - 2)(0 - 3) = (-2)(-3) = 6\)

Positive. So this interval works.

Interval 2: \(2 \lt x \lt 3\)

Choose \(x = 2.5\).

\((2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25\)

Negative. Does not work.

Interval 3: \(x \gt 3\)

Choose \(x = 4\).

\((4 - 2)(4 - 3) = (2)(1) = 2\)

Positive. Works.

Answer:

\(x \lt 2\text{ or }x \gt 3\)

Interval notation:

\((-\infty, 2) \cup (3, \infty)\)

Important: because the inequality is \(\gt 0\), not \(\ge 0\), we do not include 2 and 3.

Quadratic Inequality Example 2

Solve:

\(x^2 - 4 \le 0\)

Step 1: Factor.

\(x^2 - 4 = (x - 2)(x + 2)\)

So:

\((x - 2)(x + 2) \le 0\)

Step 2: Find roots.

\(x - 2 = 0 \Rightarrow x = 2\)
\(x + 2 = 0 \Rightarrow x = -2\)

Step 3: Split number line.

\((-\infty, -2),\quad (-2, 2),\quad (2, \infty)\)

Step 4: Test each interval.

Interval 1: \(x \lt -2\)

Choose \(x = -3\).

\((-3)^2 - 4 = 9 - 4 = 5\)

Positive. We need \(\le 0\), so it does not work.

Interval 2: \(-2 \lt x \lt 2\)

Choose \(x = 0\).

\(0^2 - 4 = -4\)

Negative. This works.

Interval 3: \(x \gt 2\)

Choose \(x = 3\).

\(3^2 - 4 = 9 - 4 = 5\)

Positive. Does not work.

Step 5: Include endpoints?

The inequality is:

\(\le 0\)

So yes, include the roots \(-2\) and \(2\), because at those points the expression equals zero.

Answer:

\(-2 \le x \le 2\)

Interval notation:

\([-2, 2]\)

Quadratic Inequality Example 3

Solve:

\(2x^2 + 3x - 2 \ge 0\)

Step 1: Factor.

\(2x^2 + 3x - 2 = (2x - 1)(x + 2)\)

So:

\((2x - 1)(x + 2) \ge 0\)

Step 2: Find roots.

\(2x - 1 = 0\)
\(2x = 1\)
\(x = \dfrac{1}{2}\)

and:

\(x + 2 = 0\)
\(x = -2\)

Step 3: Split number line.

\((-\infty, -2),\quad \left(-2, \dfrac{1}{2}\right),\quad \left(\dfrac{1}{2}, \infty\right)\)

Step 4: Test intervals.

Interval 1: \(x \lt -2\)

Choose \(x = -3\).

\((2(-3) - 1)(-3 + 2) = (-7)(-1) = 7\)

Positive. Works.

Interval 2: \(-2 \lt x \lt \dfrac{1}{2}\)

Choose \(x = 0\).

\((2(0) - 1)(0 + 2) = (-1)(2) = -2\)

Negative. Does not work.

Interval 3: \(x \gt \dfrac{1}{2}\)

Choose \(x = 1\).

\((2(1) - 1)(1 + 2) = (1)(3) = 3\)

Positive. Works.

Step 5: Include endpoints?

The inequality is:

\(\ge 0\)

So yes, include the roots.

Answer:

\(x \le -2\text{ or }x \ge \dfrac{1}{2}\)

Interval notation:

\((-\infty, -2] \cup \left[\dfrac{1}{2}, \infty\right)\)
Chapter 4
Chapter 4

Quadratic Inequalities


1
Steps to Solve

Form: ax² + bx + c > 0 (or <, ≥, ≤)

  • 1Rearrange so one side equals 0
  • 2Solve ax² + bx + c = 0 → find roots x₁ and x₂
  • 3Test a point in each interval to find the sign (+/−)
  • 4Select intervals matching the inequality; include endpoints for ≥ or ≤
💡 For > 0 / ≥ 0 → keep positive intervals.  For < 0 / ≤ 0 → keep negative intervals.
2
Worked Examples

Case 1 — x² − 5x + 6 > 0

  • 1Roots: (x − 2)(x − 3) = 0 → x = 2, x = 3
  • 2Test each interval:
IntervalTest pointSign
x < 2x = 1+ positive ✓
2 < x < 3x = 2.5− negative ✗
x > 3x = 4+ positive ✓
x < 2 or x > 3  →  (−∞, 2) ∪ (3, +∞)
Upward parabola inequality diagram

Case 2 — x² − 4x + 3 ≤ 0

  • 1Roots: (x − 1)(x − 3) = 0 → x = 1, x = 3
  • 2Test each interval:
IntervalTest pointSign
x < 1x = 0+ positive
1 < x < 3x = 2− negative ✓
x > 3x = 4+ positive
1 ≤ x ≤ 3  →  [1, 3]
Downward parabola inequality diagram
Case 3 — x² + 1 < 0  →  No solution
x² + 1 ≥ 1 always → can never be < 0 → No solution ∅
Case 4 — x² + 1 ≥ 0  →  All real numbers
x² + 1 always positive → always true → All real numbers ℝ
Quadratic inequality solution diagram Quadratic sign intervals diagram

Big difference between equations and inequalities

Equation

Example:

\(x^2 - 5x + 6 = 0\)

Answer:

\(x = 2,\quad x = 3\)

An equation usually gives exact values.

Inequality

Example:

\(x^2 - 5x + 6 \gt 0\)

Answer:

\(x \lt 2\text{ or }x \gt 3\)

An inequality usually gives ranges of values.

Linear vs Quadratic: Main Pattern

Type Form Number of solutions
Linear equation \(ax + b = 0\) usually one solution
Quadratic equation \(ax^2 + bx + c = 0\) usually two, one, or no real solutions
Linear inequality \(ax + b \gt 0\) usually one interval
Quadratic inequality \(ax^2 + bx + c \gt 0\) usually one or two intervals

Very important SAT-style tricks

1. For linear equations, isolate \(x\)

Example:

\(4x - 9 = 15\)

Add 9:

\(4x = 24\)

Divide by 4:

\(x = 6\)

Fast.

2. For quadratic equations, first try factoring

Example:

\(x^2 + 7x + 12 = 0\)

Ask:

What two numbers multiply to 12 and add to 7?

Answer:

\(3,\quad 4\)

So:

\((x + 3)(x + 4) = 0\)
\(x = -3,\quad -4\)

3. For quadratic inequalities, use sign chart

Example:

\((x - 1)(x + 4) \lt 0\)

Roots:

\(x = 1,\quad x = -4\)

A product of two factors is negative when the signs are different.

That happens between the roots:

\(-4 \lt x \lt 1\)

Quick summary

Linear equation

\(2x + 3 = 11\)

Solution:

\(x = 4\)

Quadratic equation

\(x^2 - 5x + 6 = 0\)

Solution:

\(x = 2,\quad 3\)

Linear inequality

\(2x + 3 \lt 11\)

Solution:

\(x \lt 4\)

Quadratic inequality

\(x^2 - 5x + 6 \gt 0\)

Solution:

\(x \lt 2\text{ or }x \gt 3\)

The main mental model:

Equations ask: “Where is it exactly equal?”
Inequalities ask: “Where is it bigger or smaller?”
Chapter 5
Chapter 5

Conclusion & Similarities


1
How These Topics Connect

All four topics ask the same first question: what values of x make the statement true?

1Equations use = and usually give exact solution points.
2Inequalities use >, <, ≥, ≤ and usually give intervals.
3Linear forms have x to the first power, so their graph is a straight line.
4Quadratic forms have x2, so their graph is a parabola.
Linear Equation vs. Linear Inequality
Both are solved by isolating x. The inequality has one extra rule: when you multiply or divide by a negative number, the sign flips.
Quadratic Equation vs. Quadratic Inequality
First find the roots of the related quadratic equation. Then, for inequalities, choose the intervals where the expression has the needed sign.
Big Similarity
Each topic depends on balance, inverse operations, and checking the final answer against the original statement.