Master Algebra from zero to fluent

Unit 01

Variables & expressions

Algebra is the language of patterns. Instead of writing "a number plus 5 equals 8," we write x + 5 = 8. The letter x is a variable — a placeholder for an unknown value we want to find.

An expression combines numbers, variables, and operations: 3x + 2. An equation adds an equals sign, making a complete statement: 3x + 2 = 14.

Variable

x, y, n, a, b …

A letter that stands for an unknown number. Any letter can be used.

Coefficient

3x → coefficient is 3

The number multiplied by the variable. In 7y, the coefficient is 7.

Constant

4x + 9 → constant is 9

A number with no variable attached. Its value never changes.

The golden rule

Same operation on both sides

Whatever you do to one side of an equation, you must do to the other. Always.

Combining like terms

Like terms share the same variable and power. You can add or subtract them freely. Terms with different variables or powers cannot be combined.

Example 1 — Basic: simplify 5x + 3y − 2x + 7y

1

Identify x terms: 5x and −2x

2

Identify y terms: 3y and 7y

3

Combine x terms: 5x − 2x = 3x

4

Combine y terms: 3y + 7y = 10y

✓

Answer: 3x + 10y

Example 2 — With powers: simplify 4x² + 3x − x² + 2x − 7

1

Group x² terms: 4x² − x² = 3x²

2

Group x terms: 3x + 2x = 5x

3

Constants stay: −7 (no like term to combine with)

✓

Answer: 3x² + 5x − 7

Example 3 — Distributive property: expand 3(2x + 4) − 2(x − 5)

1

Distribute 3: 6x + 12

2

Distribute −2 (note the sign): −2x + 10

3

Combine: 6x − 2x + 12 + 10 = 4x + 22

✓

Answer: 4x + 22

Unit 02

Linear equations

A linear equation has one variable raised to the power of 1. To solve it, isolate the variable by applying inverse operations in reverse order — undo addition/subtraction first, then multiplication/division.

Undo addition

x + 5 = 12 → subtract 5

The inverse of adding 5 is subtracting 5 from both sides.

Undo multiplication

4x = 20 → divide by 4

The inverse of multiplying by 4 is dividing both sides by 4.

Undo division

x/3 = 6 → multiply by 3

The inverse of dividing by 3 is multiplying both sides by 3.

Verify your answer

Substitute back & check

Always plug your answer back into the original equation to confirm it works.

Example 1 — One step: solve x + 13 = 20

1

Subtract 13 from both sides: x = 20 − 13

✓

Answer: x = 7 — Verify: 7 + 13 = 20 ✓

Example 2 — Two steps: solve 5x − 3 = 17

1

Add 3 to both sides: 5x = 20

2

Divide both sides by 5: x = 4

✓

Verify: 5(4) − 3 = 20 − 3 = 17 ✓

Example 3 — Fraction: solve x/3 + 7 = 10

1

Subtract 7 from both sides: x/3 = 3

2

Multiply both sides by 3: x = 9

✓

Verify: 9/3 + 7 = 3 + 7 = 10 ✓

Example 4 — Variables on both sides: solve 5x + 4 = 3x + 14

1

Move all x terms to one side — subtract 3x: 2x + 4 = 14

2

Subtract 4: 2x = 10

3

Divide by 2: x = 5

✓

Verify: 5(5)+4 = 29 and 3(5)+14 = 29 ✓

Example 5 — Brackets: solve 4(2x − 3) = 20

1

Expand brackets: 8x − 12 = 20

2

Add 12: 8x = 32

3

Divide by 8: x = 4

✓

Verify: 4(2×4 − 3) = 4(5) = 20 ✓

Unit 03

Quadratic equations

A quadratic equation has the form ax² + bx + c = 0. Because the variable is squared, there can be up to two solutions (roots). Three methods exist; choose the fastest one for the problem.

Factoring

(x + p)(x + q) = 0

Find two numbers that multiply to c and add to b. Fastest when the numbers work out cleanly.

Quadratic formula

x = (−b ± √(b²−4ac)) / 2a

Always works, even when factoring is impossible. Memorise this formula.

The discriminant (Δ)

Δ = b² − 4ac

Δ > 0: two real roots. Δ = 0: one root (repeated). Δ < 0: no real roots.

Completing the square

(x + n)² = k

Rearrange to this form, then take ±√ of both sides. Foundation of the quadratic formula.

Example 1 — Factoring: solve x² − 5x + 6 = 0

1

Need two numbers that multiply to 6 and add to −5: −2 and −3

2

Write factored form: (x − 2)(x − 3) = 0

3

Each bracket = 0: x = 2 or x = 3

✓

Check: (2)²−5(2)+6 = 0 ✓ (3)²−5(3)+6 = 0 ✓

Example 2 — Quadratic formula: solve 2x² + 3x − 2 = 0

1

Identify: a=2, b=3, c=−2

2

Discriminant: Δ = 9 − 4(2)(−2) = 9 + 16 = 25

3

Apply formula: x = (−3 ± √25) / 4 = (−3 ± 5) / 4

4

x₁ = (−3+5)/4 = 0.5 x₂ = (−3−5)/4 = −2

✓

Roots: x = 0.5 and x = −2

Example 3 — Completing the square: solve x² + 6x + 5 = 0

1

Move constant: x² + 6x = −5

2

Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4

3

Write as perfect square: (x + 3)² = 4

4

Take square root: x + 3 = ±2

✓

Roots: x = −1 and x = −5

Example 4 — No real roots: solve x² + x + 1 = 0

1

Compute discriminant: Δ = 1² − 4(1)(1) = 1 − 4 = −3

✓

Δ < 0 → No real roots. This parabola does not cross the x-axis.

Unit 04

Graphing functions

Every equation can be drawn as a picture. Linear equations y = mx + b make straight lines. Quadratic equations y = ax² + bx + c make parabolas — U-shapes that open up or down.

Slope (m)

rise ÷ run

Positive: line rises left to right. Negative: falls. Zero: horizontal. Undefined: vertical.

y-intercept (b)

point (0, b)

Where the line crosses the y-axis. Set x = 0 in any equation to find it.

Vertex of a parabola

x = −b / 2a

The turning point — lowest point if a > 0, highest if a < 0.

x-intercepts (roots)

Set y = 0 and solve

Where the curve crosses the x-axis. A parabola can have 0, 1, or 2 of them.

Example 1 — Plot y = 2x − 3: find key points

1

Slope: m = 2 — rises 2 units for every 1 unit right

2

y-intercept: set x=0 → y = −3 → point (0, −3)

3

x-intercept: set y=0 → 0 = 2x − 3 → x = 1.5 → point (1.5, 0)

✓

Plot (0, −3) and (1.5, 0), draw a line through them

Example 2 — Parabola y = x² − 4x + 3: find vertex and roots

1

Vertex x-coordinate: x = −b/2a = 4/2 = 2

2

Vertex y-coordinate: y = 4 − 8 + 3 = −1 → vertex (2, −1)

3

Roots: factor (x−1)(x−3) = 0 → x = 1 and x = 3

✓

Parabola opens up, vertex at (2,−1), crosses x-axis at x=1 and x=3

Unit 05

Systems of equations

A system is two or more equations sharing the same variables. The solution is the point (x, y) that satisfies all equations at once — where the lines cross on a graph.

Substitution

Isolate y, substitute

Solve one equation for y, plug into the other. Best when one variable is already isolated.

Elimination

Add/subtract equations

Multiply equations so one variable cancels when you add or subtract them.

One solution

Lines intersect once

Different slopes → lines cross at exactly one point (x, y).

No solution / infinite

Parallel / identical lines

Same slope, different intercepts → no solution. Same line → infinite solutions.

Example 1 — Substitution: y = 2x + 1 and 3x + y = 16

1

y is isolated in equation 1: y = 2x + 1

2

Substitute into equation 2: 3x + (2x + 1) = 16

3

Simplify: 5x = 15 → x = 3

4

Find y: y = 2(3) + 1 = 7

✓

Solution: (3, 7)

Example 2 — Elimination: 2x + y = 10 and 2x − y = 2

1

Subtract eq 2 from eq 1: 2y = 8 → y = 4

2

Substitute y=4 into eq 1: 2x + 4 = 10 → x = 3

✓

Solution: (3, 4)

Example 3 — Elimination with multiplication: 3x + 2y = 12 and 5x − 4y = 2

1

Multiply eq 1 by 2: 6x + 4y = 24

2

Add to eq 2: 11x = 26 → x = 26/11 ≈ 2.36

3

Substitute back: 3(26/11) + 2y = 12 → y = 21/11 ≈ 1.91

✓

Solution: (26/11, 21/11) — not all systems have integer solutions!

Unit 06

Inequalities

An inequality uses <, >, ≤, or ≥ instead of =. The solution is a range of values. You solve it exactly like an equation — with one critical rule that most students get wrong.

    The flip rule: When you multiply or divide both sides by a negative number, the inequality sign must reverse direction. Example: if −x > 3, then x < −3.
  

Example 1 — Basic: solve 3x + 5 > 14

1

Subtract 5: 3x > 9

2

Divide by 3 (positive — no flip): x > 3

✓

Solution: all values of x greater than 3. Open circle at 3, shade right.

Example 2 — Flip rule: solve −2x + 3 > 7

1

Subtract 3: −2x > 4

2

Divide by −2 → flip the sign!: x < −2

✓

Open circle at −2, shade left. All values less than −2.

Example 3 — Compound inequality: solve −3 ≤ 2x + 1 < 9

1

Subtract 1 from all three parts: −4 ≤ 2x < 8

2

Divide all parts by 2: −2 ≤ x < 4

✓

Solution: x ∈ [−2, 4) — closed circle at −2, open circle at 4.

Example 4 — Inequality with brackets: solve 2(x − 3) ≤ 4x + 2

1

Expand: 2x − 6 ≤ 4x + 2

2

Subtract 2x: −6 ≤ 2x + 2

3

Subtract 2: −8 ≤ 2x

4

Divide by 2: x ≥ −4

✓

Solution: all values of x ≥ −4. Closed circle at −4, shade right.

Final exam

20-question graded test

Answer all 20 questions, then submit. You will receive your score, a grade, and a full explanation for every question — including exactly why the wrong answers are wrong.

0 / 20 answered

Master Algebrafrom zero to fluent