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Year 10 · Algebra & Coordinates

The Algebraist's
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Chapter I · Learn

Solving Linear Equations

An equation is a balance. Whatever you do to one side, you must do to the other. Your job is to get the variable alone on one side.

The golden rule of balance
Whatever operation you perform, apply it to BOTH sides.

Worked Example — One step

Solve: x + 7 = 20

x + 7 − 7 = 20 − 7    (subtract 7 from both sides)

x = 13

Two steps

Solve: 3x + 4 = 19

3x + 4 − 4 = 19 − 4    (subtract 4 first)

3x = 15

3x ÷ 3 = 15 ÷ 3    (then divide by 3)

x = 5

Order matters: undo operations in REVERSE of how they were applied. If the equation was built by "multiply by 3, then add 4", solve by "subtract 4, then divide by 3".

Variable on both sides

Solve: 5x − 3 = 2x + 12

5x − 2x − 3 = 2x − 2x + 12    (subtract 2x from both sides)

3x − 3 = 12

3x = 15    (add 3)

x = 5    (divide by 3)

Chapter II · Learn

Substitution & Simplifying

Before solving, you often need to plug in values for variables, or simplify an expression so it's easier to work with.

Substitution

Evaluate 2a + 3b − c when a = 4, b = −2, c = 1.

= 2(4) + 3(−2) − 1

= 8 + (−6) − 1

= 1

⚠ Common mistake: forgetting brackets around negative numbers. 3 × −2 = −6, not 3−2. Always write 3(−2).

Simplifying — combining like terms

Simplify: 5x + 3 − 2x + 7

= (5x − 2x) + (3 + 7)

= 3x + 10

Like terms have the same variable raised to the same power. x and 3x are like terms. x and x² are NOT.

Expanding brackets

Simplify: 3(x + 4) − 2(x − 5)

= 3x + 12 − 2x + 10    (distribute each bracket)

= x + 22

Watch signs carefully when expanding a negative bracket: −2(x − 5) = −2x + 10, not −2x − 10.

Chapter III · Learn

Inequalities

An inequality is like an equation, but with a range of answers rather than a single value.

<  less than    ≤  less than or equal    >  greater than    ≥  greater than or equal

Reading a number line

On a number line, an open circle ○ means "not included" (< or >), and a filled circle ● means "included" (≤ or ≥).

This diagram shows 2 ≤ x < 6: x is at least 2 (filled), but strictly less than 6 (open).

Solving inequalities

Solve: 3x + 5 > 17

3x > 12    (subtract 5)

x > 4    (divide by 3)

⚠ The sign-flip rule: when you multiply or divide BOTH sides by a negative number, FLIP the inequality sign.
−2x < 10 → x > −5  (divide by −2, flip sign)

Word problem → inequality

Mia has $50. She buys pens at $3 each. She wants to have at least $15 left. How many pens can she buy?

50 − 3n ≥ 15    (remaining ≥ 15)

−3n ≥ −35

n ≤ 35/3    (divide by −3, FLIP)

n ≤ 11.67, so n ≤ 11 pens (whole pens only)

Chapter IV · Learn

Linear Graphs

A linear equation graphs as a straight line. The most useful form is y = mx + c, where m is the gradient (slope) and c is the y-intercept.

Gradient: rise over run

m = (y₂ − y₁) ÷ (x₂ − x₁)  =  rise / run

For a line through (1, 2) and (4, 8):

m = (8 − 2) ÷ (4 − 1) = 6 ÷ 3 = 2

x- and y-intercepts

The y-intercept is where the line crosses the y-axis — set x = 0.
The x-intercept is where the line crosses the x-axis — set y = 0.

For y = 2x − 6:

y-intercept: x=0 → y = −6    so (0, −6)

x-intercept: y=0 → 0 = 2x − 6 → x = 3    so (3, 0)

Parallel & perpendicular

Parallel lines: same gradient.  m₁ = m₂
Perpendicular lines: gradients multiply to −1.  m₁ × m₂ = −1
Equivalently: perpendicular gradient = negative reciprocal.

Line with m = 3 → parallel line has m = 3, perpendicular line has m = −1/3.

Distance and midpoint

Distance between (x₁, y₁) and (x₂, y₂):

d = √((x₂−x₁)² + (y₂−y₁)²)

Midpoint: average the x's, average the y's.

M = ( (x₁+x₂)/2 , (y₁+y₂)/2 )

Chapter V · Learn

Simultaneous Equations

Two equations, two unknowns. The solution is the pair (x, y) that satisfies BOTH at once. Graphically: where the two lines cross.

Method 1 — Substitution

Solve: y = 2x + 1 and 3x + y = 16

Substitute the first into the second:

3x + (2x + 1) = 16

5x + 1 = 16

x = 3

Back-substitute: y = 2(3) + 1 = 7

Solution: (3, 7)

Method 2 — Elimination

Solve: 2x + y = 11 and 3x − y = 4

Add the two equations — the y terms cancel:

(2x + 3x) + (y − y) = 11 + 4

5x = 15 → x = 3

Back-substitute into the first: 2(3) + y = 11 → y = 5

Solution: (3, 5)

When to use which?
• If a variable is already isolated (y = ...), use substitution.
• If coefficients line up to cancel cleanly, use elimination.
• If neither, multiply one or both equations to force a cancellation.

Check your answer

Always substitute (x, y) back into BOTH original equations. If either fails, you made an arithmetic error.

Chapter VI · Learn

Word Problems

The real skill is turning words into algebra. Once the equation is set, the solving is the easy part.

The three-step approach
① DEFINE a variable for the unknown.
② WRITE an equation that captures the words.
③ SOLVE and interpret — does the answer make sense?

Worked Example — Linear wage

Sam earns $180 for 8 hours of work. How much for 14 hours?

Hourly rate = 180 ÷ 8 = $22.50

Equation: Pay = 22.50 × h

For h = 14: Pay = 22.50 × 14 = $315

Pattern problems

Look for the rate of change (the gradient) and the starting value (y-intercept).

A sequence: 5, 8, 11, 14, ... The rate is +3 per step, starting from 5 (at step 1).

Rule: y = 3n + 2    (for step n)

Check: n=1 → 3(1)+2 = 5 ✓

Two-unknown problems

Three apples and two bananas cost $8. Five apples and two bananas cost $12. Find the price of each.

Let a = apple, b = banana

3a + 2b = 8    (1)

5a + 2b = 12    (2)

(2) − (1): 2a = 4 → a = 2

Sub back: 3(2) + 2b = 8 → b = 1

Apple = $2, Banana = $1

Complete

The Algebraist's Mastery

You have mastered all six chapters of the Codex.

Your algebraist's tools:
① Solve equations by balancing both sides
② Substitute values and simplify expressions
③ Solve and graph inequalities (flip the sign!)
④ Work with linear graphs, gradients, intercepts, distance & midpoint
⑤ Solve simultaneous equations by substitution or elimination
⑥ Turn word problems into algebra