IGCSE Maths · Algebra and Graphs · Core & Extended

Equations I: Linear Equations, Simultaneous Equations and Changing the Subject

An equation is a statement that two expressions are equal — solving it means finding the value (or values) that make that true. This note covers the shared foundation for both tiers: constructing equations from words, solving linear equations, solving simultaneous linear equations, and rearranging formulas. Quadratic equations, fractional equations, and non-linear simultaneous equations are Extended-only and get their own note.

Why this matters Every formula you’ve used across every topic so far — density, simple interest, speed, area — is an equation waiting to be solved for whichever quantity you don’t yet know. Being fluent at rearranging and solving is what turns “I know the formula” into “I can actually answer the question.”

1. Constructing Expressions, Equations and Formulas

Many exam questions describe a situation in words and expect you to translate it into algebra before solving anything.

Worked example A number is multiplied by 3, then 4 is added, giving a result of 19. Write an equation, and solve it.
Let the number be x: 3x + 4 = 19
3x = 15 → x = 5
Try it — Constructing Equations
  1. A number is doubled, then 7 is subtracted, giving 15. Write and solve the equation.
  2. Three consecutive whole numbers sum to 72. Write an equation using n for the smallest number, and solve it.
Show answers
  1. 2x − 7 = 15 → x = 11
  2. n + (n+1) + (n+2) = 72 → 3n + 3 = 72 → n = 23 (so the numbers are 23, 24, 25)

2. Solving Linear Equations in One Unknown

Keep the equation balanced — whatever you do to one side, do to the other. Aim to isolate the unknown on its own.

Worked example — basic Solve 3x + 4 = 10.
3x = 6 → x = 2
Worked example — brackets and the unknown on both sides Solve 5 − 2x = 3(x + 7).
Expand the bracket: 5 − 2x = 3x + 21
Collect x terms on one side: 5 − 21 = 3x + 2x
−16 = 5x → x = −3.2
Worked example — a fraction in the equation Solve (x + 2)/3 = 5.
Multiply both sides by 3: x + 2 = 15
x = 13
Try it — Linear Equations
  1. Solve 4x − 5 = 19.
  2. Solve 7 − 3x = 2(x − 4).
  3. Solve (2x − 1)/5 = 3.
  4. Solve 4(x + 3) = 2(x + 9).
Show answers
  1. x = 6
  2. 7 − 3x = 2x − 8 → 15 = 5x → x = 3
  3. 2x − 1 = 15 → x = 8
  4. 4x + 12 = 2x + 18 → 2x = 6 → x = 3

3. Simultaneous Equations — Substitution Method

When one equation is already (or easily) rearranged to make one letter the subject, substitute that expression into the other equation.

Worked example Solve: y = 2x + 1   and   3x + y = 16.
Substitute the first equation into the second: 3x + (2x + 1) = 16
5x + 1 = 16 → 5x = 15 → x = 3
Substitute back: y = 2(3) + 1 = 7
Solution: x = 3, y = 7
Try it — Substitution Method
  1. Solve: y = x + 4 and 2x + y = 19.
  2. Solve: y = 3x − 2 and 4x − y = 10.
Show answers
  1. 2x + (x+4) = 19 → 3x = 15 → x = 5, y = 9
  2. 4x − (3x−2) = 10 → x + 2 = 10 → x = 8, y = 22

4. Simultaneous Equations — Elimination Method

Multiply one or both equations so that a letter has matching (or opposite) coefficients, then add or subtract the equations to eliminate it.

Worked example — coefficients already matching Solve: 2x + 3y = 13   and   5x − 3y = 1.
The y-coefficients are already opposites (+3 and −3) — add the equations:
(2x + 5x) + (3y − 3y) = 13 + 1 → 7x = 14 → x = 2
Substitute into the first equation: 2(2) + 3y = 13 → 3y = 9 → y = 3
Worked example — scaling both equations first Solve: 2x + 5y = 26   and   3x − 2y = 1.
Match the x-coefficients: multiply the first by 3, the second by 2:
6x + 15y = 78
6x − 4y = 2
Subtract: 19y = 76 → y = 4
Substitute into 3x − 2y = 1: 3x − 8 = 1 → 3x = 9 → x = 3
Try it — Elimination Method
  1. Solve: 3x + 2y = 16 and x − 2y = 0.
  2. Solve: 4x + y = 14 and 2x + y = 8.
  3. Solve: 3x + 2y = 16 and x + 3y = 10.
Show answers
  1. Add: 4x = 16 → x = 4; sub: 4 − 2y = 0 → y = 2
  2. Subtract: 2x = 6 → x = 3; sub: 6 + y = 8 → y = 2
  3. Multiply the second equation by 3: 3x + 9y = 30. Subtract the first: 7y = 14 → y = 2. Substitute into x + 3y = 10: x + 6 = 10 → x = 4.

5. Changing the Subject of a Formula

“Making x the subject” means rearranging so x is alone on one side. Treat it exactly like solving an equation — just with letters instead of only numbers.

Worked example Make x the subject of y = 3x + 2.
y − 2 = 3x → x = (y − 2)/3
Worked example — a real formula The formula for the perimeter of a rectangle is P = 2(l + w). Make l the subject.
P = 2l + 2w
P − 2w = 2l → l = (P − 2w)/2
Try it — Changing the Subject (Basic)
  1. Make x the subject of y = 5x − 4.
  2. Make t the subject of v = u + at.
  3. Make b the subject of A = ½bh.
Show answers
  1. x = (y + 4)/5
  2. t = (v − u)/a
  3. b = 2A/h

6. Changing the Subject — Harder Cases Extended only

Extended questions may have the subject appearing twice, or hidden inside a power or a root.

Worked example — subject appears twice Make x the subject of x = ab + cx.
Get every x term on one side: x − cx = ab
Factorise out x: x(1 − c) = ab
Answer: x = ab/(1 − c)
Worked example — subject has a power Make r the subject of A = πr².
Divide by π: r² = A/π
Square root both sides: r = √(A/π)
Worked example — subject is under a root Make x the subject of y = √(x + 3).
Square both sides: y² = x + 3
Answer: x = y² − 3
Try it — Changing the Subject (Harder) Extended
  1. Make x the subject of y = mx + cx (hint: factorise x out first).
  2. Make r the subject of V = (4/3)πr³.
  3. Make x the subject of y = √(2x − 1).
  4. Make p the subject of q = 5p − kp.
Show answers
  1. y = x(m + c) → x = y/(m + c)
  2. r³ = 3V/(4π) → r = ³√[3V/(4π)]
  3. y² = 2x − 1 → x = (y² + 1)/2
  4. q = p(5 − k) → p = q/(5 − k)
Common mistakes to watch for
  • Forgetting to do the same operation to every term on both sides, not just one.
  • Losing a negative sign when moving a term across the equals sign.
  • In elimination, adding when you should subtract (or vice versa) — check whether the matched coefficients have the same sign or opposite signs first.
  • When the subject appears twice, trying to isolate it before factorising — you must collect every instance of the subject on one side and factorise it out.
  • Forgetting to square (or square root) both entire sides of an equation, not just part of one side.

7. Quick-Fire Challenge Round

True or False?
  1. To solve 2x = 10, you divide both sides by 2.
  2. In elimination, if both coefficients of y are +3, you should add the equations to eliminate y.
  3. Making x the subject of y = √x gives x = y².
  4. A formula’s subject can never appear more than once, even in Extended-level questions.
Show answers
  1. True.
  2. False — matching (same-sign) coefficients are eliminated by subtracting, not adding; opposite-sign coefficients are eliminated by adding.
  3. True.
  4. False — Extended questions can have the subject appear twice, requiring factorising to isolate it. Ext.
Spot the Error
  1. “Solve 3x + 5 = 20: 3x = 20 + 5 = 25, so x = 25/3.”
  2. “Make x the subject of y = x² + 1: x = y² − 1.”
  3. “Eliminate y from 2x + y = 9 and x − y = 3 by subtracting: (2x−x) + (y−y) = 9−3, giving x = 6.”
Show answers
  1. 5 should be subtracted from both sides, not added: 3x = 20 − 5 = 15, so x = 5.
  2. To undo squaring, take the square root, not square again: x² = y − 1, so x = √(y − 1).
  3. The y-coefficients here are +1 and −1 — opposite signs — so they’re eliminated by adding the equations, not subtracting: (2x+x) + (y−y) = 9+3 gives 3x = 12, so x = 4 (and y = 1). Subtracting was the wrong operation, and it also doesn’t actually eliminate y here — (y − (−y)) = 2y, not 0.

8. Mixed Practice — Bringing It All Together

Core & Extended
  1. Solve 6x − 7 = 2x + 9.
  2. Solve: y = 4x − 1 and 2x + y = 17, using substitution.
  3. Solve: 3x + 4y = 23 and x − y = 3, using elimination.
  4. Make w the subject of P = 2l + 2w.
Show answers
  1. 4x = 16 → x = 4
  2. 2x + (4x−1) = 17 → 6x = 18 → x = 3, y = 4(3)−1 = 11
  3. Multiply the second equation by 3: 3x − 3y = 9. Subtract from the first: 7y = 14 → y = 2. Substitute into x − y = 3: x = 5.
  4. w = (P − 2l)/2
Extended-level mixed set Extended only
  1. Make x the subject of y = ax − bx.
  2. Make r the subject of C = 2πr.
  3. Make x the subject of y = √(3x + 4).
Show answers
  1. y = x(a − b) → x = y/(a − b)
  2. r = C/(2π)
  3. y² = 3x + 4 → x = (y² − 4)/3

Practice at Home

A large set again — take it steadily.

Part A — Constructing and solving linear equations
  1. A number is tripled, then 6 is subtracted, giving 21. Write and solve the equation.
  2. Solve 5x + 3 = 2x + 18.
  3. Solve 6 − 4x = 3(2x − 1).
  4. Solve (3x + 1)/4 = 7.
Part B — Simultaneous equations, substitution
  1. Solve: y = 2x − 3 and 3x + y = 12.
  2. Solve: y = x + 6 and 5x − y = 6.
Part C — Simultaneous equations, elimination
  1. Solve: 4x + 3y = 25 and 4x − y = 5.
  2. Solve: 2x + 5y = 24 and 3x − 2y = 5.
  3. Solve: 5x + 2y = 19 and 3x + 2y = 13.
Part D — Changing the subject (basic)
  1. Make x the subject of y = 7x + 3.
  2. Make u the subject of v = u + at.
  3. Make h the subject of V = lwh.
Part E — Changing the subject (harder) Extended
  1. Make x the subject of y = px + qx.
  2. Make r the subject of A = 4πr².
  3. Make x the subject of y = √(x − 5).
Part F — Applied / challenge
  1. The sum of two numbers is 24, and their difference is 6. Write two equations and solve them simultaneously to find both numbers.
  2. A formula for kinetic energy is E = ½mv². Make v the subject.
  3. Solve simultaneously: 2x + 3y = 7 and 4x + 6y = 14. What do you notice, and what does it mean about these two equations? (Hint: try eliminating as usual and see what happens.)

Key Vocabulary Recap

Linear equation — an equation where the unknown appears only to the power 1
Simultaneous equations — two or more equations solved together, for the same unknowns
Substitution method — replacing one variable with an equivalent expression from another equation
Elimination method — adding or subtracting equations to remove one variable
Subject of a formula — the variable that stands alone on one side

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section C2.5, points 1, 2, 3 and 4 (Core subject content) and section E2.5, points 1, 2, 4 and 7 (Extended subject content) — Equations: constructing them, solving linear equations, solving simultaneous linear equations, and changing the subject of a formula. Core is limited to formulas where the subject appears once with no power or root; Extended’s harder subject-change cases are marked above. Fractional equations, quadratic equations, and simultaneous equations involving one linear and one non-linear equation (E2.5 points 3, 5 and 6) are Extended-only and covered in a separate note.

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