Equations
Section: Algebra and graphs 1 | Syllabus: Cambridge IGCSE Mathematics (0580)
Constructing and Solving Linear Equations
Turning a word problem into an equation, then solving it, is a two-stage skill: choose a letter for the unknown, translate the words into an equation, and then use inverse operations to isolate the letter.
- To construct an equation, let a letter represent the unknown quantity, then translate each phrase into algebra in the order described
- To solve a linear equation, undo each operation applied to the unknown, working outwards from the unknown one step at a time
- If there are brackets, expand them first
Worked Example: Constructing an Equation
- Question: Kofi thinks of a number, multiplies it by 3, then subtracts 7. The result is 20. Find the number.
- Step 1: Let the number be n: 3n - 7 = 20
- Step 2: Add 7 to both sides: 3n = 27
- Step 3: Divide both sides by 3: n = 9
- Answer: 9
Worked Example: Solving an Equation with Brackets
- Question: Solve.
6(2x - 3) = 42- Step 1: Expand the bracket: 12x - 18 = 42
- Step 2: Add 18 to both sides: 12x = 60
- Step 3: Divide both sides by 12: x = 5
- Answer: x = 5
Common Mistakes
MistakeTranslating the words in the wrong order, e.g. writing 7 - 3n instead of 3n - 7 for "multiply by 3, then subtract 7"
Fixbuild the expression in the exact order the words describe, one operation at a time, to avoid reversing a subtraction
MistakeExpanding a bracket by multiplying only the first term, e.g. writing 6(2x-3) as 12x - 3
Fixevery term inside the bracket must be multiplied by the number outside: 6(2x-3) = 12x - 18
Simultaneous Linear Equations
Two equations with two unknowns can be solved together by eliminating one unknown - scaling the equations so one letter's coefficients match, then adding or subtracting.
- Multiply one or both equations so that the coefficients of one letter are equal in size
- Subtract the equations if those coefficients have the same sign; add them if the coefficients have opposite signs
- Solve the resulting equation for the remaining letter, then substitute back into either original equation
Worked Example: Solving Simultaneous Equations by Elimination
- Question: Solve the simultaneous equations.
2x + 3y = 3
4x - 5y = 17- Step 1: Multiply the first equation by 5 and the second by 3, so both have a y-coefficient of size 15: 10x+15y=15, 12x-15y=51
- Step 2: The y-terms have opposite signs, so add the equations: 22x = 66 ⇒ x = 3
- Step 3: Substitute x = 3 into the first equation: 2(3) + 3y = 3 ⇒ 3y = -3 ⇒ y = -1
- Answer: x = 3, y = -1
Common Mistakes
MistakeMultiplying only one of the two equations when trying to match coefficients
Fixboth equations usually need scaling - each equation must be multiplied so the target letter's coefficient becomes the same size in both
MistakeAdding when the equations should be subtracted (or the reverse), based on the signs of the matched coefficients
Fixsubtract when the matched coefficients have the same sign; add when they have opposite signs - this is what makes that letter cancel
Changing the Subject of a Formula
Changing the subject means rearranging a formula so a different letter is isolated on one side. The same inverse-operation approach used for solving equations applies here too.
- Treat the formula like an equation: undo whatever has been done to the subject, one step at a time, working outward
- Every operation must be applied to every term on both sides, not just to the term nearest the subject
Worked Example: Changing the Subject
- Question: Make m the subject of the formula.
F = ma²- Step 1: Divide both sides by a²: m = (F/a^2)
- Answer: m = F/a²
- Question: Make x the subject of the formula.
y = 3x + 7- Step 1: Subtract 7 from both sides: y - 7 = 3x
- Step 2: Divide both sides by 3: x = (y-7/3)
- Answer: x = (y-7)/3
Common Mistakes
MistakeApplying an operation to only part of one side, e.g. rearranging y = x/4 + 3 by multiplying by 4 but only turning the +3 into +3, writing 4y = x + 3 instead of 4y = x + 12
Fixevery single term on both sides must be multiplied by 4, including the +3: 4y = x + 12, so x = 4y - 12
MistakeDividing by only part of the coefficient of the subject, e.g. in F = ma² dividing by a instead of a²
Fixdivide by the entire coefficient attached to the subject - in F = ma², that coefficient is a², not just a
Fractional Equations
Extended Only
An equation with fractions can be turned into a normal linear equation by clearing every denominator first - either by cross-multiplying two fractions, or by multiplying through by an algebraic denominator.
- For an equation with a fraction on each side, cross-multiply: multiply each numerator by the opposite denominator
- For an equation with an algebraic denominator, multiply both sides by that denominator to clear it, expanding any brackets that result
Worked Example: Cross-Multiplying a Fractional Equation
- Question: Solve.
(2x+1)/3 = (3x-2)/5- Step 1: Cross-multiply: 5(2x+1) = 3(3x-2)
- Step 2: Expand both sides: 10x+5 = 9x-6
- Step 3: Collect terms: x = -11
- Answer: x = -11
Worked Example: An Algebraic Denominator
- Question: Solve.
20/(2x+1) = 4- Step 1: Multiply both sides by (2x+1): 20 = 4(2x+1)
- Step 2: Expand: 20 = 8x+4
- Step 3: Solve: 16 = 8x ⇒ x = 2
- Answer: x = 2
Common Mistakes
MistakeCross-multiplying numerator with numerator and denominator with denominator, instead of crossing them
Fixcross-multiplying means each numerator multiplies the opposite denominator: (2x+1)/3 = (3x-2)/5 becomes 5(2x+1) = 3(3x-2)
MistakeDropping the brackets when multiplying by an algebraic denominator, e.g. writing 20 = 4 × 2x + 1 instead of 20 = 4(2x+1)
Fixthe whole denominator must stay in brackets until it is fully expanded: 4(2x+1) = 8x + 4, not 4 × 2x + 1
Simultaneous Equations: One Linear, One Non-Linear
Extended Only
When one equation is linear and the other involves a squared term, substitution turns the pair into a single quadratic equation, which usually gives two pairs of solutions.
- Rearrange the linear equation for one letter if needed, then substitute it into the non-linear equation
- Solve the resulting quadratic equation - there are usually two solutions, giving two (x, y) pairs
- Substitute each x-value back into the linear equation to find its matching y-value
Worked Example: Substitution with a Quadratic
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