Equations.
Build the relationship, solve it systematically and check it in the original statement.
Form and solve linear, fractional, simultaneous and quadratic equations · rearrange formulas
Build an equation, then keep it balanced
An equation is a balance: the left and right sides have the same value. Solving it means finding the input that makes this statement true, while preserving that equality at every step.
Adding, subtracting, multiplying or dividing both sides by the same permitted quantity keeps the balance. Use the inverse operation to undo what happened to the unknown, working backward through the calculation. Write each change on both sides rather than treating terms as objects that jump across an equals sign.
Define the unknown, translate the relationship, and perform the same operation on both sides. Check the answer in the original equation.
Solve 5 − 2x = 3(x + 7).
−5x = 16
x = −165
Two tickets and one drink cost $17. One ticket and two drinks cost $13.
Let t be a ticket price and d a drink price.
t + 2d = 13
Double the second equation and subtract the first: 3d = 9, so d = 3. Then t = 7. Check both totals.
Pause and explainTo solve x + 7 = 12, why subtract 7 from both sides?
It undoes the addition beside x while preserving equality. The result is x = 5.
Fractional equations: clear denominators
Fractions can hide a simple balance inside awkward notation. Clearing denominators means multiplying every term on both sides by a suitable common denominator, so the fractions become whole expressions.
Write restrictions before multiplying by an expression containing the unknown. Use brackets around whole numerators and apply the multiplier to every term. After solving, check the original equation: clearing denominators does not make a previously forbidden value valid.
Solve x − 13 + x + 22 = 4.
Multiply every term by 6, the LCM of 3 and 2.
5x + 4 = 24
x = 4
Solve x2x + 1 = 4.
Restriction: x ≠ −12.
−7x = 4
x = −47
Solve 2x + 2 + 32x − 1 = 1.
x ≠ −2, 12. Multiply every term by (x + 2)(2x − 1).
7x + 4 = 2x² + 3x − 2
x² − 2x − 3 = 0
(x − 3)(x + 1) = 0
Solutions: x = 3 or −1. Both satisfy the restrictions.
Pause and explainWhy check a solution against the original denominators?
A value making an original denominator zero is not allowed, even if later algebra seems to produce it.
Three routes to solve a quadratic
A quadratic equation contains a squared unknown and may have two solutions. First arrange it with zero on one side so that factorisation or the quadratic formula can be used consistently.
The zero-product idea explains factor solving: if two numbers multiply to zero, at least one must be zero. Set each factor equal to zero separately. If factorisation is not convenient, identify a, b and c from the fully rearranged equation before using the formula.
Solve x² − 5x + 6 = 0 by factorisation.
x = 2 or x = 3
A product is zero only if at least one factor is zero. Put the equation equal to zero before using this rule.
Solve x² + 6x + 2 = 0 by completing the square.
x + 3 = ±√7
x = −3 ± √7
Quadratic formula
For ax² + bx + c = 0, include the signs in a, b and c. This formula is provided in the syllabus List of formulas.
Solve 2x² − 3x − 1 = 0 exactly.
Use both signs. Keep the surd if an exact answer is requested. A negative discriminant b² − 4ac means no real solutions.
Pause and explainWhy does (x − 2)(x + 5) = 0 give two possibilities?
Either x − 2 = 0 or x + 5 = 0, so x = 2 or x = −5. Each makes the product zero.
Change the subject of a formula
A formula links several quantities. Its subject is the quantity standing alone on one side. Changing the subject uses the same balance operations as solving an equation, but other letters remain as general quantities.
Locate every occurrence of the required subject. Undo operations in a sensible order; if the subject occurs in more than one term, collect those terms and factor it out. Do not insert numerical values unless the question asks for a calculation.
Make x the subject of y = ax + bx.
x = ya + b, a + b ≠ 0
When the subject appears twice, collect its terms and factor it out.
Make x the subject of y = (x + a)(x − b).
yx − x = by + a
x(y − 1) = by + a
x = by + ay − 1
Use x ≠ b from the original formula and y ≠ 1 for this rearranged form.
Make r the subject of A = πr², for a circle radius.
Divide by π, then take the non-negative root because a radius is non-negative. In an unrestricted equation x² = k, both signs may be needed.
Pause and explainIn A = lw, how do you make w the subject?
Divide both sides by l, assuming l is not zero. Then w equals A divided by l.
Your turn. Substitute back.
Identify the equation type before choosing a method. Linear equations usually need inverse operations, while a quadratic needs all terms collected and a suitable quadratic method.
Show one balance-preserving step per line. Finally substitute into the original statement, checking denominator restrictions and whether every proposed solution makes sense in the word problem.
Solve 3x + 4 = 10.
Reveal worked answer
Solve 2x + y = 11 and 3x − y = 9.
Reveal worked answer
Solve 3x² − 2x − 1 = 0.
Reveal worked answer
Solve x² − 4x − 2 = 0 exactly.
Reveal worked answer
Solve xx + 2 = 3x − 6.