Number & Algebra — Worked Solutions (Year 10 Maths)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our Year 10 Maths course →
Worked examples for Year 10 Maths Number & Algebra. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Year 10 Maths — Number & Algebra. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Factorising and solving a quadratic
Question
Solve $x^2 - 2x - 15 = 0$ by factorising.
Solution
Factorise first, then use the null factor law.
Find two numbers that multiply to $-15$ and add to $-2$: that's $-5$ and $3$.
So $x^2 - 2x - 15 = (x - 5)(x + 3) = 0$.
A product is zero when a factor is zero, so $x - 5 = 0$ or $x + 3 = 0$.
Therefore $x = 5$ or $x = -3$. Always state both solutions — a quadratic gives two, and dropping one loses a mark.
Let's think about what factorising does — we rewrite the quadratic as two brackets multiplied together, because a product equalling zero is something we can actually solve.
We want two numbers whose product is the constant $-15$ and whose sum is the middle coefficient $-2$. Testing pairs, $-5$ and $3$ work: $-5 \times 3 = -15$ and $-5 + 3 = -2$.
So $x^2 - 2x - 15 = (x - 5)(x + 3) = 0$.
Now the key idea: if two things multiply to give zero, at least one of them must be zero. So either $x - 5 = 0$, giving $x = 5$, or $x + 3 = 0$, giving $x = -3$.
Both answers are valid solutions because the curve crosses the $x$-axis in two places.
Factorise, then null factor law.
- Product $= -15$, sum $= -2$ → numbers $-5$ and $3$
- $(x - 5)(x + 3) = 0$
- $x - 5 = 0$ → $x = 5$
- $x + 3 = 0$ → $x = -3$
$x = 5$ or $x = -3$.
Where the marks go
- 1 mark: Correct factorisation $(x - 5)(x + 3)$
- 1 mark: Applies the null factor law to both brackets
- 1 mark: States both solutions $x = 5$ and $x = -3$
Key idea
Factorise to a product of brackets, then set each bracket to zero — a quadratic has two solutions, so report both.
Example 2 — Simultaneous equations
Question
Solve the simultaneous equations $3x + 2y = 16$ and $x - 2y = -4$.
Solution
The $y$ terms are $+2y$ and $-2y$, so add the equations to eliminate $y$ straight away.
$(3x + 2y) + (x - 2y) = 16 + (-4)$, giving $4x = 12$, so $x = 3$.
Substitute $x = 3$ into $x - 2y = -4$: $3 - 2y = -4$, so $-2y = -7$ and $y = 3.5$.
Solution: $x = 3,\ y = 3.5$. Check in the other equation: $3(3) + 2(3.5) = 9 + 7 = 16$. Correct.
With simultaneous equations we're looking for the one pair $(x, y)$ that satisfies both at once. The neat trick here is elimination.
Look at the $y$ terms: one is $+2y$ and the other is $-2y$. If we add the two equations, those cancel out:
$(3x + 2y) + (x - 2y) = 16 + (-4)$, which simplifies to $4x = 12$, so $x = 3$.
Now we know $x$, we substitute back to find $y$. Using $x - 2y = -4$: $3 - 2y = -4$, so $-2y = -7$ and $y = 3.5$.
Finally — and this is why checking matters — substitute into the first equation: $3(3) + 2(3.5) = 16$. It works, so we can trust $x = 3,\ y = 3.5$.
Eliminate $y$ (coefficients $+2$ and $-2$).
- Add: $4x = 12$ → $x = 3$
- Sub into $x - 2y = -4$: $3 - 2y = -4$
- $-2y = -7$ → $y = 3.5$
- Check: $3(3) + 2(3.5) = 16$ ✓
$x = 3,\ y = 3.5$.
Where the marks go
- 1 mark: Adds the equations to eliminate $y$
- 1 mark: Solves $4x = 12$ to get $x = 3$
- 1 mark: Substitutes back to find $y = 3.5$
- 1 mark: Verifies the solution in the other equation
Key idea
When the coefficients of one variable are opposites, adding the equations eliminates it; substitute back, then check both equations.
Frequently asked questions
Step-by-step solutions to Number & Algebra questions in Year 10 Maths, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Year 10 Maths, using the methods and notation expected in exams and assessments.