Worked Solutions
Number & Algebra — Worked Solutions (Year 10 Maths)
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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.
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.