Mathematics | Section E2.5 | Answers
Mark Scheme — 15 Questions
Section A — Recall
Questions 1–5
1.
State the difference between an expression and an equation.
Answer
An expression is a group of terms with no equals sign, such as $2n + 3$. It can only be simplified, not solved.
An equation has an equals sign joining two expressions, such as $2n + 3 = 7$. It can be solved to find the value of the unknown.
Key difference: an equation has an equals sign and can be solved; an expression does not and cannot.
2.
Solve $2x + 7 = 15$.
Answer
- Subtract 7 from both sides: $\quad 2x = 8$
- Divide both sides by 2: $\quad x = 4$
$x = 4$
3.
Write an expression for the product of two consecutive even numbers.
Answer
- Any even number can be written as $2n$.
- The next even number is 2 more, so it is $2n + 2$.
- “Product” means multiply: $\quad 2n(2n + 2)$
$2n(2n + 2)$ (which expands to $4n^2 + 4n$)
Exam tip: Using $2n$ guarantees the number is even, because 2 times any whole number is always even.
4.
Write down the standard form of a quadratic equation.
Answer
$$ax^2 + bx + c = 0$$
Exam tip: A quadratic always contains an $x^2$ term. Getting it into this standard form (equal to zero) is the first step before factorising or using the quadratic formula.
5.
Make $x$ the subject of the formula $y = 2x + 1$.
Answer
- Subtract 1 from both sides: $\quad y – 1 = 2x$
- Divide both sides by 2: $\quad x = \dfrac{y – 1}{2}$
$x = \dfrac{y – 1}{2}$
Section B — Application
Questions 6–10
6.
Solve $4(x – 1) = 2x + 6$.
Answer
- Expand the bracket: $\quad 4x – 4 = 2x + 6$
- Subtract $2x$ from both sides: $\quad 2x – 4 = 6$
- Add 4 to both sides: $\quad 2x = 10$
- Divide both sides by 2: $\quad x = 5$
$x = 5$
Exam tip: Expand brackets first, then collect the $x$ terms on one side and the numbers on the other. Check: $4(5 – 1) = 16$ and $2(5) + 6 = 16$ ✓
7.
Solve the fractional equation $\dfrac{x}{3x – 2} = 2$.
Answer
- Multiply both sides by $(3x – 2)$: $\quad x = 2(3x – 2)$
- Expand the bracket: $\quad x = 6x – 4$
- Subtract $6x$ from both sides: $\quad -5x = -4$
- Divide both sides by $-5$: $\quad x = \dfrac{4}{5}$
$x = \dfrac{4}{5}$
Exam tip: Clear the fraction first by multiplying both sides by the denominator. This turns it into an ordinary linear equation.
8.
Solve the simultaneous equations $3x + y = 11$ and $2x – y = 4$.
Answer
- The $y$ terms are $+y$ and $-y$, so add the two equations to remove $y$: $$3x + y + 2x – y = 11 + 4 \quad\Longrightarrow\quad 5x = 15$$
- Divide by 5: $\quad x = 3$
- Substitute $x = 3$ into $3x + y = 11$: $\quad 9 + y = 11 \Longrightarrow y = 2$
$x = 3, \quad y = 2$
Exam tip: When one equation has $+y$ and the other has $-y$, adding them cancels $y$ straight away. Check in the other equation: $2(3) – 2 = 4$ ✓
9.
Solve $x^2 + 7x + 12 = 0$ by factorisation.
Answer
- Find two numbers that multiply to $+12$ and add to $+7$: these are $+3$ and $+4$.
- Write as two brackets: $\quad (x + 3)(x + 4) = 0$
- Set each bracket equal to zero: $\quad x + 3 = 0 \;$ or $\; x + 4 = 0$
$x = -3 \quad$ or $\quad x = -4$
Exam tip: For $x^2 + bx + c$, look for two numbers that multiply to $c$ and add to $b$.
10.
Make $x$ the subject of the formula $y = \sqrt{x – 2}$.
Answer
- Square both sides to remove the square root: $\quad y^2 = x – 2$
- Add 2 to both sides: $\quad x = y^2 + 2$
$x = y^2 + 2$
Exam tip: The opposite of a square root is squaring. Square both sides to free the letter under the root.
Section C — Challenge
Questions 11–15
11.
A cinema sells adult tickets for $x$ dollars and child tickets for $y$ dollars. 3 adult tickets and 2 child tickets cost \$23. 1 adult ticket and 2 child tickets cost \$13.
(a) Write two equations. (b) Solve to find the price of each ticket.
Answer
(a) The two equations
“3 adult and 2 child cost \$23”: $\quad 3x + 2y = 23$
“1 adult and 2 child cost \$13”: $\quad x + 2y = 13$
(b) Solving the equations
- Both equations have $+2y$, so subtract the second from the first to remove $y$: $$(3x + 2y) – (x + 2y) = 23 – 13 \quad\Longrightarrow\quad 2x = 10$$
- Divide by 2: $\quad x = 5$
- Substitute $x = 5$ into $x + 2y = 13$: $\quad 5 + 2y = 13 \Longrightarrow 2y = 8 \Longrightarrow y = 4$
Adult ticket $= \$5$, child ticket $= \$4$.
Exam tip: When both equations contain the same term ($+2y$ here), subtracting one from the other removes it. Check: $3(5) + 2(4) = 15 + 8 = 23$ ✓
12.
Solve $\dfrac{3}{x + 1} + \dfrac{2}{x – 1} = 1$.
Answer
- Multiply every term by both denominators $(x + 1)(x – 1)$: $$3(x – 1) + 2(x + 1) = (x + 1)(x – 1)$$
- Expand each part: $\quad 3x – 3 + 2x + 2 = x^2 – 1$
- Simplify the left side: $\quad 5x – 1 = x^2 – 1$
- Move everything to one side: $\quad 0 = x^2 – 5x$
- Factorise: $\quad x(x – 5) = 0$
- Set each factor to zero: $\quad x = 0 \;$ or $\; x = 5$
$x = 0 \quad$ or $\quad x = 5$
Exam tip: Adding fractions over a common denominator often produces a quadratic. Always move everything to one side and solve it like any quadratic. Check $x = 0$: $\frac{3}{1} + \frac{2}{-1} = 3 – 2 = 1$ ✓
13.
(a) Write $x^2 + 8x + 10$ in completed square form.
(b) Hence solve $x^2 + 8x + 10 = 0$, in surd form.
(c) Solve $2x^2 – 4x – 3 = 0$ using the quadratic formula, in surd form.
Answer
(a) Completed square form
- Halve the number in front of $x$ (half of 8 is 4) and write $(x + 4)^2$.
- $(x + 4)^2 = x^2 + 8x + 16$, which is 6 too big, so subtract 6: $$x^2 + 8x + 10 = (x + 4)^2 – 6$$
$x^2 + 8x + 10 = (x + 4)^2 – 6$
(b) Solving using the completed square
- Set equal to zero: $\quad (x + 4)^2 – 6 = 0$
- Rearrange: $\quad (x + 4)^2 = 6$
- Square root both sides (remember $\pm$): $\quad x + 4 = \pm\sqrt{6}$
- Subtract 4: $\quad x = -4 \pm \sqrt{6}$
$x = -4 + \sqrt{6} \quad$ or $\quad x = -4 – \sqrt{6}$
(c) Solving using the quadratic formula
- Identify the values: $\quad a = 2, \; b = -4, \; c = -3$
- Substitute into $x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a}$: $$x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-3)}}{2(2)} = \frac{4 \pm \sqrt{16 + 24}}{4} = \frac{4 \pm \sqrt{40}}{4}$$
- Simplify the surd: $\quad \sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$
- So $\quad x = \dfrac{4 \pm 2\sqrt{10}}{4} = \dfrac{2 \pm \sqrt{10}}{2}$ (dividing top and bottom by 2)
$x = \dfrac{2 + \sqrt{10}}{2} \quad$ or $\quad x = \dfrac{2 – \sqrt{10}}{2}$
Exam tip: “Surd form” means leave the answer with the square root sign — do not round to a decimal. Always simplify the surd (here $\sqrt{40} = 2\sqrt{10}$) and cancel where possible.
14.
Consider $y = x + 1$ and $y = x^2 – 1$.
(a) Solve to find all pairs of values. (b) Explain why there are two solution pairs.
Answer
(a) Solving the equations
- Both equations equal $y$, so set them equal to each other: $\quad x + 1 = x^2 – 1$
- Move everything to one side: $\quad 0 = x^2 – x – 2$
- Factorise and solve: $\quad (x – 2)(x + 1) = 0 \Longrightarrow x = 2 \;$ or $\; x = -1$
- Find $y$ using $y = x + 1$:
- When $x = 2$: $\;y = 2 + 1 = 3$
- When $x = -1$: $\;y = -1 + 1 = 0$
$(x, y) = (2,\; 3) \quad$ and $\quad (x, y) = (-1,\; 0)$
(b) Why there are two solution pairs
One equation is linear (a straight line) and the other is non-linear (it has an $x^2$ term). Solving them leads to a quadratic equation, which has two solutions for $x$. Each $x$-value gives its own $y$-value, so there are two solution pairs.
Exam tip: A linear and a non-linear equation usually give two solution pairs. Always pair each $x$ with its matching $y$ — never leave them separate.
15.
(a) Make $x$ the subject of $y = \dfrac{x + 4}{x – 2}$.
(b) Make $r$ the subject of $V = \pi r^2 h$.
Answer
(a) Subject appears twice
- Multiply both sides by $(x – 2)$: $\quad y(x – 2) = x + 4$
- Expand the bracket: $\quad yx – 2y = x + 4$
- Collect the $x$ terms on one side, the rest on the other: $\quad yx – x = 4 + 2y$
- Factorise out $x$ (it now appears once): $\quad x(y – 1) = 4 + 2y$
- Divide by $(y – 1)$: $\quad x = \dfrac{4 + 2y}{y – 1}$
$x = \dfrac{4 + 2y}{y – 1}$
(b) Power of the subject
- Divide both sides by $\pi h$: $\quad r^2 = \dfrac{V}{\pi h}$
- Square root both sides: $\quad r = \sqrt{\dfrac{V}{\pi h}}$
$r = \sqrt{\dfrac{V}{\pi h}}$
Exam tip: When the subject appears twice, the key step is to factorise it out so it appears only once. When the subject is squared, deal with the square last by taking the square root.
