Evaluating Expressions and Formulae

How to find the value of an algebraic expression by substitution, with whole numbers, negative numbers, fractions and decimals, and how to use a formula to find the value of its subject.

Before you start

In algebraic notation the $\times$ sign is left out and division is written as a fraction: $3x$ means $3 \times x$ , $ab$ means $a \times b$ , $x^{2}$ means $x \times x$ , $x^{3}$ means $x \times x \times x$ , and $\frac{x + 3}{5}$ means $(x + 3) \div 5$ .

The order of operations: brackets first, then powers, then $\times$ and $\div$ from left to right, then $+$ and $-$ from left to right. For example, $2 + 3 \times 4^{2} = 2 + 3 \times 16 = 2 + 48 = 50$ .

With negative numbers: $(-4) \times 3 = -12$ , $(-4) \times (-3) = 12$ , and subtracting a negative number is adding: $5 - (-2) = 5 + 2 = 7$ .

Substitution

An algebraic expression such as $3a + 4b$ has no single value: its value depends on the numbers that $a$ and $b$ stand for. Once we are told those numbers, we can work out the value of the expression.

Replacing each letter by the number it stands for is called substitution. Finding the value of an expression in this way is called evaluating the expression.

How to evaluate an expression by substitution

1. Write the expression with the $\times$ signs put back: $3a$ is $3 \times a$ , $ab$ is $a \times b$ , $a^{2}$ is $a \times a$ .

2. Replace each letter by its value. Writing the value in brackets is always safe. A negative value must always go in brackets, and so must a fraction that is squared or cubed: with $a = -2$ , $3a$ is $3 \times (-2)$ and $a^{2}$ is $(-2)^{2}$ .

3. Work it out in the order of operations: brackets, then powers, then $\times$ and $\div$ from left to right, then $+$ and $-$ from left to right. A fraction line acts as a bracket around the top and around the bottom.

Example 1: Substituting whole numbers

Given that $a = 5$ and $b = 2$ , find the value of
(a) $3a + 4b$ ,
(b) $7b - a$ ,
(c) $4(a - b)$ ,
(d) $\frac{a + 3}{b}$ .

Solution

(a) $3a + 4b = 3 \times 5 + 4 \times 2$ . Multiply before adding: $= 15 + 8 = 23$ . (Note that $3a$ is $3 \times 5 = 15$ , not 35.)

(b) $7b - a = 7 \times 2 - 5 = 14 - 5 = 9$ .

(d) $\frac{a + 3}{b} = \frac{5 + 3}{2}$ . The fraction line keeps the top together, so add first: $= \frac{8}{2} = 4$ .

(a) 23 (b) 9 (c) 12 (d) 4

Powers come before multiplication. In $2p^{2}$ only $p$ is squared, and then the result is doubled; in $(2p)^{2}$ the bracket is worked first, so $2p$ is found and then squared. The next example shows that the two give different values.

Example 2: Powers and the order of operations

Given that $p = 3$ and $q = 4$ , find the value of
(a) $2p^{2}$ ,
(b) $(2p)^{2}$ ,
(c) $p^{3} - q$ ,
(d) $\frac{3pq}{q + 2}$ .

Solution

(a) $2p^{2} = 2 \times 3^{2}$ . The power comes first: $3^{2} = 9$ , so $2p^{2} = 2 \times 9 = 18$ .

(b) $(2p)^{2} = (2 \times 3)^{2}$ . The bracket comes first: $2 \times 3 = 6$ , so $(2p)^{2} = 6^{2} = 36$ .

(d) Top: $3pq = 3 \times 3 \times 4 = 36$ . Bottom: $q + 2 = 4 + 2 = 6$ . So $\frac{3pq}{q + 2} = \frac{36}{6} = 6$ .

(a) 18 (b) 36 (c) 23 (d) 6

Substituting Negative Numbers

When a letter stands for a negative number, the brackets matter most. Compare these two:

$(-3)^{2}$ means $(-3) \times (-3)$ . A negative times a negative is positive, so $(-3)^{2} = 9$ .

$-3^{2}$ means $-(3^{2})$ : the power is worked first, on 3 alone, and the minus sign is taken afterwards. So $-3^{2} = -(3 \times 3) = -9$ .

Rule: a negative value goes in brackets

When you substitute a negative number, write it in brackets, so that a power or a product acts on the whole number, sign and all.

A negative number squared is positive. A negative number cubed is negative: $(-3)^{3} = (-3) \times (-3) \times (-3) = 9 \times (-3) = -27$ .

Example 3: Powers of a negative number

Given that $x = -3$ , find the value of
(a) $x^{2}$ ,
(b) $-x^{2}$ ,
(c) $x^{3}$ ,
(d) $2x^{2}$ ,
(e) $(2x)^{2}$ .

Solution

(a) $x^{2} = (-3)^{2} = (-3) \times (-3) = 9$ .

(b) $-x^{2}$ means the negative of $x^{2}$ : the power is worked first. $-x^{2} = -(-3)^{2} = -(9) = -9$ .

(d) $2x^{2} = 2 \times (-3)^{2}$ . The power first: $(-3)^{2} = 9$ , so $2x^{2} = 2 \times 9 = 18$ .

(e) $(2x)^{2} = (2 \times (-3))^{2}$ . The bracket first: $2 \times (-3) = -6$ , so $(2x)^{2} = (-6)^{2} = (-6) \times (-6) = 36$ .

(a) 9 (b) $-9$ (c) $-27$ (d) 18 (e) 36

Take particular care with a minus sign in front of a negative value. Write the subtraction and the negative number separately, then change "subtract a negative" into "add": $12 - (-13) = 12 + 13$ .

Example 4: Adding and subtracting with a negative value

Given that $x = 4$ and $y = -5$ , find the value of
(a) $3x - 2.6y$ ,
(b) $x - y^{2}$ ,
(c) $\frac{x}{2} + \frac{y}{10}$ .

Solution

(a) $3x - 2.6y = 3 \times 4 - 2.6 \times (-5)$ .

Multiply first: $3 \times 4 = 12$ and $2.6 \times (-5) = -13$ . So $3x - 2.6y = 12 - (-13)$ .

Subtracting $-13$ is adding 13: $12 - (-13) = 12 + 13 = 25$ .

(b) $x - y^{2} = 4 - (-5)^{2}$ . The power first: $(-5)^{2} = 25$ . So $x - y^{2} = 4 - 25 = -21$ .

Adding a negative number is subtracting: $2 + \left(-\frac{1}{2}\right) = 2 - \frac{1}{2} = \frac{3}{2} = 1\frac{1}{2}$ .

(a) 25 (b) $-21$ (c) $1\frac{1}{2}$

Example 5: Three letters, two of them negative

Given that $a = -2$ , $b = 3$ and $c = -1$ , find the value of
(a) $3a^{2}$ ,
(b) $ab - c$ ,
(c) $\frac{a - b}{c}$ .

Solution

(a) $3a^{2} = 3 \times (-2)^{2}$ . Only $a$ is squared: $(-2)^{2} = 4$ , so $3a^{2} = 3 \times 4 = 12$ . (Multiplying first, $(3 \times (-2))^{2} = 36$ , would be wrong: that is $(3a)^{2}$ .)

(b) $ab - c = (-2) \times 3 - (-1)$ . Multiply first: $(-2) \times 3 = -6$ . So $ab - c = -6 - (-1) = -6 + 1 = -5$ .

(c) Top: $a - b = (-2) - 3 = -5$ . Bottom: $c = -1$ . So $\frac{a - b}{c} = \frac{-5}{-1} = 5$ , since a negative divided by a negative is positive.

(a) 12 (b) $-5$ (c) 5

Substituting Fractions and Decimals

The steps are the same when a letter stands for a fraction or a decimal: put the value in brackets, then follow the order of operations, using what you know about fractions and decimals.

When the answer is not exact, give it correct to 3 significant figures, unless the question asks for something else. Keep the full value on your calculator until the last step and round only at the end.

Example 6: Substituting fractions

Given that $a = \frac{1}{2}$ and $b = -2$ , find the value of
(a) $4a - 3b$ ,
(b) $a^{2}b$ ,
(c) $\frac{a + b}{a}$ .

Solution

(a) $4a - 3b = 4 \times \frac{1}{2} - 3 \times (-2) = 2 - (-6) = 2 + 6 = 8$ .

(b) $a^{2}b$ means $a \times a \times b$ . So $a^{2}b = \left(\frac{1}{2}\right)^{2} \times (-2) = \frac{1}{4} \times (-2) = -\frac{2}{4} = -\frac{1}{2}$ .

So $\frac{a + b}{a} = \left(-\frac{3}{2}\right) \div \frac{1}{2} = \left(-\frac{3}{2}\right) \times 2 = -3$ .

(a) 8 (b) $-\frac{1}{2}$ (c) $-3$

Example 7: Substituting decimals, with an answer to 3 significant figures

Given that $m = 2.5$ and $n = -0.4$ , find the value of
(a) $m^{2} + 3n$ ,
(b) $\frac{mn}{m + n}$ , giving your answer correct to 3 significant figures.

Solution

(a) $m^{2} + 3n = 2.5^{2} + 3 \times (-0.4)$ . Power first, then multiplication: $2.5^{2} = 6.25$ and $3 \times (-0.4) = -1.2$ .

So $m^{2} + 3n = 6.25 + (-1.2) = 6.25 - 1.2 = 5.05$ .

(b) Top: $mn = 2.5 \times (-0.4) = -1$ . Bottom: $m + n = 2.5 + (-0.4) = 2.1$ .

So $\frac{mn}{m + n} = \frac{-1}{2.1} = -0.476190\ldots$

The answer is not exact, so round it: $-0.476190\ldots = -0.476$ (3 s.f.).

(a) 5.05 (b) $-0.476$ , correct to 3 significant figures

Formulae

You already know rules such as "the perimeter of a rectangle is twice the sum of its length and its breadth". Written with letters, such a rule is a formula.

Definition: formula

A formula is a rule, written with letters, that connects two or more quantities. The plural of formula is formulae.

In the formula $P = 2(l + b)$ , the letter $P$ stands alone on one side of the equals sign. It is called the subject of the formula.

To find the value of the subject, substitute the values of the other letters into the right-hand side and evaluate it.

Before using a formula, make sure you know what each letter stands for and in what unit. The answer takes the unit of the subject.

In this lesson a formula is used only to find its subject. Finding one of the other letters when the subject is known needs an equation, which is taught in the chapter on equations.

Example 8: Perimeter and area of a rectangle

A rectangle has length $l$ cm and breadth $b$ cm. Its perimeter, $P$ cm, and its area, $A$ cm², are given by the formulae
$P = 2(l + b)$ and $A = lb$ .
Find $P$ and $A$ when $l = 12.5$ and $b = 8$ .

Solution

Perimeter: $P = 2(l + b) = 2 \times (12.5 + 8)$ . The bracket first: $12.5 + 8 = 20.5$ . So $P = 2 \times 20.5 = 41$ .

Area: $A = lb = 12.5 \times 8 = 100$ .

$P$ is a number of centimetres and $A$ a number of square centimetres.

The perimeter is 41 cm and the area is 100 cm².

Example 9: The average of three numbers

The average, $A$ , of three numbers $a$ , $b$ and $c$ is given by the formula $A = \frac{a + b + c}{3}$ .
(a) Find $A$ when $a = 15$ , $b = 20$ and $c = 13$ .
(b) The temperatures at midnight in Seoul on three nights in winter were $-4$ °C, $2$ °C and $-7$ °C. Use the formula to find the average midnight temperature.

Solution

(a) $A = \frac{15 + 20 + 13}{3} = \frac{48}{3} = 16$ .

(b) Substitute $a = -4$ , $b = 2$ and $c = -7$ , each negative value in brackets:

$A = \frac{(-4) + 2 + (-7)}{3}$ .

The top first, from left to right: $(-4) + 2 = -2$ , and $(-2) + (-7) = -9$ .

So $A = \frac{-9}{3} = -3$ . The temperatures are in °C, so the average is in °C.

(a) $A = 16$ (b) The average midnight temperature was $-3$ °C.

Example 10: Distance, speed and time

An object moving at a constant speed of $s$ km/h for $t$ hours travels $d$ km, where $d = st$ .
(a) Ravi cycles at a constant speed of 18 km/h for 2.5 hours. Find the distance he travels.
(b) A coach travels on the expressway at a constant speed of 90 km/h for 40 minutes. Find the distance it travels.

Solution

(a) $s = 18$ and $t = 2.5$ . So $d = st = 18 \times 2.5 = 45$ .

(b) The speed is in km/h, so the time must be in hours before it is substituted. 40 minutes is $\frac{40}{60} = \frac{2}{3}$ hour, so $t = \frac{2}{3}$ .

$s = 90$ , so $d = st = 90 \times \frac{2}{3} = 60$ .

Check: at 90 km/h the coach goes 1.5 km each minute, and $1.5 \times 40 = 60$ .

(a) Ravi travels 45 km. (b) The coach travels 60 km.

Example 11: The circumference of a circle

The circumference, $C$ m, of a circle of radius $r$ m is given by the formula $C = 2\pi r$ . A circular pond has a radius of 35 m. Find its circumference
(a) as a multiple of $\pi$ ,
(b) correct to 3 significant figures.

Solution

(a) $r = 35$ , so $C = 2 \times \pi \times 35 = 70\pi$ .

(b) Using the calculator's value of $\pi$ : $70\pi = 219.911\ldots$

Correct to 3 significant figures, $219.911\ldots = 220$ (3 s.f.).

(a) The circumference is $70\pi$ m. (b) The circumference is 220 m, correct to 3 significant figures.

Example 12: A cost with a fixed charge

A company hires out vans. It charges a fixed fee of $40 for the day and $0.60 for each kilometre driven. The cost of hiring a van for a day and driving it $n$ km is $C$ dollars, where
$C = 40 + 0.6n$ .
Find the cost when the van is driven (a) 85 km, (b) 137 km.

Solution

The 40 is the fixed fee, paid however far the van goes; $0.6n$ is the charge for $n$ km at $0.60 a kilometre.

(a) $n = 85$ : $C = 40 + 0.6 \times 85$ . Multiply first: $0.6 \times 85 = 51$ . So $C = 40 + 51 = 91$ .

(b) $n = 137$ : $C = 40 + 0.6 \times 137 = 40 + 82.2 = 122.2$ .

$C$ is a number of dollars, so 122.2 is written $122.20.

(a) The cost is $91. (b) The cost is $122.20.

Common Mistakes

Leaving out the brackets round a negative value. With $x = -3$ , $x^{2}$ is $(-3)^{2} = 9$ . Writing $-3^{2}$ gives $-9$ , which is wrong.
Multiplying before squaring. With $x = -3$ , $2x^{2} = 2 \times 9 = 18$ , not $(2 \times (-3))^{2} = 36$ . Only the letter next to the power is squared.
Reading $3a$ as a two-digit number. With $a = 5$ , $3a = 3 \times 5 = 15$ , not 35.
Losing the sign when subtracting a negative. $12 - (-13) = 12 + 13 = 25$ , not $12 - 13 = -1$ . Write the step $12 - (-13)$ before you change it.
Working from left to right without the order of operations. With $x = 4$ , $2 + 3x = 2 + 12 = 14$ , not $5 \times 4 = 20$ .
Mixing units in a formula. In $d = st$ with $s$ in km/h, $t$ must be in hours: 40 minutes is $\frac{2}{3}$ hour, not 40 and not 0.4.
Rounding too early. Keep the full value until the last step, then give the answer to 3 significant figures (money to the nearest cent).

Practice

Try each question before you open its solution. The questions get harder as you go.

Question 1

Given that $x = 6$ , find the value of
(a) $5x - 8$ ,
(b) $20 - 2x$ ,
(c) $3(x + 4)$ ,
(d) $\frac{x^{2}}{4}$ .

ShowHide solution

(a) $5x - 8 = 5 \times 6 - 8 = 30 - 8 = 22$ .

(b) $20 - 2x = 20 - 2 \times 6 = 20 - 12 = 8$ .

(d) $\frac{x^{2}}{4} = \frac{6^{2}}{4} = \frac{36}{4} = 9$ .

(a) 22 (b) 8 (c) 30 (d) 9

Question 2

Given that $a = 3$ and $b = 5$ , find the value of
(a) $2a^{2} + b$ ,
(b) $(a + b)^{2}$ ,
(c) $a^{2} + b^{2}$ ,
(d) $\frac{ab}{b - a}$ .

ShowHide solution

(a) $2a^{2} + b = 2 \times 3^{2} + 5 = 2 \times 9 + 5 = 18 + 5 = 23$ .

(b) $(a + b)^{2} = (3 + 5)^{2} = 8^{2} = 64$ .

(c) $a^{2} + b^{2} = 3^{2} + 5^{2} = 9 + 25 = 34$ . Parts (b) and (c) show that $(a + b)^{2}$ and $a^{2} + b^{2}$ are not the same.

(d) Top: $ab = 3 \times 5 = 15$ . Bottom: $b - a = 5 - 3 = 2$ . So $\frac{ab}{b - a} = \frac{15}{2} = 7\frac{1}{2}$ .

(a) 23 (b) 64 (c) 34 (d) $7\frac{1}{2}$

Question 3

Given that $x = -2$ , find the value of
(a) $x^{2}$ ,
(b) $-x^{2}$ ,
(c) $x^{3}$ ,
(d) $3x^{2}$ ,
(e) $(3x)^{2}$ ,
(f) $-x^{3}$ .

ShowHide solution

(a) $x^{2} = (-2)^{2} = (-2) \times (-2) = 4$ .

(b) $-x^{2} = -(-2)^{2} = -(4) = -4$ .

(d) $3x^{2} = 3 \times (-2)^{2} = 3 \times 4 = 12$ .

(e) $(3x)^{2} = (3 \times (-2))^{2} = (-6)^{2} = 36$ .

(f) $-x^{3} = -(-2)^{3} = -(-8) = 8$ .

(a) 4 (b) $-4$ (c) $-8$ (d) 12 (e) 36 (f) 8

Question 4

Given that $p = -4$ and $q = 3$ , find the value of
(a) $5q - p$ ,
(b) $pq^{2}$ ,
(c) $(p - q)^{2}$ ,
(d) $p^{2} - q^{2}$ .

ShowHide solution

(a) $5q - p = 5 \times 3 - (-4) = 15 - (-4) = 15 + 4 = 19$ .

(b) $pq^{2} = (-4) \times 3^{2}$ . Only $q$ is squared: $3^{2} = 9$ , so $pq^{2} = (-4) \times 9 = -36$ .

(d) $p^{2} - q^{2} = (-4)^{2} - 3^{2} = 16 - 9 = 7$ .

(a) 19 (b) $-36$ (c) 49 (d) 7

Question 5

Given that $a = 2$ , $b = -3$ and $c = -4$ , find the value of
(a) $\frac{abc}{a + b + c}$ ,
(b) $b^{2} - 4ac$ .

ShowHide solution

(a) Top: $abc = 2 \times (-3) \times (-4) = (-6) \times (-4) = 24$ .

Bottom: $a + b + c = 2 + (-3) + (-4) = -1 + (-4) = -5$ .

So $\frac{abc}{a + b + c} = \frac{24}{-5} = -4.8$ .

(b) $b^{2} - 4ac = (-3)^{2} - 4 \times 2 \times (-4)$ .

Power and multiplication first: $(-3)^{2} = 9$ and $4 \times 2 \times (-4) = -32$ .

So $b^{2} - 4ac = 9 - (-32) = 9 + 32 = 41$ .

(a) $-4.8$ (b) 41

Question 6

Given that $x = \frac{2}{3}$ and $y = -\frac{1}{2}$ , find the value of
(a) $3x + 4y$ ,
(b) $x^{2} - y$ ,
(c) $\frac{x}{y}$ .

ShowHide solution

(a) $3x + 4y = 3 \times \frac{2}{3} + 4 \times \left(-\frac{1}{2}\right) = 2 + (-2) = 0$ .

(b) $x^{2} - y = \left(\frac{2}{3}\right)^{2} - \left(-\frac{1}{2}\right) = \frac{4}{9} + \frac{1}{2}$ .

With the common denominator 18: $\frac{4}{9} + \frac{1}{2} = \frac{8}{18} + \frac{9}{18} = \frac{17}{18}$ .

(c) $\frac{x}{y} = \frac{2}{3} \div \left(-\frac{1}{2}\right) = \frac{2}{3} \times (-2) = -\frac{4}{3} = -1\frac{1}{3}$ .

(a) 0 (b) $\frac{17}{18}$ (c) $-1\frac{1}{3}$

Question 7

Given that $m = 1.5$ and $n = -0.8$ , find the value of $\frac{m^{2} - n}{m + n}$ , giving your answer correct to 3 significant figures.

ShowHide solution

Top: $m^{2} - n = 1.5^{2} - (-0.8) = 2.25 + 0.8 = 3.05$ .

Bottom: $m + n = 1.5 + (-0.8) = 0.7$ .

So $\frac{m^{2} - n}{m + n} = \frac{3.05}{0.7} = 4.357142\ldots = 4.36$ (3 s.f.).

4.36, correct to 3 significant figures

Question 8

The sum, $S$ , of the whole numbers from 1 to $n$ , $1 + 2 + 3 + \ldots + n$ , is given by the formula $S = \frac{n(n + 1)}{2}$ .
(a) Check the formula for $n = 4$ by adding the numbers.
(b) Find the sum of the whole numbers from 1 to 25.
(c) Find the sum of the whole numbers from 1 to 100.

ShowHide solution

(a) Adding: $1 + 2 + 3 + 4 = 10$ . By the formula: $S = \frac{4 \times (4 + 1)}{2} = \frac{4 \times 5}{2} = \frac{20}{2} = 10$ . They agree.

(b) $n = 25$ : $S = \frac{25 \times 26}{2} = \frac{650}{2} = 325$ .

(a) Both give 10. (b) 325 (c) 5050

Question 9

Three consecutive even numbers are $x - 2$ , $x$ and $x + 2$ , where $x$ is an even number. Their product, $P$ , is given by $P = (x - 2)x(x + 2)$ . Find $P$ when
(a) $x = 10$ ,
(b) $x = -4$ .

ShowHide solution

(a) The numbers are $10 - 2 = 8$ , $10$ and $10 + 2 = 12$ . So $P = 8 \times 10 \times 12 = 80 \times 12 = 960$ .

(b) The numbers are $(-4) - 2 = -6$ , $-4$ and $(-4) + 2 = -2$ .

So $P = (-6) \times (-4) \times (-2) = 24 \times (-2) = -48$ .

(a) 960 (b) $-48$

Question 10

Siti buys 3 sweets at $x$ cents each at a provision shop and has $6.50 left. The amount of money she had at first is $p$ dollars, where $p = 6.50 + 0.03x$ .
(a) Explain why the cost of the sweets is $0.03x$ dollars.
(b) Find how much money she had at first if each sweet cost 15 cents.
(c) Find how much money she had at first if each sweet cost 40 cents.

ShowHide solution

(a) 3 sweets at $x$ cents each cost $3x$ cents. There are 100 cents in a dollar, so $3x$ cents is $\frac{3x}{100}$ dollars, which is $0.03x$ dollars.

(b) $x = 15$ : $p = 6.50 + 0.03 \times 15 = 6.50 + 0.45 = 6.95$ .

(a) See the working. (b) She had $6.95 at first. (c) She had $7.70 at first.

Question 11

A temperature of $C$ °C is the same as $F$ °F (degrees Fahrenheit), where $F = 1.8C + 32$ .
(a) A typical afternoon in Singapore is 30 °C. Write this temperature in °F.
(b) A freezer is set to $-15$ °C. Write this temperature in °F.

ShowHide solution

(a) $C = 30$ : $F = 1.8 \times 30 + 32 = 54 + 32 = 86$ .

(b) $C = -15$ , in brackets: $F = 1.8 \times (-15) + 32$ .

Multiply first: $1.8 \times (-15) = -27$ . So $F = -27 + 32 = 5$ .

(a) 86 °F (b) 5 °F

Question 12

A circle has radius $x$ cm. A triangle has sides of $2x$ cm, $2x$ cm and $3x$ cm. The circumference of a circle of radius $r$ cm is $2\pi r$ cm. When $x = 4$ , which has the longer perimeter, the circle or the triangle, and by how much? Give your answers correct to 3 significant figures.

ShowHide solution

Circle: $r = x = 4$ , so the circumference is $2 \times \pi \times 4 = 8\pi = 25.132\ldots$ cm.

Triangle: the sides are $2 \times 4 = 8$ cm, 8 cm and $3 \times 4 = 12$ cm, so its perimeter is $8 + 8 + 12 = 28$ cm.

28 is greater than $25.132\ldots$ , so the triangle has the longer perimeter.

The difference is $28 - 25.132\ldots = 2.867\ldots = 2.87$ cm (3 s.f.). The circumference itself is 25.1 cm (3 s.f.).

The triangle has the longer perimeter: 28 cm, against 25.1 cm for the circle, longer by 2.87 cm (to 3 significant figures).

Key Points

Substitution is replacing each letter in an expression by its value; working out the result is evaluating the expression.
Put the $\times$ signs back before substituting: $3a$ is $3 \times a$ , $ab$ is $a \times b$ , $a^{2}b$ is $a \times a \times b$ .
Work in the order of operations: brackets, then powers, then $\times$ and $\div$ from left to right, then $+$ and $-$ from left to right.
A fraction line groups its top and its bottom: work each out, then divide.
Put a negative value in brackets: $(-3)^{2} = 9$ , but $-3^{2} = -9$ .
With $x = -3$ : $x^{2} = 9$ , $-x^{2} = -9$ , $x^{3} = -27$ , $2x^{2} = 18$ and $(2x)^{2} = 36$ .
Subtracting a negative is adding: $12 - (-13) = 12 + 13$ .
A formula is a rule connecting two or more quantities; the letter standing alone on one side is its subject.
To use a formula, know what each letter stands for and its unit, substitute, and give the subject's value with its unit.
An answer that is not exact is given correct to 3 significant figures (money to the nearest cent); round only at the end.