Lesson 3: BIDMAS and substitution

Introduction

Last lesson, each expression contained only one 'type' of operation. Addition and subtraction is one type of operation (two sides of the same coin, you might say), and multiplication and division is one type of operation. Now we're going to see expressions which mix different types of operations and it's going to be more difficult to decide which bits to work out first. Unless ... you know about the order of operations (BIDMAS)!

I'm also going to start to explain what a term is. It's a tricky one but I want to be able to use this word in future to explain other things so it's important.

SECTION A

The order of operations tells us to work out operations in this order:

B  ⇒ Work out stuff in Brackets first

I  ⇒ Then work out Indices (powers & roots)

DM ⇒ Division and Multiplication come next

AS ⇒ Finally Addition and Subtraction

We're interested in the DM and AS parts in Section A. We work out Division and Multiplication before Addition and Subtraction.

Example 1

Evaluate \(~8+7x~\),   if \(~x=2\)

Substitute: \(~8+7\times 2\)

Identify what to do first: \(~8+(7\times 2)\)

Do it: \(~8+14\)

Finish off: \(~22\)

When substituting and when working bits out, keep everything in the same position. In this example, \(\times ~2~\) is positioned exactly where the \(~x~\) was within the expression, and \(~14~\) is placed exactly where \(~(7\times 2)~\) was.

The 'Identify what to do first' line just shows that you've thought about BIDMAS and you're using brackets to show the reader which bit you're going to work out next.

Example 2

Evaluate \(~9-\Large\frac{a}{4}\normalsize~+b~\),   if \(~a=12\)  and  \(~b=7\)

Substitute: \(~9-\Large\frac{12}{4}\normalsize~+7\)

Identify what to do first: \(~9-\Large(\frac{12}{4}~)\normalsize~+7\)

Do it: \(~9-3+7\)

Finish off: \(~13\)

Example 3

Evaluate \(~2x-\Large\frac{15}{y}~\),   if \(~x=10~\)  and  \(~y=3\)

Substitute: \(~2\times 10~-\Large\frac{15}{3}\)

Identify what to do first: \(~(2\times 10)-\Large(\frac{15}{3}~)\)

Do it: \(~20-5\)

Finish off: \(~15\)

Multiplication and division are level in BIDMAS so just do them together.

Example 4

Evaluate \(~3a+\Large\frac{5b}{2c}~\),   if \(~a=13~\),  \(~b=6~\)  and  \(~c=3\)

Substitute: \(~3\times 13~+\Large\frac{5\times 6}{2\times 3}\)

Identify what to do first: \(~(3\times 13)~+~\Large(\frac{5\times 6}{2\times 3}~)\)

Work out top and bottom of fraction: \(~(3\times 13)~+~\Large(\frac{30}{6}~)\)

Do multiplication and division: \(~39+5\)

Do addition: \(~44\)

Practise to master

SECTION A

Evaluate the following expressions using your knowledge of the order of operations.

01)  \(8+5x~\)   (\(x=7\))

\(8+(5\times 7)\)
\(8+35\)
\(43\)
I start by writing my first line of workings \(~8+5\times 7~\) and then draw brackets on the same line to show which bit I'm going to work out next. That way, I save having to write \(~8+(5\times 7)~\) out separately. This makes the workings a bit shorter but still keeps them perfectly clear. Genius!

02)  \(8-pq~\)   (\(p=1\) ,  \(q=1\))

\(8-(1\times 1)\)
\(8-1\)
\(7\)

03)  \(4x+7y~\)   (\(x=3\) ,  \(y=2\))

\((4\times 3)+(7\times 2)\)
\(12+14\)
\(26\)

04)  \(4c-3de~\)   (\(c=14\) ,  \(d=3\) ,  \(e=5\))

\((4\times 14)-(3\times 3\times 5)\)
\(56-45\)
\(11\)

05)  \(6x+7y-5z~\)   (\(x=5\) ,  \(y=2\) ,  \(z=4\))

\((6\times 5)+(7\times 2)-(5\times 4)\)
\(30+14-20\)
\(24\)

06)  \(1+\Large\frac{a}{7}~\)   (\(a=42\))

\(1+\large(\frac{42}{7}~)\)
\(1+6\)
\(7\)

07)  \(6x-\Large\frac{48}{y}~\)   (\(x=12\) ,  \(y=4\))

\((6\times 12)-\large(\frac{48}{4}~)\)
\(72-12\)
\(60\)

08)  \(\Large\frac{a}{6}\normalsize-4b~\)   (\(a=60\) ,  \(b=2\))

\(\large(\frac{60}{6}~)\normalsize-(4\times 2)\)
\(10-8\)
\(2\)

09)  \(\Large\frac{36}{x}~\normalsize+\Large\frac{y}{2}~\)   (\(x=4\) ,  \(y=12\))

\(\large(\frac{36}{4}~)\normalsize+\large(\frac{12}{2}~)\)
\(9+6\)
\(15\)

10)  \(\Large\frac{r}{6}~\normalsize+5s-\Large\frac{t}{7}~\)   (\(r=72\) ,  \(s=3\) ,  \(t=21\))

\(\large(\frac{72}{6}~)\normalsize+(5\times 3)-\large(\frac{21}{7}~)\)
\(12+15-3\)
\(24\)

11)  \(8x-\Large\frac{2y}{3}~\)   (\(x=7\) ,  \(y=15\))

\((8\times 7)-\large(\frac{2\times 15}{3}~)\)
\((8\times 7)-\large(\frac{30}{3}~)\)
\(56-10=46\)

12)  \(\Large\frac{50}{pq}~\normalsize +3r\)   (\(p=2\) ,  \(q=5\) ,  \(r=6\))

\(\large(\frac{50}{2\times 5}~)~\normalsize +(3\times 6)\)
\(\large(\frac{50}{10}~)~\normalsize +(3\times 6)\)
\(5+18=23\)

13)  \(\Large\frac{42}{x}~\normalsize-~\Large\frac{3y}{2z}\)   (\(x=2\) ,  \(y=14\) ,  \(z=3\))

\(\large(\frac{42}{2}~)~\normalsize-~\large(\frac{3\times 14}{2\times 3}~)\)
\(\large(\frac{42}{2}~)~\normalsize-~\large(\frac{42}{6}~)\)
\(21-7=14\)

14)  \(5+\Large\frac{8a}{3bc}~\)   (\(a=6\) ,  \(b=2\) ,  \(c=8\))

\(5+\large(\frac{8\times 6}{3\times 2\times 8}~)\)
\(5+\large(\frac{48}{48}~)\)
\(5+1=6\)

15)  \(\Large\frac{xy}{2z}~\normalsize -11\)   (\(x=6\) ,  \(y=11\) ,  \(z=3\))

\(\large(\frac{6\times 11}{2\times 3}~)~\normalsize -11\)
\(\large(\frac{66}{6}~)~\normalsize -11\)
\(11-11=0\)

16)  \(\Large\frac{3p}{q}~\normalsize -4+3r\)   (\(p=4\) ,  \(q=2\) ,  \(r=7\))

\(\large(\frac{3\times 4}{2}~)~\normalsize -4+(3\times 7)\)
\(\large(\frac{12}{2}~)~\normalsize -4+(3\times 7)\)
\(6-4+21=23\)

SECTION B

Expressions in brackets

An expression is a collection of terms separated by \(~+~\) and/or \(~-~\) signs.
So terms are the parts of an expression separated by \(~+~\) and/or \(~-~\) signs.

A term can be a number or letter.
The expressions in Lesson 2, Section A were made up of terms like this.

A term can be a combination of number(s) and letter(s) multiplied together and/or divided.
The terms in Section A of this lesson (that weren't simply a number or a letter) were like this.

But wait!

A term can also contain little expressions. What's that you say? You can have little expressions within terms within bigger expressions?! Don't worry, it just sounds complicated!

The term \(~5(x+4)~\) is a combination of a number and something multiplied together. But that something is not a simple letter, it's a whole mini expression, which we have to put in brackets to keep it separate from the \(~5~\). We evaluate the expression in brackets before we multiply by 5 because brackets are always worked out first!

The term \(~\Large\frac{a+4}{3(7-b)}~\)  is a combination of three things multiplied or divided.
Two of those things are whole mini expressions! The little expression on the bottom has to be put in brackets to keep it apart from the \(~3~\), while the expression on the top doesn't need brackets because it's on its own.

Work carefully through the examples and you'll be fine!

Example 1

Evaluate \(~2(x+5)~\),   if \(~x=16~\)

Substitute: \(~2\times (16+5)\)

Brackets first: \(~2\times 21\)

Finish off: \(~42\)

Without brackets, you'd have \(~2x+5~\), you'd do multiplication first \(~(2\times 16=32)~\) and then addition \(~(32+5=37)~\).

Example 2

Evaluate \(~5(x-2)+6y~\),   if \(~x=4~\) and \(~y=3~\)

Substitute: \(~5\times (4-2)+6\times 3\)

Brackets first: \(~5\times 2+6\times 3\)

Do multiplication: \(~10+18\)

Finish off: \(~28\)

Previously, I would have put brackets in \(~5\times 2+6\times 3~\) to show which bits I'm going to work out next. I'm going to stop doing that now because there are already brackets in the question and it might get confusing!

Example 3

Evaluate \(~5(x-3)(y+2)~\),   if \(~x=7~\) and \(~y=1~\)

Substitute: \(~5\times (7-3)\times (1+2)~\)

Brackets first: \(~5\times 4\times 3~\)

Finish off: \(~60\)

Brackets right next to each other with nothing in between means they're multiplied together.

Example 4

Evaluate \(~\Large\frac{32-p}{5+q}~\),   if \(~p=4\)  and \(~q=2\)

Substitute: \(~\Large\frac{32-4}{5+2}~\)

Top and bottom of fraction: \(~\Large\frac{28}{7}~\)

Finish off: \(~4\)

Example 5

Evaluate \(~\Large\frac{2(a-5)}{1+b}~\),   if \(~a=11\)  and \(~b=3\)

Substitute: \(~\Large\frac{2(11-5)}{1+3}~\)

Brackets first: \(~\Large\frac{2\times 6}{1+3}~\)

Top and bottom of fraction: \(~\Large\frac{12}{4}~\)

Finish off: \(~3\)

Practise to master

SECTION B

Evaluate the following expressions.

01)  \(3(x-1)~\) ,   if \(~x=7~\)

\(3\times (7-1)\)
\(3\times 6\)
\(18\)
Without brackets, we'd do multiplication before subtraction. But brackets force us to do the subtraction first!

02)  \(4(n+9)-5~\) ,   if \(~n=3~\)

\(4\times (3+9)-5\)
\(4\times 12-5\)
\(48-5\)
\(43\)
Brackets first then multiplication then subtraction.

03)  \(7x-2(y+4)~\) ,   if \(~x=6~\) and \(~y=2~\)

\(7\times 6-2\times(2+4)\)
\(7\times 6-2\times 6\)
\(42-12\)
\(30\)
Brackets first then two multiplications then subtraction.

04)  \(2(3x+1)+8~\) ,   if \(~x=2~\)

\(2\times (3\times 2+1)+8~\)
\(2\times (6+1)+8~\)
\(2\times 7+8~\)
\(14+8~\)
\(22~\)
Here there are two operations within the brackets and they have to be worked out in the right order, of course!

05)  \(7a+3(19-4b)~\) ,   if \(~a=4~\) and \(~b=3~\)

\(7\times 4+3\times (19-4\times 3)~\)
\(7\times 4+3\times (19-12)~\)
\(7\times 4+3\times 7~\)
\(28+21~\)
\(49\)

06)  \(3\Large(\frac{x}{2}\normalsize-9\Large)\normalsize~+~5y~\) ,   if \(~x=26~\) and \(~y=4~\)

\(3\times \large(\frac{26}{2}\normalsize-9\large)\normalsize~+~5\times 4~\)
\(3\times (13-9)+5\times 4~\)
\(3\times 4+5\times 4~\)
\(12+20~\)
\(32~\)

07)  \(2(x+1)(y-2)~\) ,   if \(~x=4~\) and \(~y=8~\)

\(2\times (4+1)\times (8-2)\)
\(2\times 5\times 6\)
\(60\)

08)  \(3x(2y-3)(z+5)~\) ,   if \(~x=2~\), \(~y=3~\) and \(~z=1~\)

\(3\times 2\times (2\times 3-3)\times (1+5)~\)
\(3\times 2\times (6-3)\times (1+5)~\)
\(3\times 2\times 3\times 6~\)
\(108\)

09)  \(\Large\frac{x+5}{y-2}~\) ,   if \(~x=13~\) and \(~y=5~\)

\(\Large\frac{13+5}{5-2}\normalsize~=~\Large\frac{18}{3}\normalsize~=~6\)

10)  \(\Large\frac{3a-4}{b+1}~\) ,   if \(~a=12~\) and \(~b=7~\)

\(\Large\frac{3\times 12-4}{7+1}\normalsize~=~\Large\frac{36-4}{7+1}\normalsize~=~\Large\frac{32}{8}\normalsize~=~4\)

11)  \(\Large\frac{x+9}{10-2y}~\) ,   if \(~x=11~\) and \(~y=3~\)

\(\Large\frac{11+9}{10-2\times 3}\normalsize~=~\Large\frac{11+9}{10-6}\normalsize~=~\Large\frac{20}{4}\normalsize~=~5\)

12)  \(\Large\frac{31-5p}{3q-2}~\) ,   if \(~p=2~\) and \(~q=3~\)

\(\Large\frac{31-5\times 2}{3\times 3-2}\normalsize~=~\Large\frac{31-10}{9-2}\normalsize~=~\Large\frac{21}{7}\normalsize~=~3\)

13)  \(\Large\frac{6(x+4)}{y-9}~\) ,   if \(~x=8~\) and \(~y=17~\)

\(\Large\frac{6\times (8+4)}{17-9}\normalsize~=~\Large\frac{6\times 12}{17-9}\normalsize~=~\Large\frac{72}{8}\normalsize~=~9\)

14)  \(\Large\frac{a-2}{2(b+3)}~\) ,   if \(~a=82~\) and \(~b=2~\)

\(\Large\frac{82-2}{2\times (2+3)}\normalsize~=~\Large\frac{82-2}{2\times 5}\normalsize~=~\Large\frac{80}{10}\normalsize~=~8\)

15)  \(\Large\frac{3(x+7)+4}{y+5}~\) ,   if \(~x=5~\) and \(~y=3~\)

\(\Large\frac{3\times (5+7)+4}{3+5}\normalsize~=~\Large\frac{3\times 12+4}{3+5}\normalsize~=~\Large\frac{36+4}{3+5}\normalsize~=~ \Large\frac{40}{8}\normalsize~=~5\)

16)  \(\Large\frac{p+14}{6(3q-16)-9}~\) ,   if \(~p=19~\) and \(~q=6~\)

\(\Large\frac{19+14}{6\times (3\times 6-16)-9}\normalsize~=~\Large\frac{19+14}{6\times (18-16)-9}\normalsize~=~ \Large\frac{19+14}{6\times 2-9}\normalsize~=~\Large\frac{19+14}{12-9}\normalsize~=~\Large\frac{33}{3}\normalsize~=~11\)

17)  \(\Large\frac{4(x-9)}{y+5}\normalsize~+3z\),   if \(~x=13~\) , \(~y=3~\) and \(~z=6~\)

\(\Large\frac{4(13-9)}{3+5}\normalsize~+3\times 6~=~\Large\frac{4\times 4}{3+5}\normalsize~+3\times 6~=~ \Large\frac{16}{8}\normalsize~+3\times 6~=~2+18~=~20\)

18)  \(\Large\frac{2(3x+5)}{y-3}\normalsize~+\Large\frac{35}{z}\) ,   if \(~x=5~\), \(~y=7~\) and \(~z=5~\)

\(\Large\frac{2\times (3\times 5+5)}{7-3}\normalsize~+\Large\frac{35}{5}\)
\(\Large\frac{2\times (15+5)}{7-3}\normalsize~+\Large\frac{35}{5}\)
\(\Large\frac{2\times 20}{7-3}\normalsize~+\Large\frac{35}{5}\)
\(\Large\frac{40}{4}\normalsize~+\Large\frac{35}{5}\)
\(10+7~=~17\)

SECTION C

The only part of BIDMAS left to deal with is the I (indices) so let's get to it! There's no need to involve powers higher than 2 at this stage so we'll just stick to squares and square roots.

Just to refresh, \(~1^2=1\times 1=1~\), \(~2^2=2\times 2=4~\), \(~3^2=3\times 3=9~\), etc. In other words, something squared means that thing multiplied by itself. Square root is the opposite of square, so \(~\sqrt{16}=4~\) because \(~4^2=16~\).

The following examples should teach you everything else you need to know.

Example 1

Evaluate \(~3x^2~\),   if \(~x=4\)

Substitute: \(~3\times 4^2~\)

Indices first: \(~3\times 16~\)

Then multiplication: \(~48\)

You'd be amazed how many times I've seen \(~3\times 4=12~\) then \(~12^2=144~\). Follow the rules of BIDMAS and you won't make this common mistake. Compare this with the following example.

Example 2

Evaluate \(~(3x)^2~\),   if \(~x=4\)

Substitute: \(~(3\times 4)^2~\)

Brackets first: \(~12^2~\)

Then indices: \(~144\)

It's really important that you understand the difference between these first two examples.

Example 3

Evaluate \(~2x^2+4x-5~\),   if \(~x=3\)

Substitute: \(~2\times 3^2+4\times 3-5~\)

Indices first: \(~2\times 9+4\times 3-5~\)

Multiplication: \(~18+12-5~\)

Addition and subtraction: \(~25\)

Example 4

Evaluate \(~3\sqrt {x}~\),   if \(~x=25\)

Substitute: \(~3\times \sqrt{25}~\)

Indices first: \(~3\times 5~\)

Multiplication: \(~15\)

Example 5

Evaluate \(~\sqrt {3x}~\),   if \(~x=12\)

Substitute: \(~\sqrt{3\times 12}~\)

Evaluate inside square root sign: \(~\sqrt{36}~\)

Square root: \(~6\)

Whatever the root sign covers has to be evaluated before the actual square rooting can be done! A root sign acts like a bracket, holding everything underneath it together.

Here are some practice questions. No need to go OTT with these. We'll meet indices again next lesson and develop things a bit further together with negative numbers.

Practise to master

SECTION C

Evaluate the following expressions.

01)  \(5x^2\) ,   if \(~x=6~\)

\(5\times 6^2\)
\(5\times 36\)
\(180\)

02)  \((5x)^2\) ,   if \(~x=6~\)

\((5\times 6)^2\)
\(30^2\)
\(900\)

03)  \(2+3a^2\) ,   if \(~a=5~\)

\(2+3\times 5^2\)
\(2+3\times 25\)
\(2+75\)
\(77\)

04)  \(2+(3a)^2\) ,   if \(~a=5~\)

\(2+(3\times 5)^2\)
\(2+15^2\)
\(2+225\)
\(227\)

05)  \((2+3a)^2\) ,   if \(~a=5~\)

\((2+3\times 5)^2\)
\((2+15)^2\)
\((17)^2\)
\(289\)

06)  \(\large\frac{p^2}{2}\) ,   if \(~p=8~\)

\(\large\frac{8^2}{2}\normalsize~=~\large\frac{64}{2}\normalsize~=~32\)

07)  \(\Big(\large\frac{p}{2}\normalsize~\Big)^2\) ,   if \(~p=8~\)

\(\Big(\large\frac{8}{2}\normalsize~\Big)^2~=~4^2~=~16\)

08)  \(4x^2y\) ,   if \(~x=3~\) and \(~y=10~\)

\(4\times 3^2\times 10\)
\(4\times 9\times 10\)
\(360\)

09)  \(3x^2-4y^2\) ,   if \(~x=4~\) and \(~y=2~\)

\(3\times 4^2-4\times 2^2\)
\(3\times 16-4\times 4\)
\(48-16\)
\(32\)

10)  Are \(~\Large\frac{x^2}{y^2}~\)  and \(~\Big(\Large\frac{x}{y}\normalsize~\Big)^2~\) equivalent?

Yes!
Pick a value of \(~x~\), substitute it into both terms and evaluate. Not convinced?
Try it with a different value of \(~x~\). Still not convinced?
\(~\Big(\Large\frac{x}{y}\normalsize~\Big)^2~=~\Large\frac{x}{y}~\normalsize\times \Large\frac{x}{y}\normalsize~=~\Large\frac{x\times x}{y\times y}\normalsize~=~\Large\frac{x^2}{y^2}\)
Remember how to multiply fractions? Times the tops and times the bottoms!

11)  \(2x^2+5x-9\) ,   if \(~x=3~\)

\(2\times 3^2+5\times 3-9\)
\(2\times 9+5\times 3-9\)
\(18+15-9\)
\(24\)
You'll see loads like this when you start plotting quadratic graphs!

12)  \(3x^2-2x+7\) ,   if \(~x=4~\)

\(3\times 4^2-2\times 4+7\)
\(3\times 16-2\times 4+7\)
\(48-8+7\)
\(47\)

13)  \(6\sqrt{x}\) ,   if \(~x=64~\)

\(6\times \sqrt{64}\)
\(6\times 8\)
\(48\)

14)  \(\sqrt{x+27}\) ,   if \(~x=22~\)

\(\sqrt{22+27}\)
\(\sqrt{49}\)
\(7\)

15)  \(\sqrt{b^2-4ac}\) ,   if \(~a=2~\), \(~b=5~\) and \(~c=3~\)

\(\sqrt{5^2-4\times 2\times 3}\)
\(\sqrt{25-4\times 2\times 3}\)
\(\sqrt{25-24}\)
\(\sqrt{1}\)
\(1\)
This is another really important expression (or part of one)!
It'll be a big help with solving quadratic equations.

16)  Do you think \(~\Large\frac{\sqrt{x}}{\sqrt{y}}~\)  and \(~\sqrt{\Large\frac{x}{y}}~\)  are equivalent?

They are!
By all means test this out by substituting values for \(~x~\) and \(~y~\).

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