Lesson 2: BIDMAS and linear equations

Introduction

The topic Substitution & formulae introduced the important role in algebra of the order of operations (also known as BIDMAS or BODMAS). Its role is just as important when it comes to Forming & solving equations. Again this lesson continues to lay solid foundations for future study so I'd advise spending some time on it.

SECTION A

A builder hires a floor sander. He pays a one-off delivery charge of £5 plus a daily charge of £12. The bill is £41 in total. How many days did he hire the sander for?

Form an equation

£12 times the number of days plus another £5 equals £41  ⇒  \(12x+5=41\)
(where \(x\) is the unknown number of days)

Solve the equation

Subtract 5 from both sides  ⇒  \(12x=36\)

Divide both sides by 12  ⇒  \(x=3\)

Answer  ⇒  The builder hired the sander for 3 days.

The importance of the order of operations

If we multiply the value of \(x\) by \(12\) and then add \(5\), we get \(41\). We have to do it this way around as multiplication comes before addition in the order of operations! We can make this a bit more visual by drawing something called a function machine.

\(x\)   →   \(\boxed{\times 12}\)   →   \(\boxed{+5}\)   →   \(41\)

When we solve an equation, everything gets reversed, including the order! Follow the next function machine from right to left and you'll see what I mean.

\(3\)   ←   \(\boxed{\div 12}\)   ←   \(\boxed{-5}\)   ←   \(41\)

The GREATER importance of keeping your balance

The absolute golden rule of solving equations:

Make a change to the whole of one side of an equation, then make exactly the same change to the whole of the other side.

To solve the equation \(12x+5=41\), we subtracted \(5\) from the whole of both sides then divided the whole of both sides by \(12\).

But what would happen if we divided the whole of both sides by \(12\) first? In other words, stuck to the golden rule but didn't worry too much about the order of operations?

\(\large\frac{12x+5}{12}\normalsize~=~\large\frac{41}{12}\)

\(\large\frac{12x}{12}\normalsize~+~\large\frac{5}{12}\normalsize~=~\large\frac{41}{12}\)

 \(~x~+~\large\frac{5}{12}\normalsize~=~\large\frac{41}{12}\)

 \(~x=\large\frac{41}{12}\normalsize~-~\large\frac{5}{12}\)

 \(~x=\large\frac{36}{12}\)

 \(~x=3~\)  Phew!

So keeping your balance (sticking to the golden rule) is the most important thing. But if you want to make solving equations quick and easy, choose the change that is easiest to make. In other words, follow the order of operations backwards!

Another example

Solve the equation \(~7+5x=42~\)

Subtract 7 from both sides  ⇒  \(5x=42-7\)

Simplify  ⇒  \(5x=35\)

Divide both sides by 5  ⇒  \(x=35\div 5\)

Simplify  ⇒  \(x=7\)

Practise to master

SECTION A

Solve the following equations.

01)  \(2x+8=14\)

Subtract 8 from both sides
\(2x=6\)
Divide both sides by 2
\(x=3\)

02)  \(5x+2=12\)

Subtract 2 from both sides
\(5x=10\)
Divide both sides by 5
\(x=2\)

03)  \(6+8x=46\)

Subtract 6 from both sides
\(8x=40\)
Divide both sides by 8
\(x=5\)

04)  \(7+3x=10\)

Subtract 7 from both sides
\(3x=3\)
Divide both sides by 3
\(x=1\)

05)  \(9x-5=31\)

Add 5 to both sides
\(9x=36\)
Divide both sides by 9
\(x=4\)

06)  \(7x-3=39\)

Add 3 to both sides
\(7x=42\)
Divide both sides by 7
\(x=6\)

07)  \(-2+3x=7\)

Add 2 to both sides
\(3x=9\)
Divide both sides by 3
\(x=3\)

08)  \(-5+4x=23\)

Add 5 to both sides
\(4x=28\)
Divide both sides by 4
\(x=7\)

09)  \(6x+4=16\)

Subtract 4 from both sides
\(6x=12\)
Divide both sides by 6
\(x=2\)

10)  \(8x-3=37\)

Add 3 to both sides
\(8x=40\)
Divide both sides by 8
\(x=5\)

11)  \(-9+2x=7\)

Add 9 to both sides
\(2x=16\)
Divide both sides by 2
\(x=8\)

12)  \(6+5x=21\)

Subtract 6 from both sides
\(5x=15\)
Divide both sides by 5
\(x=3\)

13)  \(7x-1=6\)

Add 1 to both sides
\(7x=7\)
Divide both sides by 7
\(x=1\)

14)  \(4x+4=40\)

Subtract 4 from both sides
\(4x=36\)
Divide both sides by 4
\(x=9\)

15)  \(9+6x=57\)

Subtract 9 from both sides
\(6x=48\)
Divide both sides by 6
\(x=8\)

16)  \(-7+11x=15\)

Add 7 to both sides
\(11x=22\)
Divide both sides by 11
\(x=2\)

Problem 1
I pick a number, multiply it by 3, subtract 8 and I get 13. Form and solve an equation to find the number I picked.

\(3x-8=13\)
Add 8 to both sides
\(3x=21\)
Divide both sides by 3
\(x=7\)
The number I picked was 7.

Problem 2
A plumber charges a call out fee of £18 plus an additional £22 per hour. The total bill is £106. How many hours did he work?

\(18+22x=106\)
Subtract 18 from both sides
\(22x=88\)
Divide both sides by 22
\(x=4\)
The plumber worked for 4 hours.

SECTION B

Questions in this section just involve an extra bit of simplification. They do allow for slightly more interesting problems, though.

Example

A triangle has sides of length \(3\) cm, \(2x\) cm and \(3x\) cm. Its perimeter is \(38\) cm. Work out its actual side lengths.

Form an equation  ⇒  \(3+2x+3x=38\)

Add terms to simplify  ⇒  \(3+5x=38\)

Subtract 3 from both sides  ⇒  \(5x=35\)

Divide both sides by 5  ⇒  \(x=7\)

Hence the side lengths are \(3\) cm, \(14\) cm and \(21\) cm
These add to make 38 so we can be confident that our answer is correct.

Practise to master

SECTION B

Solve these linear equations by first simplifying (collecting) the \(~x~\) terms.

01)  \(3x+5x+7=31\)

Collect terms to simplify
\(8x+7=31\)
Subtract 7 from both sides
\(8x=24\)
Divide both sides by 8
\(x=3\)

02)  \(7x-2x+3=58\)

Collect terms to simplify
\(5x+3=58\)
Subtract 3 from both sides
\(5x=55\)
Divide both sides by 5
\(x=11\)

03)  \(9x+6x-8=52\)

Collect terms to simplify
\(15x-8=52\)
Add 8 to both sides
\(15x=60\)
Divide both sides by 15
\(x=4\)

04)  \(3x-x-5=13\)

Collect terms to simplify
\(2x-5=13\)
Add 5 to both sides
\(2x=18\)
Divide both sides by 2
\(x=9\)

05)  \(6x-9+x+1=27\)

Collect terms to simplify
\(7x-8=27\)
Add 8 to both sides
\(7x=35\)
Divide both sides by 7
\(x=5\)

06)  \(4x+7-x-2=8\)

Collect terms to simplify
\(3x+5=8\)
Subtract 5 from both sides
\(3x=3\)
Divide both sides by 3
\(x=1\)

07)  \(5x+3+4x+14=71\)

Collect terms to simplify
\(9x+17=71\)
Subtract 17 from both sides
\(9x=54\)
Divide both sides by 9
\(x=6\)

08)  \(8x-4-4x-7=21\)

Collect terms to simplify
\(4x-11=21\)
Add 11 to both sides
\(4x=32\)
Divide both sides by 4
\(x=8\)

Problem 1
A salesman gets paid £50 per day plus an additional £6 for every sale he makes. Yesterday he was paid £98. How many sales did he make yesterday?

\(50+6x=98\)
Subtract 50 from both sides
\(6x=48\)
Divide both sides by 6
\(x=8\)
He made 8 sales yesterday.

Problem 2
A quadrilateral has sides of length \(11\) cm, \(5\) cm, \(3x+1\) cm and \(4x\) cm. Its perimeter is \(59\) cm. Work out its side lengths.

\(11+5+3x+1+4x=59\)
Collect terms to simplify
\(17+7x=59\)
Subtract 17 from both sides
\(7x=42\)
Divide both sides by 7
\(x=6\)
Substitute for \(x\)
\(3x+1=3(6)+1=18+1=19\)
\(4x=4(6)=24\)
Side lengths are \(11\) cm, \(5\) cm, \(19\) cm and \(24\) cm.

Problem 3
The angles of a triangle are \(~3x+11~\), \(~2x-7~\) and \(~x+20\). Work out the size of the largest angle.

\(3x+11+2x-7+x+20=180\)
Collect terms to simplify
\(6x+24=180\)
Subtract 24 from both sides
\(6x=156\)
Divide both sides by 6
\(x=26\)
Substitute for \(x\)
\(3x+11=3(26)+11=78+11=89\)
\(2x-7=2(26)-7=52-7=45\)
\(x+20=26+20=46\)
The size of the largest angle is \(89^{\circ}\).

Problem 4
Two decorators work together, one charges £15 per hour and the other £20 per hour. The total cost of paint and brushes is £43. The final bill including labour and materials is £183. How long did the decorators take?

\(15x+20x+43=183\)
Collect terms to simplify
\(35x+43=183\)
Subtract 43 from both sides
\(35x=140\)
Divide both sides by 35
\(x=4\)
The decorators worked together for 4 hours.

SECTION C

Remember learning how to deal with negative unknowns last lesson?

Example

Solve the equation \(~8-3x=2~\)

Subtract 8 from both sides  ⇒  \(~-3x=-6~\)

Divide both sides by ‒3  ⇒  \(~x=2~\)

I could have divided both sides by positive 3 instead of negative 3 but then I'd have \(~-x=-2~\) and I would have to multiply both sides by \(-1\) to get \(x\) on its own.

The following questions combine what we've done so far this lesson with some good old negative number work.

Practise to master

SECTION C

Solve these equations using your knowledge of negative numbers.

01)  \(2x+3=1\)

Subtract 3 from both sides
\(2x=-2\)
Divide both sides by 2
\(x=-1\)

02)  \(8x+19=-5\)

Subtract 19 from both sides
\(8x=-24\)
Divide both sides by 8
\(x=-3\)

03)  \(3x-13=-1\)

Add 13 to both sides
\(3x=12\)
Divide both sides by 3
\(x=4\)

04)  \(6x-5=-35\)

Add 5 to both sides
\(6x=-30\)
Divide both sides by 6
\(x=-5\)

05)  \(-4x-13=11\)

Add 13 to both sides
\(-4x=24\)
Divide both sides by ‒4
\(x=-6\)

06)  \(-2x-19=-1\)

Add 19 to both sides
\(-2x=18\)
Divide both sides by ‒2
\(x=-9\)

07)  \(-5x-7=-42\)

Add 7 to both sides
\(-5x=-35\)
Divide both sides by ‒5
\(x=7\)

08)  \(17-2x=5\)

Subtract 17 from both sides
\(-2x=-12\)
Divide both sides by ‒2
\(x=6\)

09)  \(-7x+26=12\)

Subtract 26 from both sides
\(-7x=-14\)
Divide both sides by ‒7
\(x=2\)

10)  \(3-4x=31\)

Subtract 3 from both sides
\(-4x=28\)
Divide both sides by ‒4
\(x=-7\)

11)  \(-12x+3=51\)

Subtract 3 from both sides
\(-12x=48\)
Divide both sides by ‒12
\(x=-4\)

12)  \(41-9x=-4\)

Subtract 41 from both sides
\(-9x=-45\)
Divide both sides by ‒9
\(x=5\)

SECTION D

Time to explore decimals in equations and learn a neat way of getting rid of them, if you don't like them. I don't!

Example

Solve the equation \(~0.5x+0.2=1.2~\)

Multiply the whole of both sides by 10  ⇒  \(~5x+2=12~\)

Subtract 2 from both sides  ⇒  \(~5x=10~\)

Divide both sides by 5  ⇒  \(~x=2~\)

Multiplying the whole of both sides by 10 at the start get's rid of the decimals. I just prefer whole numbers, I guess.

Practise to master

SECTION D

Solve the following equations.

01)  \(0.2x+0.8=1.4\)

Multiply the whole of both sides by 10
\(2x+8=14\)
Subtract 8 from both sides
\(2x=6\)
Divide both sides by 2
\(x=3\)

02)  \(0.4x+2=5.6\)

Multiply the whole of both sides by 10
\(4x+20=56\)
Subtract 20 from both sides
\(4x=36\)
Divide both sides by 4
\(x=9\)

03)  \(10x-5.9=7.9\)

Multiply the whole of both sides by 10
\(100x-59=79\)
Add 59 to both sides
\(100x=138\)
Divide both sides by 100
\(x=1.38\)

04)  \(5x+0.8=4.3\)

Multiply the whole of both sides by 10
\(50x+8=43\)
Subtract 8 from both sides
\(50x=35\)
Divide both sides by 5
\(10x=7\)
Divide both sides by 10
\(x=0.7\)
Instead of dividing both sides by 5, I could have multiplied both sides by 2. That would also have made the next step nice and easy and that's what its all about!

05)  \(3x-1.3=2.9\)

Multiply the whole of both sides by 10
\(30x-13=29\)
Add 13 to both sides
\(30x=42\)
Divide both sides by 3
\(10x=14\)
Divide both sides by 10
\(x=1.4\)

06)  \(1.2x+1.1=8\)

Multiply the whole of both sides by 10
\(12x+11=80\)
Subtract 11 from both sides
\(12x=69\)
Divide both sides by 12
\(x=\frac{69}{12}\)
Simplify the fraction
\(x=\frac{23}{4}\)
You can also write your answer as \(5.75\).

07)  \(0.9-0.8x=1.9\)

Multiply the whole of both sides by 10
\(9-8x=19\)
Subtract 9 from both sides
\(-8x=10\)
Divide both sides by ‒8
\(x=-\frac{10}{8}\)
Simplify the fraction
\(x=-\frac{5}{4}\)
You can also write your answer as \(-1.25\).

08)  \(-0.3x+1.2=-5.2\)

Multiply the whole of both sides by 10
\(-3x+12=-52\)
Subtract 12 from both sides
\(-3x=-64\)
Divide both sides by ‒3
\(x=\frac{64}{3}\)
You could also write your answer as \(~21\frac{1}{3}~\).

SECTION E

This section should reinforce your understanding of the role of the order of operations in solving equations. Work your way carefully through these examples and make sure you fully understand them.

Example 1

Solve the equation \(~\large\frac{3x}{5}\normalsize~+1=7\)

Subtract 1 from both sides  ⇒  \(~\large\frac{3x}{5}\normalsize~=6\)

Multiply both sides by 5  ⇒  \(~3x=30\)

Divide both sides by 3  ⇒  \(~x=10\)

The importance of the order of operations

Here's the original equation in the form of a function machine.

\(x\)   →   \(\boxed{\times 3}\)   →   \(\boxed{\div 5}\)   →   \(\boxed{+1}\)   →   \(7\)

Follow the reverse function machine from right to left to see how the equation was solved.

\(10\)   ←   \(\boxed{\div 3}\)   ←   \(\boxed{\times 5}\)   ←   \(\boxed{-1}\)   ←   \(7\)

The challenge is to get the first function machine right using your knowledge of the order of operations.

By the way, we could have swapped around the final two steps (they have equal priority in the order of operations AND they are next to each other in the function machine) but we like to get rid of denominators as soon as we can!

The GREATER importance of keeping your balance

You could also start this by multiplying throughout by 5 to give \(~3x+5=35~\). I talked about this in Section A but it's also related to what you learned in Section D. I'll leave you to think about this, if you feel like it!

Example 2

Solve the equation \(~\large\frac{2x+7}{3}\normalsize~=5\)

Multiply both sides by 3  ⇒  \(~2x+7=15\)

Subtract 7 from both sides  ⇒  \(~2x=8\)

Divide both sides by 2  ⇒  \(~x=4\)

The importance of the order of operations

Here's the original equation in the form of a function machine.

\(x\)   →   \(\boxed{\times 2}\)   →   \(\boxed{+7}\)   →   \(\boxed{\div 3}\)   →   \(5\)

Follow the reverse function machine from right to left to see how the equation was solved.

\(x\)   ←   \(\boxed{\div 2}\)   ←   \(\boxed{-7}\)   ←   \(\boxed{\times 3}\)   ←   \(5\)

Here the multiplication and division steps could not be swapped around as they are not next to each other in the function machine.

Practise to master

SECTION E

Solve these equations paying particular attention to the order of operations.

01)  \(\large\frac{2x}{3}\normalsize~+4=10\)

Subtract 4 from both sides
\(\large\frac{2x}{3}\normalsize~=6\)
Multiply both sides by 3
\(2x=18\)
Divide both sides by 2
\(x=9\)

02)  \(\large\frac{3}{4}\normalsize~x+8=17\)

Subtract 8 from both sides
\(\large\frac{3}{4}\normalsize~x=9\)
Multiply both sides by 4
\(3x=36\)
Divide both sides by 3
\(x=12\)

03)  \(\large\frac{4}{x}\normalsize~+3=5\)

Subtract 3 from both sides
\(\large\frac{4}{x}\normalsize~=2\)
Multiply both sides by \(x\)
\(4=2x\)
Divide both sides by 2
\(x=2\)

04)  \(\large\frac{2x+1}{3}\normalsize~=9\)

Multiply both sides by 3
\(2x+1=27\)
Subtract 1 from both sides
\(2x=26\)
Divide both sides by 2
\(x=13\)

05)  \(\large\frac{5x-9}{7}\normalsize~=3\)

Multiply both sides by 7
\(5x-9=21\)
Add 9 to both sides
\(5x=30\)
Divide both sides by 5
\(x=6\)

06)  \(\large\frac{x+12}{4}\normalsize~=11\)

Multiply both sides by 4
\(x+12=44\)
Subtract 12 from both sides
\(x=32\)

07)  \(\large\frac{3x}{5}\normalsize~+7=1\)

Subtract 7 from both sides
\(\large\frac{3x}{5}\normalsize~=-6\)
Multiply both sides by 5
\(3x=-30\)
Divide both sides by 3
\(x=-10\)

08)  \(3-\large\frac{2}{3}\normalsize~x=-1\)

Subtract 3 from both sides
\(-\large\frac{2}{3}\normalsize~x=-4\)
Multiply both sides by ‒1
\(\large\frac{2}{3}\normalsize~x=4\)
Multiply both sides by 3
\(2x=12\)
Divide both sides by 2
\(x=6\)

09)  \(\large\frac{3}{x}\normalsize~+10=-2\)

Subtract 10 from both sides
\(\large\frac{3}{x}\normalsize~=-12\)
Multiply both sides by \(x\)
\(3=-12x\)
Divide both sides by ‒12
\(x=-\large\frac{3}{12}\)
Simplify the fraction
\(x=-\large\frac{1}{4}\)

10)  \(\large\frac{1-2x}{3}\normalsize~=5\)

Multiply both sides by 3
\(1-2x=15\)
Subtract 1 from both sides
\(-2x=14\)
Divide both sides by ‒2
\(x=-7\)

11)  \(\large\frac{5-4x}{9}\normalsize~=5\)

Multiply both sides by 9
\(5-4x=45\)
Add \(4x\) to both sides
\(5=4x+45\)
Subtract 45 from both sides
\(-40=4x\)
Divide both sides by 4
\(x=-10\)
I dealt with the negative \(x\) term a bit differently here than in the previous question. Either way is fine.

12)  \(\large\frac{2x}{5}\normalsize~+\large\frac{1}{5}\normalsize~=\large\frac{3}{5}\)

Multiply throughout by 5
\(2x+1=3\)
Subtract 1 from both sides
\(2x=2\)
Divide both sides by 2
\(x=1\)
Yes, that's allowed, I kept my balance!

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