Lesson 1: Simple linear equations

Introduction

An equation is just an equals sign with an expression on either side. So a formula like \(E=mc^2~\) is a type of equation. However, when we talk about solving equations, it usually implies that there is only one variable involved and we have to work out the value of that variable.

A linear equation is an equation in which the variable, say \(~x~\), is not raised to a power other than 1. So there's no \(~x^2~\), \(~x^3~\), \(~x^4~\), etc.

Importantly, this lesson doesn't just teach you how to form and solve simple linear equations. It lays the foundations for solving all linear equations.

SECTION A

Understanding the basic method

You will become very good at forming and solving equations if you learn the basics well. You're going to learn the all-important basics with the help of four mathematical problems. Please, please don't skip over these because you can work out the answer in your head. That would be missing the point completely!

Problem 1
I think of a number, add 5 and the answer is 12. What number am I thinking of?

Problem 2
I am 11 years younger than my brother. I am 7 years of age. How old is my brother?

Problem 3
Jenny's phone bill is four times higher than Ben's. Jenny's bill is £24. How high is Ben's?

Problem 4
Owen ran one-fifth of the distance that Matthew ran. Owen ran 4km. How far did Matthew run?

Form an equation
► If it helps, try to make the question sound a bit more mathematical, starting with the thing you're asked to work out.

The number I'm thinking of add 5 equals 12.

My brother's age take away 11 equals 7.

Ben's phone bill multiplied by 4 equals 24

Matthew's distance divided by 5 equals 4.

► Then write this using mathematical symbols and using \(~x~\) (or another letter, if you wish) to represent the unknown quantity (the thing you're asked to work out).

\(x+5=12\)

\(x-11=7\)

\(4x=24\)

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

Solve the equation
► The balance method is used to solve all linear equations.
► An equation is like a set of weighing scales, with the equals sign being the pivot in the middle. The scales must always be in balance; so whenever you make a change to one side, you must make exactly the same change to the other side.
► The aim is to make identical changes to both sides of the equation until the unknown is on its own on one side, with a single number on its own on the other side.

STEP 1: Look at the side that \(~x~\) is on. Decide what you need to get rid of on that side so that \(~x~\) is on its own. In problems 1-4, the operations are \(+5~\), \(-11~\), \(\times 4~\) and \(\div 5~\). Get rid of these and \(~x~\) will be on its own on one side of the equals sign.

STEP 2: To get rid of an operation, you have to reverse it (undo it, if you like).

subtract 5 to get rid of the \(+5~\)

add 11 to get rid of the \(-11~\)

divide by 4 to get rid of the \(\times 4~\)

multiply by 5 to get rid of the \(\div 5~\)

BUT remember, whatever you do to one side, you must also do to the other side!

\(x=12-5\)

\(x=7\)

Problem solved!
The number I'm thinking of is 7.

\(x=7+11\)

\(x=18\)

Problem solved!
My brother is 18 years old.

\(x=24\div 4\)

\(x=6\)

Problem solved!
Ben's phone bill is £6.

\(x=4\times 5\)

\(x=20\)

Problem solved!
Matthew ran a distance of 20km.

You'll see from the practice questions that there's a lot of emphasis on solving equations at this stage and not so much on forming them. It's a bit like learning how to eat before learning how to cook!

Practise to master

SECTION A

Solve these simple linear equations. Your workings should clearly show what identical changes you've made to both sides of the equation.

01)  \(x+4=7\)

Subtract 4 from both sides
\(x=7-4\)
\(x=3\)

02)  \(x-3=5\)

Add 3 to both sides
\(x=5+3\)
\(x=8\)

03)  \(2x=18\)

Divide both sides by 2
\(x=18\div 2\)
\(x=9\)

04)  \(\large\frac{x}{4}\normalsize~=3\)

Multiply both sides by 4
\(x=3\times 4\)
\(x=12\)

05)  \(x+14=21\)

Subtract 14 from both sides
\(x=21-14\)
\(x=7\)

06)  \(x-8=1\)

Add 8 to both sides
\(x=1+8\)
\(x=9\)

07)  \(5x=30\)

Divide both sides by 5
\(x=30\div 5\)
\(x=6\)

08)  \(\large\frac{x}{5}\normalsize~=6\)

Multiply both sides by 5
\(x=6\times 5\)
\(x=30\)

09)  \(x+9=14\)

Subtract 9 from both sides
\(x=14-9\)
\(x=5\)

10)  \(x-4=11\)

Add 4 to both sides
\(x=11+4\)
\(x=15\)

11)  \(9x=27\)

Divide both sides by 9
\(x=27\div 9\)
\(x=3\)

12)  \(\large\frac{x}{8}\normalsize~=1\)

Multiply both sides by 8
\(x=1\times 8\)
\(x=8\)

13)  \(a+19=30\)

Subtract 19 from both sides
\(a=30-19\)
\(a=11\)

14)  \(y-10=9\)

Add 10 to both sides
\(y=9+10\)
\(y=19\)

15)  \(6p=42\)

Divide both sides by 6
\(p=42\div 6\)
\(p=7\)

16)  \(\large\frac{c}{3}\normalsize~=12\)

Multiply both sides by 3
\(c=12\times 3\)
\(c=36\)

Problem 1
I think of a number, subtract 8 and get the answer 5. Form an equation, solve it and interpret the solution.

\(x-8=5\)
Add 8 to both sides
\(x=5+8\)
\(x=13\)
The number I picked was 13.

Problem 2
The cost of 3 identical pairs of jeans is £48. Form and solve an equation to find the cost of one pair.

\(3x=48\)
Divide both sides by 3
\(x=48\div 3\)
\(x=16\)
The cost of one pair of jeans is £16.

SECTION B

This section only slightly extends what we've done so far but it does allow for more interesting problems.

Example

Andrew, Burt and Carl shared £40 between them. Burt got twice as much as Andrew, and Carl got five times more than Andrew. Form and solve an equation to work out how much each got.

Let \(~x~\) be the amount Andrew got. That means Burt got \(~2x~\) and Carl got \(~5x~\). What they got together must add up to £40.

Form an equation  ⇒  \(x+2x+5x=40\)

Add terms to simplify  ⇒  \(8x=40\)

Divide both sides by 8  ⇒  \(x=\large\frac{40}{8}\)

Solution  ⇒  \(x=5\)

So Andrew got £5, Burt got £10 and Carl got £25.

Practise to master

SECTION B

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

01)  \(4x+2x=24\)

Collect terms to simplify
\(6x=24\)
Divide both sides by 6
\(x=24\div 6\)
\(x=4\)

02)  \(5x+3x=64\)

Collect terms to simplify
\(8x=64\)
Divide both sides by 8
\(x=64\div 8\)
\(x=8\)

03)  \(8x-3x=45\)

Collect terms to simplify
\(5x=45\)
Divide both sides by 5
\(x=45\div 5\)
\(x=9\)

04)  \(11x-9x=22\)

Collect terms to simplify
\(2x=22\)
Divide both sides by 2
\(x=22\div 2\)
\(x=11\)

05)  \(3x+2x+x=36\)

Collect terms to simplify
\(6x=36\)
Divide both sides by 6
\(x=36\div 6\)
\(x=6\)

06)  \(9x+4x-6x=28\)

Collect terms to simplify
\(7x=28\)
Divide both sides by 7
\(x=28\div 7\)
\(x=4\)

07)  \(12b-9b+b=40\)

Collect terms to simplify
\(4b=40\)
Divide both sides by 4
\(b=40\div 4\)
\(b=10\)

08)  \(7e-3e-e=99\)

Collect terms to simplify
\(3e=99\)
Divide both sides by 3
\(e=99\div 3\)
\(e=33\)

Problem 1
The angles \(~2x~\), \(~x~\) and \(~3x~\) make a right angle. Form and solve an equation to work out the size of each angle.

There are 90° in a right angle so the equation is
\(2x+x+3x=90\)
Collect terms to simplify
\(6x=90\)
Divide both sides by 6
\(x=90\div 6\)
\(x=15\)
The angles are 30°, 15° and 45°.
Working out the value of \(~x~\) did not complete our solution here. We then had to substitute that value into \(~2x~\) and \(~3x~\) to find the other two angles.

Problem 2
The angles in a triangle are \(~3x~\), \(~2x~\) and \(~4x~\). Form and solve an equation to work out the size of each angle.

The angles in a triangle add up to 180° so the equation is
\(3x+2x+4x=180\)
Collect terms to simplify
\(9x=180\)
Divide both sides by 9
\(x=180\div 9\)
\(x=20\)
The angles are 60°, 40° and 80°.
Problems often require you to know stuff from other areas of maths, in this case geometry and the angle sum of a triangle.

Problem 3
Ed, Frank and Graham lay 120 bricks. Frank lays 6 times more bricks than Ed and Graham lays half as many as Frank. Form and solve an equation to work out how many bricks each fella laid.

The equation is
\(x+6x+3x=120\)
Collect terms to simplify
\(10x=120\)
Divide both sides by 10
\(x=120\div 10\)
\(x=12\)
So Ed lays 12 bricks, Frank lays 72 bricks and Graham lays 36 bricks.
I checked to make sure these added up to 120, which they did.

Problem 4
A rectangle is \(~2x~\)cm wide and \(~3x~\)cm long. Its perimeter is 50cm. Form and solve an equation to work out the dimensions of the rectangle.

The equation is
\(2x+3x+2x+3x=50\)
Collect terms to simplify
\(10x=50\)
Divide both sides by 10
\(x=50\div 10\)
\(x=5\)
So the rectangle is 10cm wide and 15cm long.

SECTION C

The remaining three sections of this lesson teach you how to recognise and deal with tricky bits in linear equations. This section covers the case of the sneaky negative unknown!

Example

Solve the equation \(~5-x=11~\)

Subtract 5 from both sides  ⇒  \(-x=11-5\)

Simplify  ⇒  \(-x=6\)

Multiply both sides by \(~-1\)  ⇒  \(x=-6\)

The 5 in the equation is positive so we subtract 5 (from both sides) to get rid of it.
What makes this tricky is the negative \(~x~\) that is left.
So we multiply by \(-1\) to get \(~x~\) on its own (rather than \(-x~\) on its own).

Alternatively

Solve the equation \(~5-x=11~\)

Add \(x\) to both sides  ⇒  \(5=11+x\)

Subtract 11 from both sides  ⇒  \(5-11=x\)

Simplify  ⇒  \(-6=x\)

Which is the same as  ⇒  \(x=-6\)

The \(x\) in the equation is negative so we add \(x\) (to both sides) to get rid of it.
You should understand both methods before settling on the one you prefer.

Practise to master

SECTION C

Use your negative number skills to help you solve these linear equations. Keep an eye out for any tricky negative unknowns!

01)  \(x+3=1\)

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

02)  \(x-9=-10\)

Add 9 to both sides
\(x=-10+9\)
\(x=-1\)

03)  \(2x=-12\)

Divide both sides by 2
\(x=-12\div 2\)
\(x=-6\)

04)  \(\large\frac{h}{4}\normalsize~=-5\)

Multiply both sides by 4
\(h=-5\times 4\)
\(h=-20\)

05)  \(x+21=-3\)

Subtract 21 from both sides
\(x=-3-21\)
\(x=-24\)

06)  \(x-1=-6\)

Add 1 to both sides
\(x=-6+1\)
\(x=-5\)

07)  \(10x=-80\)

Divide both sides by 10
\(x=-80\div 10\)
\(x=-8\)

08)  \(\large\frac{h}{-4}\normalsize~=5\)

Multiply both sides by −4
\(h=5\times -4\)
\(h=-20\)

09)  \(1-x=25\)

Subtract 1 from both sides
\(-x=25-1\)
\(-x=24\)
Multiply both sides by −1
\(x=-24\)

10)  \(-13-x=25\)

Add 13 to both sides
\(-x=25+13\)
\(-x=38\)
Multiply both sides by −1
\(x=-38\)

11)  \(-4x=44\)

Divide both sides by −4
\(x=44\div -4\)
\(x=-11\)

12)  \(\large\frac{-h}{4}\normalsize~=5\)

Multiply both sides by 4
\(-h=5\times 4\)
\(-h=20\)
Multiply both sides by −1
\(h=-20\)

13)  \(3-x=-8\)

Subtract 3 from both sides
\(-x=-8-3\)
\(-x=-11\)
Multiply both sides by −1
\(x=11\)

14)  \(-7-x=-3\)

Add 7 to both sides
\(-x=-3+7\)
\(-x=4\)
Multiply both sides by −1
\(x=-4\)

15)  \(-3x=-21\)

Divide both sides by −3
\(x=-21\div -3\)
\(x=7\)

16)  \(-\large\frac{h}{4}\normalsize~=5\)

Multiply both sides by −1
\(\large\frac{h}{4}\normalsize~=-5\)
Multiply both sides by 4
\(h=-5\times 4\)
\(h=-20\)

Note: Questions 8, 12 and 16 in particular show three ways of writing the same thing. If you recognise one of these types in Sections D and E, you may change the position of the minus sign in the fraction, if it makes it easier for you.

SECTION D

The second tricky bit that often catches people out is the unknown as a denominator case.

Example

Solve the equation \(~\large\frac{12}{x}\normalsize~=3\)

Multiply both sides by \(x\)  ⇒  \(12=3x\)

Divide both sides by 3  ⇒  \(12\div 3=x\)

Simplify  ⇒  \(4=x\)

Which is the same as  ⇒  \(x=4\)

Multiplying both sides by \(x\) makes the \(x\) disappear from the left side and reappear on the right BUT NOT as a denominator. Just dividing by 12 at the start would leave \(\large\frac{1}{x}~\) on its own, not \(x\)!

Practise to master

SECTION D

Solve these linear equations, making sure you deal correctly with the unknown as a denominator.

01)  \(\large\frac{12}{x}\normalsize~=6\)

Multiply both sides by \(x\)
\(12=6x\)
Divide both sides by 6
\(12\div 6=x\)
\(2=x\)
\(x=2\)

02)  \(\large\frac{20}{x}\normalsize~=4\)

Multiply both sides by \(x\)
\(20=4x\)
Divide both sides by 4
\(20\div 4=x\)
\(5=x\)
\(x=5\)

03)  \(\large\frac{27}{x}\normalsize~=9\)

Multiply both sides by \(x\)
\(27=9x\)
Divide both sides by 9
\(27\div 9=x\)
\(3=x\)
\(x=3\)

04)  \(\large\frac{18}{x}\normalsize~=3\)

Multiply both sides by \(x\)
\(18=3x\)
Divide both sides by 3
\(18\div 3=x\)
\(6=x\)
\(x=6\)

05)  \(\large\frac{-4}{x}\normalsize~=1\)

Multiply both sides by \(x\)
\(-4=1x\)
\(x=-4\)

06)  \(\large\frac{10}{x}\normalsize~=-5\)

Multiply both sides by \(x\)
\(10=-5x\)
Divide both sides by −5
\(10\div -5=x\)
\(-2=x\)
\(x=-2\)

07)  \(\large\frac{11}{-x}\normalsize~=2\)

Multiply both sides by \(-x\)
\(11=-2x\)
Divide both sides by −2
\(11\div -2=x\)
\(-5.5=x\)
\(x=-5.5\)
In algebra, we generally prefer fractions to decimals but 5.5 is nice and short so it doesn't bother us too much!

08)  \(-\large\frac{17}{x}\normalsize~=3\)

Multiply both sides by \(x\)
\(-17=3x\)
Divide both sides by 3
\(-\large\frac{17}{3}\normalsize~=x\)
\(x=-\large\frac{17}{3}\)
We're definitely keeping this as a fraction. As a decimal it would be recurring!

SECTION E

This is not so much a tricky bit as ... well, a bit. Anyway, it'll make you more algebraically flexible so it's definitely worth exploring. Let's talk about two different ways of writing little fractions in algebra.

Example

Solve the equation \(~\large\frac{3x}{4}\normalsize~=6\)

Multiply both sides by 4  ⇒  \(3x=6\times 4\)
Simplify  ⇒  \(3x=24\)
Divide both sides by 3  ⇒  \(x=24\div 3\)
Simplify  ⇒  \(x=8\)

Alternatively, the same question might be written like this

Solve the equation \(~\large\frac{3}{4}\normalsize~x=6\)

Divide both sides by \(\frac{3}{4}\)  ⇒  \(x=6\div \frac{3}{4}\)
This is the same as  ⇒  \(x=6\times \frac{4}{3}\)
Simplify  ⇒  \(x=8\)

If I was given the second version with the fraction in front, I would treat it like the first version (since they are equivalent) and solve it like the first version. It just seems easier! You can even rewrite the question, if you like.

Practise to master

SECTION E

Solve these equations (rewrite the fraction part, if you like). Do you prefer \(\frac{3x}{4}\) or \(\frac{3}{4}~x\) ?

01)  \(\large\frac{2x}{3}\normalsize~=8\)

Multiply both sides by 3
\(2x=8\times 3\)
\(2x=24\)
Divide both sides by 2
\(x=24\div 2\)
\(x=12\)

02)  \(\large\frac{3}{5}\normalsize~x=6\)

Multiply both sides by 5
\(3x=6\times 5\)
\(3x=30\)
Divide both sides by 3
\(x=30\div 3\)
\(x=10\)
I didn't bother rewriting the question but I did solve it as though I had rewritten it.

03)  \(\large\frac{4x}{3}\normalsize~=60\)

Multiply both sides by 3
\(4x=60\times 3\)
\(4x=180\)
Divide both sides by 4
\(x=180\div 4\)
\(x=45\)

04)  \(\large\frac{3}{8}\normalsize~x=6\)

Multiply both sides by 8
\(3x=6\times 8\)
\(3x=48\)
Divide both sides by 3
\(x=48\div 3\)
\(x=16\)

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

Multiply both sides by −5
\(2x=1\times -5\)
\(2x=-5\)
Divide both sides by 2
\(x=-5\div 2\)
\(x=-\large\frac{5}{2}\)
I could also have written my answer as the decimal −2.5

06)  \(-\large\frac{5}{6}\normalsize~x=10\)

Multiply both sides by 6
\(-5x=10\times 6\)
\(-5x=60\)
Divide both sides by −5
\(x=60\div -5\)
\(x=-12\)

07)  \(\large\frac{8}{3x}\normalsize~=16\)

Multiply both sides by \(3x\)
\(8=16\times 3x\)
\(8=48x\)
Divide both sides by 48
\(x=8\div 48\)
\(x=\large\frac{1}{6}\)

08)  \(\large\frac{1}{5}\normalsize~x=4\)

Multiply both sides by 5
\(x=4\times 5\)
\(x=20\)

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