3. Linear equations
Linear equations are the easiest type of equation to solve because the unknown is not raised to any power other than 1.
We can solve linear equations by rearrangement. We must do the same operation on both sides of the equals sign.
For example:
x – 19 = – 8
Add 19 to both sides:
x = 11
7x = 42
Example 2:
Divide both sides by 7:
x = 6

4. Equation solving 1 Teacher notes
This activity allows you to practice solving equations.
Type 1 equations all just involve a single step to solve.

5. Linear equations
When more than one operation is performed on the unknown we need to solve the equation in several steps.
For example,
4x + 5 = 29
subtract 5 from both sides:
4x = 24
divide both sides by 4:
x = 6
Teacher notes
Establish that for more complex equations a more rigorous method is required.
Remind pupils that we are trying to get the unknown x on its own on the left hand side of the equals sign.
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Check that 4 × is equal to 29 in the original equation.

6. Equation solving 2 Teacher notes
This activity allows you to practice solving equations.
Type 2 equations involve multiple steps to solve.

7. Equations with unknowns on both sides
In some cases the unknown appears on both sides of the equals sign.
For example:
8x – 2 = 2x + 1
We need to work systematically to get the unknowns on the left and the numbers on the right.
Remember to perform the same operations on both sides.
unknowns
numbers
8x – 2 = 2x + 1
Teacher notes
Start by explaining that we only want terms containing x on the left-hand side. Ask pupils how we could ‘get rid of’ the – 2 from the left-hand side (by adding 2 to both sides of the equation).
Next, ask how we could ‘get rid of’ the 2x from the right-hand side (by adding 2x to both sides of the equation).
We could write 3 ÷ 6 as a fraction, 3/6. This then cancels to 1/2.
It is usually better to write the solution as a vulgar fraction when the equivalent decimal cannot be written exactly.
The solution should be substituted into the original equation to make sure the solution is correct.
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add 2 to both sides:
8x = 2x + 3
subtract 2x from both sides:
6x = 3
divide both sides by 6:
x = 0.5

8. Equation solving 3 Teacher notes
This activity allows you to practice solving equations.
Type 3 equations involve the unknown on both sides of the equation.

9. Scales

11. Equations with brackets
Equations can contain brackets.
3(2x – 1) = 7x
For example:
multiply out the brackets: 6x
– 3
= 7x
subtract 7x from both sides:
–x – 3 = 0
add 3 to both sides:
–x = 3
multiply by –1:
x = –3
Teacher notes
Talk through this method for solving an equation containing brackets.
Check the answer –3 by substituting it back into the original equation and verifying that the left hand side of the equation equals the right hand side.
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12. Equations with brackets
Example:
2(3x – 5) = 4x
multiply out the brackets: 6x
–10
= 4x
add 10 to both sides:
6x = 4x + 10
subtract 4x from both sides:
2x = 10
divide both sides by 2:
x = 5
Alternatively:
divide both sides by 2:
3x – 5 = 2x
add 5 to both sides:
3x = 2x + 5
Teacher notes
Talk through these methods for solving an equation containing brackets.
Check the answer 5 by substituting it back into the original equation and verifying that the left hand side of the equation equals the right hand side.
Note that we can only use the second method if the term on the right-hand side is divisible by the number in front of the bracket on the left-hand side.
Point out that using the second method in this example involves fewer steps and is therefore more efficient. However, if pupils are unsure of what to do it may be best to multiply out any brackets first.
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subtract 2x from both sides:
x = 5
In this example, dividing first means that there are fewer steps.
We can only do this because both sides are divisible by 2.

13. Equation solving 4 Teacher notes
This activity allows you to practice solving equations.
Type 4 equations involve fractions and brackets. Press “new” if a question is at the wrong level.

14. A word of warning! (x – 1) = 2(x – 1) Example:
divide both sides by (x – 1):
1 = 2
This is false, so something has gone wrong.
Notice that the solution to this equation is x = 1, so when we divided by (x – 1) we in fact divided by zero. This is not allowed.
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Next time someone “proves” to you that 1 = 0, watch out! They probably divided by zero.

16. Equations with fractional coefficients
Sometimes the coefficient of an unknown is a fraction. 3 4
x – 5 = 9 – x
For example,
We can remove the 4 from the denominator by multiplying both sides of the equation by 4. 3 4
4( x – 5) = 4(9 – x)
expand the brackets:
3x – 20 = 36 – 4x
add 4x to both sides:
Teacher notes
Explain that we multiply by the number in the denominator to cancel it out.
Point out that 3/4 x can also be written as 3x/4.
Ask pupils to check the solution by substituting x = 8 into the original equation.
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7x – 20 = 36
add 20 to both sides:
7x = 56
divide both sides by 7:
x = 8

17. Equations with fractional coefficients
If an equation contains more than one fraction, these can be removed by multiplying throughout by the lowest common multiple of the two denominators. 2 3
x = x + 1 1
For example,
The lowest common multiple of 3 and 2 is 6, so we need to multiply both sides by 6. 2 3
6( x) = 6( x + 1) 1
Teacher notes
Talk through the multiplication of each term by 6. for example we can think of 6 × 2/3 as 6 × 2 ÷ 3 or as 6 ÷ 3 × 2.
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expand the brackets:
4x = 3x + 6
subtracting 3x from both sides:
x = 6

18. Equations with fractional coefficients
2x + 7 5
= x – 1
Example,
In this example the whole of one side of the equation is divided by 5.
To remove the 5 from the denominator we multiply both sides of the equation by 5.
2x + 7 = 5(x – 1)
expand the brackets:
2x + 7 = 5x – 5
swap sides:
Teacher notes
Talk through the equation on the board. Explain that when we have (2x + 7)/ 5 the dividing line acts as a bracket – that means that we add before we divide. To get rid of the fraction we must start by multiplying by 5. If we multiply the left hand side by 5 we must also multiply the right hand side by 5.
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5x – 5 = 2x + 7
add 5 to both sides:
5x = 2x + 12
subtract 2x from both sides:
3x = 12
divide both sides by 3:
x = 4

19. Equation solving 4 Teacher notes
This activity allows you to practice solving equations.
Type 4 equations involve fractions and brackets. Press “new” if a question is at the wrong level.

21. Solving equations involving division
When both sides of an equation are divided by number, those numbers must be removed.
5x – 3 4 =
2x – 1 3
For example,
We do this by multiplying both sides by the lowest common multiple of the two denominators.
The lowest common multiple of 4 and 3 is 12. Multiplying every term by 12 gives us: 3 4
12(5x – 3) 4 =
12(2x – 1) 3 1 1
which simplifies to: (5x – 3) = 4(2x – 1)

22. Solving equations involving division
We can then solve the equation as usual.
3(5x – 3) = 4(2x – 1)
expand the brackets:
15x – 9 = 8x – 4
add 9 to both sides:
15x = 8x + 5
subtract 8x from both sides:
7x = 5
x = 5 7
divide both sides by 7:
Although the answer could be written as a rounded decimal, it is more exact left as a fraction.

23. Solving equations involving division
We have seen that
5x – 3 4 =
2x – 1 3
simplifies to (5x – 3) = 4(2x – 1)
How could we perform this simplification in one step?
Multiplying both sides by 4 cancels out the 4 on the left hand side and multiplies the expression on the right-hand side by 4.
Teacher notes
Cross-multiplication is an efficient way to rearrange an equation or formula in this form. Make sure that pupils understand why it works in terms of doing the same thing to both sides of the equation.
Multiplying both sides by 3 cancels out the 3 on the right hand side and multiplies the expression on the left-hand side by 3.
Doing this in one step in often called cross-multiplication.

24. Solving equations involving division
Sometimes the unknowns appear in the denominator. 4
(x + 3) 5
(3x – 5) =
For example,
In this example, we can multiply both sides by (x + 3) and (3x – 5) in one step to give:
4(3x – 5) = 5(x + 3)
expand the brackets:
12x – 20 = 5x + 15
Teacher notes
Ensure that all pupils are happy with the cross-multiplication step before moving on to solve the equation.
subtract 5x from both sides:
7x – 20 = 15
add 20 to both sides:
7x = 35
divide both sides by 7:
x = 5

25. Equation solving 4 Teacher notes
This activity allows you to practice solving equations.
Type 4 equations involve fractions and brackets. Press “new” if a question is at the wrong level.

27. Equivalent fractions Teacher notes
Ask pupils to tell you which equations they believe to be equivalent to the one in the question.
Clicking on an expression will reveal whether of not it is equivalent to the required expression. Each time an equivalent expression is found ask pupils to tell you what has been done to both sides of the equation to make the selected equivalent expression.
Ask pupils to tell you which form of the equation is the easiest to solve (the form with no denominators).
Ask pupils to solve the equation in the window and substitute this solution into the equivalent equations found to verify the solution.

28. Constructing an equation
I’m thinking of a number.
When I subtract 9 from the number and double it, I get the same answer as dividing the number by 5.
What number am I thinking of?
Let’s call the unknown number n.
Teacher notes
If we let n be the unknown number, the statement implies:
2(n – 9) = n/5
Solving for n:
10(n – 9) = n
10n – 90 = n
9n = 90
n = 10
Photo credit: © beltsazar, Shutterstock.com
We can solve this problem by writing the equation: n 5
2(n – 9) =
The number with 9 subtracted and doubled
is the same as
the number divided by 5.

29. In one year’s time Ben will be three times Alice’s age.
Happy family
The sum of the ages of Ben and his daughter, Alice, is 66 years.
In one year’s time Ben will be three times Alice’s age.
How old is Alice now?
Check your answer.
Teacher notes:
Let a be Alice’s age, and b be Ben’s age.
To solve this problem we can write Bens age in two different ways using the information given. Equating these two expressions will give us an equation in a.
The first piece of information gives us that a + b = 66, so b = 66 – a.
The second piece of information gives us that 3(a + 1) = b + 1, which rearranges to b = 3a + 2
Equating the expressions:
3a + 2 = 66 – a
So:
a = 16
To check:
b = 66 – 16 = 50
So we need (16 + 1) × 3 = , which is true, so the answer is correct.
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30. Happy family
Alice gets £7 a week pocket money and already owns £48. She spends £2 a week on sweets.
Her brother Charlie also gets £7 a week. He has no money at the moment, but spends nothing.
Teacher notes
Let n be the number of weeks.
Then in week n, Alice has n pounds.
Charlie has 7n pounds.
So solve n = 7n.
So n = 24.
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In how many weeks will they both have the same amount of money?

31. Find the value of x Teacher notes
Discuss the equation that needs to be written to find x.
Write this equation on the board using the pen tool and ask a volunteer to solve it.
Choose level 2 to give the perimeter in term of x.
Reveal the solution and ask pupils to use it to find the actual width and height of the rectangle and so check that the solution is correct.

32. Roast times Teacher notes
Fill in in the missing details of the cook book. Write linear equations to describe the cooking times on the left hand side, and use these to fill in the table on the right, or vice-versa.
To find the details for the lamb, we know that the cooking time of the lamb will be T = a + bw, with w the weight of the lamb. Therefore, from the details in the sentence we deduce that 60 = a + b, and that 105 = a + 2.5b. Solving the linear equations gives a = 30, b = 30. Therefore, Lamb = 30 mins + 30 mins/lb.
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Photo credit (turkey): © Adam Filipowicz, Shutterstock.com