成人快手

What are equations?

Equations are made up of two expressions on either side of an equals sign, such as:

\(x + 1 = 2\)

To solve an equation, you aim to find the value of the missing number.

Example

'I think of a number, add four, and the answer is seven.'

Written algebraically, this statement is \(x + 4 = 7\), where \(x\) represents the number you thought of.

\(x + 4 = 7\) is an example of an algebraic equation where \(x\) represents the unknown number.

The number you first thought of must be three (\(3 + 4 = 7\)).

Therefore, \(x = 3\) is the solution to the equation \(x + 4 = 7\).

Writing an equation

Let's turn the problem below into an equation.

3 loaves of bread and a 拢4 pack of cheese cost 拢10. How much is each loaf?

In this equation we want to work out the cost of one loaf of bread.

As the cost of bread is the unknown, we will represent this with a x.

We need 3 loaves of bread, so in our equation we will represent this with 3x

We know the cheese costs 拢4 and the total spend is 拢10.

So now we can write our equation:

3x + 4 = 10

Solving an equation

When solving an equation it is important to find the value of one letter.

Let's solve the equation 3x + 4 = 10

  • To remove the + 4 we complete the inverse (opposite) operation by subtracting the 4.

    3x = 10 鈥 4

  • That gives us 3x = 6 (3 loaves cost 拢6)

  • To work out the cost of one loaf we divide both sides of the equation by 3.

  • x = 6 梅 3

  • So x = 2 (one loaf costs 拢2)

Balancing the equation

An equation is like a weighing scale - both sides balance because they represent the same amount.

To solve the equation you need to find the value of the missing number by performing the same operation on each side.

Example 1

Suppose you are trying to find out how many sweets are in the bag shown here.

Each of the sweets weighs the same amount.

Let \({x}\) equal the amount of sweets in the bag.

We can ignore the weight of the bag.

Bag of sweets

By subtracting three sweets from each side, the scales remain balanced.

You can now see that one bag is equivalent to two sweets. Written algebraically, this is:

\(x + 3 = 5\)

Subtract \(3\) from both sides, to give:

\(x = 2\)

There are 2 sweets in the bag.

Bag of sweets minus three.

Example 2

In this case, two bags of sweets are equivalent to six sweets.

Remember - the weight of the bag is zero, so you鈥檒l only need to consider the weight of the sweets inside the bag.

Bag of sweets

To find the equivalent of one bag, divide both sides by two.

Written algebraically, this is:

\(2x = 6\)

Divide both sides by \(2\) to give:

\(x = 3\)

Bag of sweets

Question

Solve these equations:

  1. \(a + 4 = 9\)

  2. \(5b = 35\)

Find out how to solve simple algebraic equations with one unknown

Have a go

Image gallerySkip image gallerySlide 1 of 7, A whiteboard with a maths problem written on it, Click here to see a step by step slideshow.

Multiples of an unknown

Sometimes an equation will have multiples of an unknown, eg \(5y = 20\).

To solve this you need to get the unknown on its own. To do this, divide both sides by \({5}\).

\(5y = 20\)

\(5y \div 5 = 20 \div 5\)

\(y = 4\)

Check your answers

When solving algebraic equations, always check your answers.

For example, if you think that the answer to the equation \(x + 5 = 12\) is \(x = 7\), then check it by replacing \(x\) with \(7\).

\(7 + 5\) does equal \(12\), so your answer is correct.

Solving more complex equations

Sometimes an equation will have multiples of an unknown and other numbers, eg \(3x + 2 = 8\).

In equations of this type, your aim is to get all the \({x}\)s (or unknowns) on one side and all the numbers on the other.

Example

Let's solve the equation:

\(3x + 2 = 8\).

We can represent this in the diagram opposite, where each bag represents the unknown value \( x \).

Therefore:

\(3x + 2 = 8\)

Sweets

We want to get the \(x\) on its own.

Start by subtracting \(2\) from both sides.

\(3x + 2 - 2 = 8 - 2\)

\(3x = 6\)

Sweets

Then divide by \(3\) on both sides:

\(x = 2\)

Sweets

Question

Solve the equation:

\(4x + 5 = 13\)

Solving equations with x on both sides

Example

Solve the equation:

\(2x + 2 = x + 4\)

The equation \(2x + 2 = x + 4\) is represented by the diagram opposite.

As before, the bags represent the unknown value (\(x\)) and the sweets represent the numbers in the equation.

Two bags and two sweets on one side of the scale and one bag and four sweets on the other side.

Aim to get the unknown value on just one side of the equation, so begin by subtracting \(x\) (taking one bag away) from each side.

One bag and two sweets on one side of the scale balanced with four sweets on the other side of the scale.

Now you have the type of equation that you recognise, so all you need to do is subtract \(2\) from both sides.

One bag on one side of the scale and two sweets on the other side of the scale.

Written algebraically, this becomes:

\(2x + 2 = x + 4\)

Subtract \(x\) from both sides to give:

\(x + 2 = 4\)

Subtract \(2\) from both sides to give:

\(x = 2\)

Solving equations when the term in x is negative

Key Point

If the term in \(x\) is negative, tackle the equation in the same way - aim to get all \(x\)s on one side of the equation.

Example 1

Solve the equation:

\(5x - 2 = 12 - 2x\)

Your aim is to get all the unknown \(x\) terms on one side of the equation, so start by adding \(2x\) to both sides:

\(7x - 2 = 12\)

Next add \(2\) to both sides:

\(7x = 14\)

And finally, divide by \(7\) to give:

\(x = 2\)

Again, you can check your answer in the original equation.

So substitute \(x = 2\) back into:

\(5x - 2 = 12 - 2x\)

\((5 \times 2) - 2 = 12 - (2 \times 2)\)

\(10 - 2 = 12 - 4\)

\(8 = 8\)

This makes sense, so the value \(x = 2\) is correct.

Example 2

Solve the equation:

\(4x = 10 - x\)

Add \(x\) to both sides to give:

\(5x = 10\)

To find \(x\), divide both sides by \(5\):

\(x = 2\)

Question

Solve the equation:

\(3x + 1 = 16 - 2x\)

Solving equations with brackets

To expand a bracket, multiply everything inside the bracket by the number outside.

Example 1

Expand the bracket:

\(2(3a + 5)\)

Multiply everything inside the bracket by the number outside to give:

\((2\times3a)+(2\times5)\)

Simplify this to:

\(6a + 10\)

Example 2

Solve the equation: \(2(b-2) = -3(2b-4)\)

First expand the brackets: \(2b-4= -6b+12\)

Then add \(6b\) to both sides: \(8b-4=12\)

Add 4 to both sides: \(8b=16\)

Finally divide both sides by \(8\): \(b=2\)

Question

Solve the equation:

\(3(5x - 4) = 2(2x + 5)\)

Solving equations written as fractions

To solve equations written as fractions, you must do the same thing to each side of the equation to get the answer.

It is useful to know that \(\frac{1}{2}x\) is the same as \(\frac{x}{2}\), and \(\frac{1}{4}x\) is the same as \(\frac{x}{4}\), etc.

For example look at the equation:

\(\frac{1}{2}x = 3\)

It can be rewritten as:

\(\frac{x}{2} = 3\)

The number at the bottom is \(2\), so multiply both sides by \(2\).

\(x = 6\)

Question

Solve the equation:

\(2x - 3 = \frac {1}{2}\)

Equations with fractions on both sides

If an equation has fractions on both sides, multiply both sides by the lowest common multiple (LCM) of the denominators鈥

Example

Solve this equation:

\(\frac {(x - 3)} {2} = \frac {(x - 2)} {3}\)

Solution

One side is divided by \(2\).

The other side is divided by \(3\).

The lowest common multiple of \(2\) and \(3\) is \(6\), so multiply both sides by \(6\).

\(\frac {(x - 3)} {2} \times 6 = \frac {(x - 2)} {3} \times 6\)

\(3(x - 3) = 2(x - 2)\)

Expand the brackets:

\(3x - 9 = 2x - 4\)

Take \(2x\) from both sides:

\(x - 9 = - 4\)

Add \(9\) to both sides:

\(x = 5\)

Test section

Question 1

Solve the equation: \({c}-{7}={1}{2}\)

Question 2

Solve the equation: \({5}-{x}={9}\)

Question 3

What's the value of \({c}\) if \({7}{c}={63}\)?

Question 4

Solve the equation: \({3}{x}+{7}={10}\)

Question 5

Solve the equation: \({5}{x}+{24} ={9}\)

Question 6

Solve the equation: \({6}{x}+{5}={19}-{x}\)

Question 7

Solve the equation: \({4}({2}{y}-{3})={3}({y}+{6})\)

Question 8

Solve the equation: \({4}{x}-{5}=\frac{1}{3}\)

Question 9

Solve the equation: \(\frac{({y}-{4})}{2}=\frac{({2}-{y})}{3}\)

Question 10

Solve the equation: \(\frac{({x}+{7})}{6}=\frac{x}{3}\)

Where next?

Discover more about this topic from Bitesize.

Algebra