Forming and solving simultaneous equations

Sometimes, the simultaneous equations we need to solve a problem will not be given to us. Instead, we will be given some information and need to form the two equations for ourselves.

Starting simply, suppose 2 numbers have a sum of 20 and a difference of 8. Let 1 of the numbers be \(x\) and the other be \(y\).

Then \(x + y = 20\) and \(x - y = 8\). We can then solve these simultaneously.

Example

100 tickets were sold for the school talent show. Adult tickets cost £5 each and concessionary tickets cost £2 each. £260 was collected in ticket sales.

Let \(x\) = the number of adult tickets sold and \(y\) = the number of concessionary tickets sold.

If 100 tickets were sold, then the total of adult tickets and concessionary tickets must be 100.

So \(x + y = 100\).

If £260 was collected in ticket sales, then (\(£5 \times x\)) + (\(£2 \times y\)) is equal to £260.

So we have:

\(x + y = 100\)

\(5x + 2y = 260\)

The information is now in a form that we know how to solve.

Question

Form a pair of equations for the following situation:

An infographic showing 4 apples and a banana labelled £1.70, and 2 apples and a banana labelled 90p.

4 apples and 1 banana cost £1.70. 2 apples and 1 banana cost 90p.

Example

Mr and Mrs Smith take their 2 children to the cinema. The total cost is £33. Mr Jones takes his 3 children to the cinema and the total cost is £27.50. Calculate the price of a child's ticket and an adult's ticket.

There are 2 adults and 2 children in the Smith family:

\(2a + 2c = 33\)

There is 1 adult and 3 children in the Jones family:

\(a + 3c = 27.5\)

Double the second equation to give a common of 2 for \(a\).

\(\begin{array}{rrrrr} \mathbf{2a} & + & 2c & = & 33 \\ \mathbf{2a} & + & 6c & = & 55 \end{array}\)

Decide whether to add or subtract the two equations by using Different Add Same Subtract (DASS).

\(\begin{array}{ccccc} 2a & + & 6c & = & 55 \\ - && - && - \\ 2a & + & 2c & = & 33 \\ = && = && = \\ && 4c & = & 22 \\ && \div 4 && \div 4 \\ && c & = & 5.5 \end{array}\)

To find the cost of an adult ticket, substitute the cost of a child ticket, £5.50, into one of the original equations:

\(a + 3c = 27.5\)

\(a + 3~\mathbf{\times~5.5} = 27.5\)

\(a + 16.50 = 27.50\)

\(a = 11\)

Check:

\(2a + 2c = 33\)

\(2~\mathbf{\times~11} + 2~\mathbf{\times~5.5} = 33\)

\(22 + 11 = 33\)

\(33 = 33\)

A child's ticket costs £5.50 and an adult's ticket costs £11.

Simultaneous equations - Intermediate & Higher tier – WJECForming and solving simultaneous equations