Creating and solving simultaneous equations

Simultaneous equations can be created to solve problems.

Example

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

Let \(a\) be the cost of an adult ticket and \(c\) be the cost of a child’s ticket. There are two adults and two children in the Smith family, so the total cost can be described by the equation:

\(2a + 2c = 33\)

There is one adult and three children in the Jones family. The equation for the total cost is:

\(a + 3c = 27.5\)

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:

Using \(a + 3c = 27.5\) with \(c = 5.5 \) gives \(a + 16.5 = 27.5\), so \(a = 11\).

Therefore, the solution is \(a = 11 c = 5.5\)

To complete the question, refer back to the context: a child’s ticket costs £5.50 and an adult’s ticket costs £11.

It is a good idea to check this by using the other equation, \(2a + 2c = 33\): \(2a + 2c\) with \(a = 11\) and \(c = 5.5\) gives \(22 + 11 = 33\), which is correct.

Creating and solving simultaneous equations

Example

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