Creating and solving simultaneous equations

Simultaneous equations can be created to solve problems algebraically.

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:

\(a + 3c = 27.5\)

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

\(a + 16.50 = 27.50\)

\(a = 11\)

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.

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Solving simultaneous equations - EduqasCreating and solving simultaneous equations