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Solving simultaneous equations - OCRCreating and solving simultaneous equations

Simultaneous equations require algebraic skills to find the values of letters within two or more equations. They are called simultaneous equations because the equations are solved at the same time.

Part of MathsAlgebra

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 拢33. Mr Jones takes his three children to the cinema and the total cost is 拢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鈥檚 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, 拢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鈥檚 ticket costs 拢5.50 and an adult鈥檚 ticket costs 拢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.