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.
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 coefficientThe amount that a letter has been multiplied by. In the example of 3a, the coefficient of a is 3 because 3 x a = 3a. of 2 for \(a\).
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.