**'Some people went to the theatre, where they bought adult tickets and child tickets. Mrs Mayor took her two parents and her four children and paid £203, whereas Mr and Mrs Clarke took their three children and paid £143. Can you work out how much Mr Sheriff and his two daughters would pay?'**

If you have completed previous 'Simultaneous Equations' activities then you will recognise this sort of problem. This activity builds on ideas established in two previous activities so if you have not completed those activities yet then you should go back and complete them before continuing with this.

In order to find out how much Mr and Mrs Sheriff and their daughters will pay we first need to know the cost of an adult ticket and the cost of a child ticket. These are known as variables and we use a = cost of an adult and c = cost of a child. So, the Mayor family becomes 3**a + 4c = 203, **whereas the Clarke family becomes 2**a + 3c = 143. **We now have our two equations, and we need to solve them simultaneously.

**3a + 4c = 203 (1)**

**2a + 3c = 143 (2)**

At this point, we want to eliminate one of the variables by adding or subtracting the equations, but before we do this we need to have the same coefficient of a or c. There is no whole number we can multiply equation (2) by that will do this, so this time we need to multiply equation (1) by one number and equation (2) by a different number that will achieve this. If we want to equate the coefficient of 'c' we can multiply (1) by 3 and (2) by 4 which will give us '12c' in both.

**9a + 12c = 609 (3)**

**8a + 12c = 572 (4)**

Since both equations have been changed we label the new equations (3) and (4). Now we have the same number of a's in both equations and the signs in front are the same so we can subtract (4) from (3).

**9a + 12c = 609 (3)**

**8a + 12c = 572 (4)**

** a = 37 (3) - (4)**

**So, an adult ticket costs £37.**

Now, to find the cost of a child ticket, substitute a = 37 into any of the four equations, usually the 'easiest' one.

**2 x 37 + 3c = 143 [sub into (2)]**

**74 + 3c = 143**

**3c = 69**

**c = 23 **

**So a child ticket costs £23.**

Now we can find the cost for the Sheriff family which is one adult and two children, **37 + 2 x 23 = 83** so they would pay £83.

If you now compare what we have done here with the previous activity, you will see that we multiply both equations in order to equate the coefficients of one variable. Other than that, everything else is the same. We could have equally multiplied equation (1) by 2 and equation (2) by 3 which would have given us '6a' in both and then we could have eliminated 'a'. This would have been just as good a method.

Now, let's look at another pair of simultaneous equations. This time just the equations without the real-life problem.

Solve **2x**** + 5y = -2 **** (1)**

and **3x**** - 4y = 20 **** (2)**

To equate coefficients of 'y' multiply (1) by 4 and (2) by 5

**2x**** + 5y = -2 **** (1) x 4**

**3x**** - 4y = 20 **** (2) x 5**

**8x**** + 20y = -8 **** (3)**

**15x**** - 20y = 100 **** (4)**

Since the signs in front of the y's are different we add equations (3) and (4).

** ****8x**** + 20y = -8 **** (3)**

**15x**** - 20y = 100 **** (4)**

**23x = 92 ****(3) + (4)**

**x = 4 ****[divide by 23]**

**2 x 4 + 5y = -2 ****[sub into (1)]**

**8 + 5y = -2**

**5y = -10 ****[subtract 8]**

**y = -2 ****[divide by 5]**

**3 x 4 - 4 x -2 = 20 ****[check in (2)]**

**12 - -8 = 20 ****[correct]**

**So x = 4 and y = -2.**

Now you should be ready to try some. Remember, its the same as before but just look for what are the best numbers to multiply each equation by.