**'Some people went to the cinema, where they bought adult tickets and child tickets. Mr and Mrs Baker took their daughters Molly, Polly and Dolly and paid £33, whereas Mr Butcher took his two sons Ron and Don and paid £19. Can you work out how much Mr and Mrs Fisher and their daughter April would pay?'**

If you have completed the activity 'Simple Simultaneous Equations' then you will recognise this sort of problem as one which can be solved using **simultaneous equations**. If you have not completed that activity yet then you should go back and complete that one before continuing with this.

In order to find out how much Mr and Mrs Fisher and their daughter 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, Mr and Mrs Baker and their daughters becomes **2a + 3c = 33**. Mr Butcher and his 2 sons becomes **a + 2c = 19. **We now have our two equations, and we need to solve them simultaneously.

**2a + 3c = 33 (1)**

**a + 2c = 19 (2)**

In the previous activity, at this point, we would either add or subtract our equations to eliminate one of the variables. However, that will not help us here since neither the coefficients of a nor of c are the same. So what we have to do is make them the same by multiplying the whole of one equation. So, if we take equation (2) and multiply it by 2 we get **2a + 4c = 38** (it means that adults and 4 children would pay £38). We can call this equation (3) and now place it above equation (1)

**2a + 4c = 38 (3)**

**2a + 3c = 33 (1)**

Now we have the same number of a's in both equations and the signs in front are the same so we can subtract (1) from (3).

**2a + 4c = 38 (3)**

**2a + 3c = 33 (1)**

** c = 5 (3) - (1)**

**So, a child ticket costs £5.**

Now, to find the cost of an adult ticket, substitute c = 5 into any of the three equations, usually the 'easiest' one.

**a + 2 x 5 = 19 [sub into (2)]**

**a + 10 = 19**

**a = 10. **

So an adult ticket costs £10.

Now we can find the cost for the Fisher family which is two adults and one child, 2 x 9 + 5 = 23, so they would pay £23.

If you now compare what we have done here with the previous activity, you will see that we have just added in one extra stage where we multiply one equation by a number in order to equate the coefficients of one variable. Other than that, everything else is the same.

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

Solve **3x + 2y = 10 **** (1)**

and **7x - y = 29 **** (2)**

Notice the coefficients of x are different, as are the coefficients of y. But if we double equation (2) we will have 2y in each

**3x + 2y = 10 **** (1)**

**7x - y = 29 **** (2) x 2**

**14x - 2y = 58**** (3)**

Since the signs in front of the y's are different we add equations (1) and (3).

**3x + 2y = 10 ****(1)**

**14x - 2y = 58 ****(3)**

** 17x = 68**

** x = 4 ****[divide by 17]**

**3 x 4 + 2y = 10 ****[sub into (1)]**

**12 + 2y = 10**

**2y = -2 ****[subtract 12]**

**y = -1 ****[divide by 2]**

**7 x 4 - -1 = 29 ****[check in (2)]**

**28 + 1 = 29 ****[correct]**

**So x = 4 and y = -1.**

Now you should be ready to try some. Remember, its the same as before with just one extra step.