Some people went to London Zoo, where they bought adult tickets and child tickets. Mr and Mrs Brown took their daughter Libby and paid £68, whereas Mr and Mrs White took their three sons, Tom, Dick and Harry and paid £100. Can you work out how much Ms Black and her children Jack and Maisy would have to pay?

Problems like this are common in maths books and exams. There are a number of ways you can solve them, such as trial and improvement, but here we want to look at a method where we turn the information into two equations and solve them at the same time. This method is known as 'solving **simultaneous equations'**.

In order to find out how much Ms Black and her two children 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 letter to stand for each one. In many simultaneous equations, x and y are used but you can use any letter and here we will use a = cost of an adult and c = cost of a child. So, Mr and Mrs Brown and their daughter becomes 2 adults and one child paying £68 or **2a + c = 68**. Mr and Mrs White and their 3 sons becomes 2 adults and 3 children paying £100 or **2a + 3c = 100. **We now have our two equations, and we need to find values for a and c which work in both equations at the same time (simultaneously).

If we line up the equations, one above the other with the one with the biggest values on top it will help.

**2a + 3c = 100**

**2a + c = 68**

We can see that both equations have 2a but the top one has 2c more than the bottom one and the cost is £32 more, so 2c = 32. If 2 children cost £32 then one must cost £16, so c =16. Knowing this we can know work out a. If 2c + c = 68 and c = 16 then 2c + 16 = 68. We can solve this equation easily by subtracting 16 from both sides giving 2a = 52, and dividing by 2 gives a = 26. So an adult ticket costs £26. Now we have our two ticket costs it is a good idea to check they work for the other family (equation). We had 2a + 3c = 100 and substituting a = 26 and c= 16 gives 2 x 26 + 3 x 16 = 52 + 48 = 100, which is correct.

Now we can find the cost for the Black family which is one adult and 2 children. 26 + 2 x 16 = 58, so they would pay £58.

Now, let's look at another pair of simultaneous equations. This time just the equations without the real-life problem and we will look at a shorter way of showing the working.

Solve **5x + y = 17**

and **3x + y = 11**

Let's number our equations (1) and (2) with the one with bigger values on top then subtract equation (2) from (1)

**5x + y = 17 ****(1)**

**3x + y = 11 ****(2)**

**2x = 6 ****(1) - (2)**

** x = 3 ** (divide by 2)

Substitute x = 3 into (2) and solve the equation [you could use equation (1) if you prefer]

**3 x 3 + y = 11**

**9 + y = 11**

**y = 2**

**So x = 3 and y = 2**

Check these work with equation (1) [use the equation you did __ not__ use to find the value of y]

**5 x 3 + 2 = 17** (which is correct)

Let's look at one more problem.

**5x - y = 17 ** **(1)**

**2x + y = 11** ** (2)**

Notice that the y's have the same coefficients (number in front) and so this is the variable which will be eliminated when we combine the equations. But because the **signs in front of them are different **(one positive and one negative) this time we **add** the equations.

**7x = 28 ****(1) + (2)**

** x = 4 ****[divide by 7]**

**2 x 4 + y = 11 ****[substitute into (2)]**

**8 + y = 11**

**y = 3**

**Check in (1)**

**5 x 4 - 3 = 17 [****correct]**

**So x = 4 and y = 3.**

Remember the variable with **equal coefficients** will be **eliminated** when the equations are combined. If the **signs** in front are the **same **you **subtract** if they are** different** you **add**. (Remember **Same Sign Subtract - SSS**)