We will call sandwiches S and packets of crisps C. These are the 'unknowns'. From the information we can write down two equations. Equation 1: 3S + 2C = 9 and Equation 2: 5S + 6C = 19. We need two equations to solve a problem with two unknowns which we have. We need to try and eliminate one of the unknowns. To do this we must multiply either Equation 1, Equation 2, or both by constants to try and make the coefficient of one of the unknowns the same in each equation.In this example we will multiply Equation 1 by 3. This gives us a new equation that we will call Equation 3. Equation 3: 9S + 6C = 27 (3 x Equation 1). As you can now see the coefficients of C in Equation 2 and 3 are the same (6). This means they can now be eliminated. We will rearrange Equations 2 & 3 to make 6C the subject. Equation 2: 6C = 19 - 5S and Equation 3: 6C = 27 - 9S. Now as both 19 - 5S and 27 - 9S equal 6C, it means they are also equal. We can now write 19 - 5S = 27 - 9S. After some rearranging we get 4S = 8 so S = 2 or a sandwich costs £2. We now substitute S = 2 back into either Equation 1,2 or 3. We will use equation 1: (3 x 2) + 2C = 9 so 2C = 3 and C = 1.5. This means a packet of crisps costs £1.50. The question asks how much does 2 sandwiches and 5 packets of crisps cost. We can getan equation from this. Equation 4 is 2S + 5C = ?. Now sub in: (2 x 2) + (5 x 1.5) = 11.5 or £11.50.