Economics, 3rd Edition

Qd = 20 − 2p

   

11

= 20 − 2p

 2p = 20 − 11

 2p = 9

  

  p = 4.5

It is always worth checking your answer to make sure you have made no mistakes along the way so 

in this case we will substitute our known value of Q into the second equation to check we get the same 

answer (p

= 4.5). So:

   

 

   

 

 

Qs = 2 + 2p

      

   

  

11

= 2 + 2p

 

1

 

1

− 2 = 2p

       

 

  

 

9

= 2p

        

   

 

p = 4.5

Sometimes we may have equations where the p and Q values are both on the same side of the equa-

tion. In this case we have to use a different technique – the elimination method.

Take the following two equations:

 

−3p + 4Q = 5

  

(1)

 

2p − 5Q = −15

 

(2)

We have labelled these two equations (1) and (2) to allow us to keep track of what we are doing and 

reduce the risk of making an error.

Remember above when we noted the fact that having a nice convenient equation where the coefficient 

was equal but the signs were opposite, enabling us to be able to eliminate one of the values to help solve 

the equation for the other unknown? That is what we need to do with these two equations. We have to 

choose to manipulate the two equations to make either the ‘p’ terms or the ‘Q’ terms have the same 

coefficient but opposite signs. A knowledge of factors and lowest common denominators is useful here.

In this example we are going to manipulate the equations to get rid of the ‘p’ terms. This allows us to 

isolate the ‘Q’ terms and thus solve for Q and then find p.

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62  PART 2  SUPPLY AND DEMAND: HOW MARKETS WORK

This is how we do this: