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Simultaneous Equations

  

Do you remember your maths from years ago? How about your Simultaneous Equations? Here's a question with full solution. If two apples and three pears cost 130 pence...

If two apples plus three pears costs 130 pence and four apples plus two pears cost 140 pence, how much does one apple plus one pear cost?

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Write out the question in mathematical format. A stands for apples and P for pears.

2A + 3P = 130

4A + 2P = 140

Now find a way to remove either the variable A or P from the equations to leave you the cost of a certain number of only one type of fruit.

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50 pence

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Writing out the question in mathematical format we have:

Eqn 1:         2A + 3P = 130

Eqn 2:         4A + 2P = 140

 

A stands for apples, P for pears and the figure on the right of the equals sign is the cost of the items in pence.

We now need to remove either the variable A or P from the equations to leave you the cost of a certain number of only one type of fruit. This can be done by multiplying equation 1 by 2.

 

Eqn 1 * 2:         4A + 6P = 260        (every number multiplied by 2)

 

Now we can subtract eqn. 2 from the multiplied eqn.1 (Be sure to keep each number on the correct side of the equals sign.)

 

Eqn. 1 *2 - Eqn 2:     

       (4A - 4A) + (6P - 2P) = (260 - 140)    

 

=>       0              +        4P        =       120

=>                                   P        =        30

 

We now have the cost of one pear; 30 pence. Knowing this we can substitute the P in equation 1 for the figure 30 and thus work out the cost of one apple.

 

Eqn 1.         2A + 3P   = 130

=>               2A + 3*30 =130

=>               2A + 90    =130

=>               2A            = 130 - 90

=>               2A            = 40

=>                 A            = 20

 

 

i.e. the cost of one apple costs 20 pence.

The cost of one apple and one pear is therefore:

20 + 30 = 50 pence.

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