Question: A teacher spent $64 buying two types of games. One type cost $12 each and the other $14 each. How many of the $12 games did she buy?

what are the options?

Equations

Let the number of 12 dollar games be x.

Let the number of 14 dollar games be y

12x + 14y = 64 Divide this equation by 2 (It makes the number smaller)

6x + 7y = 32 You only have this equation. There is a unique equation that answers this problem, but it is a bit of a guess. These equations are called Diophantine equations. Only integer solutions are allowed and on this occasion, only numbers greater than or equal to zero are allowed.

Solution

You should work with y. The largest number you can have for y is 4

y = 4

6x + 7 (4) = 32

6x + 28 = 32 Subtract 28 from both sides.

6x = 32 - 28

6x = 4 When you do the division you do not get a whole number. y = 4 is not the answer.

Y = 3

6x + 3*7 = 32

6x + 21 = 32 Subtract 21 from both sides

6x = 32 - 21

6x = 11 When dividing by 6, you do not get a whole number answer. y = 3 is not the answer.

y = 2

6x + 7y = 32

6x + 7 (2) = 32

6x + 14 = 32 Subtract 14 from each side.

6x = 32 - 14

6x = 18 Divide by 6

x = 3

So she bought three 12 dollar games (and two 14 dollar games).

Comment

The thing about Diophantine Equations is that they are usually constructed in such a way as not to give more than 1 possible answer. You can try y = 1 and 0 but if you get a decimal answer, those answers are not allowed.