Two sorts

creatures

On Mars there are two sorts of creatures

Zerts have 4 legs and Zios have 7 legs.

Jacob the astronaut looked out of his space

ship and saw 55 legs.

• How many Zerts might there be?

• How many Zios might there be?

•How many creatures might there be altogether?

There are two solutions:

There might be either 1 Zert and 12 Ziots or 5 Zerts and 5 Ziots. There might me either 13 or 10 creatures altogether.

Explanation:

This problem has a second question that is missing: Do you think there are any different answers?

You can create a system of two equations.

1. Name the variables:

x: number of of Zerts y: number of Zios

2. Translate the verbal statements into algebraic expressions:

Zerts have 4 legs: 4x Zios have 7 legs: 7y Jacob saw 55 legs: 4x + 7y = 55

3. Solution

Since there is one equation and two variables there could be many possible solutions, to the equation. But since x and y are restricted to positive integer numbers, the situation may be different.

You can test some numbers.

First, realize that the greatest possible value of y is when x = 0. Thus 7y has to be less than 55. The multiples of 7 that are less than 55 are 14, 21, 28, 35, 42, and 49.

Thus, the only possible values of y are 1, 2, 3, 4, 5, 6, and 7.

If y = 1: 4x = 55 - 7 (1) = 48 ⇒ x = 48/4 = 12. Thus, this is your first solution: (12,5).

If y = 2: 4x = 55 - 7 (2) = 41. Since 41 is not muliple of 4. This is not a solution.

If y = 3: 4x = 55 - 21 = 34. 34 is not a multiple of 4. Not a solution.

If y = 4: 4x = 55 - 28 = 27. Not a solution, either.

If y = 5: 4x = 55 - 35 = 20 ⇒ x = 20/4 = 5. This is a second solution: (5,5)

If y = 6: 4x = 55 - 42 = 13. Not a multiple of 4.

If y = 7: 4x = 55 - 49 = 6. Not a multiple of 6.

Conclusion

There are two solutions:

There might be either 1 Zert and 12 Ziots or 5 Zerts and 5 Ziots. There might me either 13 or 10 creatures altogether.