The correct answer to this question is B.
The solutions of the equations of the situation can be:
z = 1 , y = 5, x = 4
z= 2 , y = 3, x = 5
z=3 , y = 1, x = 6
z = 0 , y = 7, x =3
The question can be expressed as a equation
6 x + 8 y + 10 z = 84
also, x + y + z = 10
⇒ x = 10- y - z
Putting it in first equation,
6(10 - y - z ) +8y + 10z = 84
⇒ 60 +2y + 4z = 84
⇒2(y + 2z ) = 14
⇒ y + 2z = 7
Now putting
z = 1 , we get y = 5,
z= 2 , y = 3
z=3 , y = 1
z = 0 , y = 7
So, only 4 possible solutions.
Therefore, there can only be 4 possible solutions for the equations.
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The answer is Abecause if you plug in the value es for each function A is the only one the makes it trye
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