The sum of three numbers is fourteen. The first number minus three times the third number is the second number. The second number is six more than the first number. Find the three numbers.

1st number (x) = 5

2nd number (y) = 11

3rd number (z) = -2

Step-by-step explanation:

Let the generic solution for this problem be x + y + z = 14.  

The first number minus three times the third number equals the second number, so x - 3z = y.  The second number is 6 more than the first number, so y = 6 + x.  

x - 3z = y, we know that y = 6 + x, so the equation becomes x - 3z = 6 + x.

After some arithmetic, we find that z = -2.  

Plugging our knowns back into the generic solution becomes:

x + x - 3z + z = 14

2x - 3(-2) - 2 = 14

2x + 6 - 2 = 14

2x + 4 = 14

2x = 10

x = 5

So we know that z = -2, and x = 5, it's just simple substitution from there.  

5 + y + (-2) = 14

5 + y = 16

y = 11

answer:

sometimes

*i think

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6