Find three numbers such that their sum is 12, the sum of the first, twice the second, and three times the third is 31, and the sum of the third and nine times the second is 1

3,-1,10

Step-by-step explanation:

let the numbers be x, y, z.

x+y+z=12

x+2y+3z=31

z+9y=1

z=1-9y

subtract 1st from 2nd

y+2z=19

y+2 (1-9y) = 19

y+2-18y=19

-17 y=17

y=-1

z=1-9 (-1) = 1+9=10

x-1+10=12

x=12+1-10=3

a = 3, b = - 1, c = 10

Step-by-step explanation:

Let the three numbers be a, b and c.

Equation 1: a + b + c = 12

Equation 2: a + 2b + 3c = 31

Equation 3: 9b + c = 1

Equation 2 - Equation 1:

Equation 4: b + 2c = 19

Equation 3 times by the number 2

Equation 5: 18b + 2c = 2

Equation 5 - Equation 4

17b = - 17

b = - 1

Substitute into Equation 4:

2c - 1 = 19

2c = 20

c = 10

Substitute into Equation 1:

a + b + c = 12

a - 1 + 10 = 12

a = 3