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Rufina [12.5K]
3 years ago
13

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 s

um of the third and nine times the second is 1
Mathematics
2 answers:
Tpy6a [65]3 years ago
8 0

Answer:

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

KatRina [158]3 years ago
3 0

Answer:

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

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