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
2 answers:
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
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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<h3>HOPE THIS HELPS :)</h3>