Question 1053040: Find three numbers such that: their sum is 12;the sum of the first, twice the second, and three times the third is 31;the sum of the third and nine times the second is 1.
Answer by CubeyThePenguin(3113) (Show Source):
You can put this solution on YOUR website! x + y + z = 12
x + 2y + 3z = 31
9y + z = 1 ----> z = 1 - 9y
Substitute the expression for z into the first two equations.
x + y + (1 - 9y) = 12 ---> x - 8y = 11
x + 2y + 3(1 - 9y) = 31 ---> x - 25y = 28
Subtract the equations to get
x - 8y = 11
-x + 25y = -28
----------------------
17y = -17
y = -1
9y + z = 1
9(-1) + z = 1 ---> z = 10
x + (-1) + 10 = 12
x = 3
solution: (x, y, z) = (3, -1, 10)
|
|
|