Question 1089003: The sum of three numbers is 11. The sum of twice the first number,
4 times the second number, and 5 times the third number is 30. The
difference between 6 times the first number and the second number
is 47. Find the three numbers.
Found 3 solutions by Boreal, josmiceli, Edwin McCravy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x+y+z=11
2x+4y+5z=30
6x-y=47
therefore, y=6x-47
substitute into the equations
x+6x-47+z=11 or 7x-47+z=11
2x+4(6x-47)+5z=30 or 26x-188+5z=30
7x+z=58, z=-7x+58
substitute that into the equation 26x+5z=218 so that 26x-35x+290=218
-9x=-72
x=8
26x+5z=218 so that 208+5z=218 and z=2
Going back to the first equation, 8+y+2=11 and y=1
2(8)+4(1)+5(2)=30
{8, 1, 2}
Answer by josmiceli(19441)  (Show Source):
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
The sum of three numbers is 11.
X + Y + Z = 11
The sum of twice the first number, 4 times the second number,
and 5 times the third number is 30.
2X + 4Y + 5Z = 30
The difference between 6 times the first number and the second number
is 47.
6X - Y = 47
So the system is
(eq. 1) X + Y + Z = 11
(eq. 2) 2X + 4Y + 5Z = 30
(eq. 3) 6X - Y = 47
Since Z is already eliminated from (eq. 3), we
eliminate Z from the other two equations by
multiplying (eq. 1) by -5 and adding to (eq. 2)
-5X - 5Y - 5Z = -55
(eq. 2) 2X + 4Y + 3Z = 30
----------------------------
(eq. 4) -3X - Y = -25
Now multiply (eq. 3) by -1 and add (eq. 4)
(eq. 3) 6X - Y = 47
-6X + Y = -47
(eq. 4) -3X - Y = -25
----------------------------
(eq. 5) -9X = -72
X = 8
Substitute 8 for X in (eq. 4)
(eq. 4) -3X - Y = -25
-3(8) - Y = -25
-24 - Y = -25
-Y = -1
Y = 1
Substitute 8 for X and 1 for Y in (eq. 1)
(eq. 1) X + Y + Z = 11
8 + 1 + Z = 11
9 + Z = 11
Z = 2
So the solution is (X,Y,Z) = (8,1,2)
Edwin
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