SOLUTION: 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 num

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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): You can put this solution on YOUR website!
Let the numbers be a, b, and c
--------------
(1)
(2)
(3)
---------------------------
(3)
Plug this into (1)
(1)
(1)
Plug this into (2)
(2)
(2)
(2)
Multiply bth sides of (3) by and
subtract (3) from (2)
(2)
(3)
--------------------------


and
(3)
(3)
(3)
(3)
and
(1)
(1)
(1)
The numbers are 8, 1, and 2
--------------------------
check:
(2)
(2)
(2)
(2)
OK
-------------
You can check (3)



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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