Question 1089003
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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</pre></b></font>