Question 1203611
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You are given this system of three linear equations 

    x  + 4y - 6z = -1    (1)

   2x  -  y + 2z = -7    (2)

   -x  + 2y - 4z =  5    (3)


Quick inspection shows that the part  "-y + 2z"  in equation (2) and the part "2y - 4z" in equation (3)
fit very well for mutual destruction. So, you multiply equation (2) by 2 and add it with equation (3).

The modified equations are

    4x - 2y + 4z = -14   (2')

   -x  + 2y - 4z =  5    (3')


After adding these equations, you get

    3x           = -14 + 5 = 9,  which implies  x = -9/3 = -3, so one unknown is just found.


Now you plug in x= -3 into equations (1) and (3). You get a system of two equations

    -3    + 4y - 6z = -1    (1'')

    -(-3) + 2y - 4z = 5     (3'')


You simplify these equations further

     4y - 6z = 2            (1''')

     2y - 4z = 2            (3''')


Next you divide all the term in (1''') by 2, keeping (3''') as is.  You get

     2y - 3z = 1            (1'''')
 
     2y - 4z = 2            (3'''')


Next you subtract equation (3'''') from equation (1'''').   You get then

           z = 1 - 2 = -1.    Thus, one more unknown is found.


To find the last unknown, y, use equation (1''''). Substitute there z= -1 to get

    2y - 3*(-1) = 1,  and  obtain  then  2y + 3 = 1,  2y = 1 - 3 = -2,  y= -2/2 = -1.


<U>ANSWER</U>.  x= -3;  y= -1;  z= -1.
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Solved.