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Solve 2x + y – z = 3 4x – y + 4z = 0 -3y + 2z = 6
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x=2 y=−4 z=−3
step1. {2x+y-z=3}*2 since in 2nd equation x has co-efficient 4. {4x-y+4z=0}*1 result will be 4x+2y-2z=6 4x-y+4z=0 step2. now subtract both equation both sides. (4x+2y-2z)-(4x-y+4z)=6-0 => 3y-6z=6---------let it be equation(1) again we have two equations -3y+2z=6 3y-6z=6 add both the equetions both side i.e. (-3y+2z)+(3y-6z)=6+6 =>-4z=12 =>z=-3 Now, Put value of z in equattion 3y-6z=6 or -3y+2z=6 you'll get, 3y-6*(-3)=6 =>3y+18=6 =>3y=-12 =>y=-4 Now put values of z and y in any of your equetions given you'll get x. :)
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