Question 56375: please solve the system
x-y+2z=7
2x+z=4
x+5y+z=9
Answer by AgusKwan(7)   (Show Source):
You can put this solution on YOUR website! x-y+2z=7 --> Equation1
2x+z=4 --> Equation2
x+5y+z=9 --> Equation3
Since Equation2 only have x and z variable, we'll eliminate y from the other 2 equation
Equation1 --> x-y+2z=7 (times 5) --> 5x - 5y + 10z = 35
Equation3 --> x+5y+z=9 ------------> x + 5y + z = 9 (add both equations)
will have 6x + 11z = 44 --> Equation 4
Equation 4 --> 6x + 11z = 44------------> 6x + 11z = 44
Equation 2 --> 2x + z = 4 (times 3)---> 6x + 3z = 12 (subtract equations)
8z = 32
z = 4
Substitute z on Equation2 --> 2x + 4 =4 --> x=0
Substitute z and x to Equation1 --> 0-y+2(4)=7 --> y=1
Therefore x=0 , y=1, z=4
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