SOLUTION: hi I am a student in algebra 2 and we are solving linear equalities for the values of x, y, z and I am extremely confused on how to solve this problem x+z=3 and x+2y-z=1 and 2x-y+z
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-> SOLUTION: hi I am a student in algebra 2 and we are solving linear equalities for the values of x, y, z and I am extremely confused on how to solve this problem x+z=3 and x+2y-z=1 and 2x-y+z
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Question 853991: hi I am a student in algebra 2 and we are solving linear equalities for the values of x, y, z and I am extremely confused on how to solve this problem x+z=3 and x+2y-z=1 and 2x-y+z=3 please help me I am super confused Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! What method do you want to use? Try substitution if this is allowed for your class right now.
z=3-x from the first equation. The next two equations using this substitution are:
x+2y-(3-x)=1 and 2x-y+(3-x)=3;
2x+2y-3=1 and x-y+3=3;
2x+2y=4 and x-y=0;
x+y=2 and x-y=0;
Using simple elimination, adding left members and adding right members gives:
2x=2 which means x=1. This also means, using x-y=0 that y=1.
Use the first equation now to find the value for z.
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