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What is the solution of the system of equations? 3x+2y+z=7 5x+5y+4z=3 3x+2y+3z=1
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@SolomonZelman
3x+2y+z=7 5x+5y+4z=3 3x+2y+3z=1 take the first and last equations and subtract them from each other 3x+2y+z=7 - 3x+2y+3z=1 3x-3x 2y-2y z-3z = 7-1 so -2z=6 therefore z=-3 plug the z into the all of the equation 3x+2y+(-3)=7 5x+5y+4(-3)=3 3x+2y+3(-3)=1 3x+2y+=10 5x+5y+(-12)=3 3x+2y+(-9)=1 3x+2y+=10 5x+5y+=15 3x+2y+=10 the 1st and last equations are the same so take out one of them, remains 3x+2y+=10 5x+5y+=15 Can you solve it from there?
@caitlynnbaby, can you solve it from there on your own?
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