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determine if the system has solution, if yes then find it. 2x+3y+7z=15 3x-4y+2z=1 x+2z=7
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from the last equation you get x=7-2z and then insert it in the second, you get: 3(7-2z)-4y+2z=1 21-6z-4y+2z=1 -4z=1+4y-21 -4z=4y-20|:(-4) z=y-5 Insert this in the first eq 2(7-2z)+3y+7(y-5)=15 14-4z+3y+7y-35=15 14-4(y-5)+3y+7y-35=15 14-4y+20+10y=50 solve for y;]
kamille it has no solution actualy :D confusing yes.its supposedto be solved by matrices
oh,sorry!!
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