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Question

To solve: x+2y+3z=1;2x+5y+6z=2;3x+7y+8z=4 using reduction method.

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Solution

x+2y+32=1...(i)2x+5y+6z=2...(ii)3x+7y+8z=4...(iii)Multiplyeq(i)by(ii)subtractfrom(ii)y=0equationbecomes2x+6z+2x+3z=23x+82=4z=2x=5

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