How do you solve the system #2x-y+3z=2#, #x-2y+3z=1#, and #4x-y+5z=5#?

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Answer 1 · 2017-02-05T14:34:40

The system is inconsistent , and has no solution.

We will solve the Problem using Method of Substitution :

From the #2^(nd)" eqn., "x=2y-3z+1...(2')#

Substituting in the #1^(st) eqn., 2(2y-3z+1)-y+3z=2#

#rArr 3y-3z=0 rArr y=z..........(1')#

Using #(2')" in the "3^(rd) eqn., 4(2y-3z+1)-y+5z=5#

#rArr 7y-7z=1..................(3')#

#(1'), and, (3') rArr 0=1," which is impossible."#

Clearly, the system is inconsistent , and has no solution.