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.

How do you solve the system 2x-y+3z=22x-y+3z=2, x-2y+3z=1x-2y+3z=1, and 4x-y+5z=54x-y+5z=5?