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

Feb 5, 2017

The system is inconsistent , and has no solution.

#### Explanation:

We will solve the Problem using Method of Substitution :

From the ${2}^{n d} \text{ eqn., } x = 2 y - 3 z + 1. . . \left(2 '\right)$

Substituting in the ${1}^{s t} e q n . , 2 \left(2 y - 3 z + 1\right) - y + 3 z = 2$

$\Rightarrow 3 y - 3 z = 0 \Rightarrow y = z \ldots \ldots \ldots . \left(1 '\right)$

Using $\left(2 '\right) \text{ in the } {3}^{r d} e q n . , 4 \left(2 y - 3 z + 1\right) - y + 5 z = 5$

$\Rightarrow 7 y - 7 z = 1. \ldots \ldots \ldots \ldots \ldots . . \left(3 '\right)$

$\left(1 '\right) , \mathmr{and} , \left(3 '\right) \Rightarrow 0 = 1 , \text{ which is impossible.}$

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