Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + y - z = 2, \;\; 2x - y + 4z = 1, \;\; -x + 6y + z = 5$


Given system of equations:

$x + y - z = 2$ $\;\;\; \cdots \; (1)$

$2x - y + 4z = 1$ $\;\;\; \cdots \; (2)$

$-x + 6y + z = 5$ $\;\;\; \cdots \; (3)$

Adding equations $(1)$ and $(3)$ gives

$7y = 7$ $\implies$ $y = 1$

Substituting the value of $y$ in equation $(1)$ gives

$x - z = 1$ $\;\;\; \cdots \; (4)$

Substituting the value of $y$ in equation $(2)$ gives

$2x + 4z = 2$ $\;\;$ i.e. $\;$ $x + 2z = 1$ $\;\;\; \cdots \; (5)$

Subtracting equations $(4)$ and $(5)$ gives

$-3z = 0$ $\implies$ $z = 0$

Substituting the value of $z$ in equation $(4)$ gives

$x = 1$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y, z\right) = \left\{\left(1, 1, 0\right) \right\}$