1. Let's start by stating the problem: solving a system of equations with 3 variables means finding values for $x$, $y$, and $z$ that satisfy all three equations simultaneously. 2. A common method is substitution or elimination. The general form is: $$\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}$$ 3. Important rule: You need as many independent equations as variables to find a unique solution. 4. Step example: Suppose we have $$\begin{cases} x + y + z = 6 \\ 2x - y + 3z = 14 \\ -x + 4y - z = -2 \end{cases}$$ 5. First, solve one equation for one variable, e.g., from the first: $$x = 6 - y - z$$ 6. Substitute $x$ into the other two equations: $$2(6 - y - z) - y + 3z = 14$$ $$-(6 - y - z) + 4y - z = -2$$ 7. Simplify each: $$12 - 2y - 2z - y + 3z = 14$$ $$-6 + y + z + 4y - z = -2$$ 8. Combine like terms: $$12 - 3y + z = 14$$ $$-6 + 5y = -2$$ 9. Solve the second for $y$: $$-6 + 5y = -2$$ $$5y = -2 + 6$$ $$5y = 4$$ $$y = \frac{4}{5}$$ 10. Substitute $y$ back into the first: $$12 - 3\left(\frac{4}{5}\right) + z = 14$$ $$12 - \frac{12}{5} + z = 14$$ 11. Simplify: $$z = 14 - 12 + \frac{12}{5}$$ $$z = 2 + \frac{12}{5} = \frac{10}{5} + \frac{12}{5} = \frac{22}{5}$$ 12. Finally, find $x$: $$x = 6 - \frac{4}{5} - \frac{22}{5} = 6 - \frac{26}{5} = \frac{30}{5} - \frac{26}{5} = \frac{4}{5}$$ 13. So the solution is: $$\boxed{\left(\frac{4}{5}, \frac{4}{5}, \frac{22}{5}\right)}$$ This means $x = \frac{4}{5}$, $y = \frac{4}{5}$, and $z = \frac{22}{5}$ satisfy all three equations.