Find the general solution to the given system. \[ \begin{array}{l}-2 x_{1}+4 x_{3}=6 \\ 4 x_{2}=8\end{array} \] \( \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{l}\text { Ex: 6-3x_2 } \\ \text { Enter } x_{1} \text { as: } \times 1 \\ \text { Enter } 2 x_{1} \text { as: } 2 x_{-} 1\end{array}\right. \)

To solve the given system of equations, let's start by simplifying each equation. 1. The first equation is: \[-2x_1 + 4x_3 = 6\] We can isolate \(x_1\) by rearranging it: \[x_1 = 2x_3 - 3\] 2. The second equation is: \[4x_2 = 8\] This simplifies to: \[x_2 = 2\] Now, since \(x_3\) can take any real value (let's denote it as a free variable \(t\)), we can express the general solution as: \[ \left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{l} 2t - 3 \\ 2 \\ t \end{array}\right] \] where \(t\) is any real number. Now the solution can be represented with parameter \(t\): \[ \left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{l} 2t - 3 \\ 2 \\ t \end{array}\right] \quad \text{for } t \in \mathbb{R}. \] This yields the general solution to the system!