## Linear Algebra: A Modern Introduction

$\begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix}=\begin{bmatrix}\dfrac{1}{4}-\dfrac{3}{4}s-\dfrac{1}{2}t \\ s \\t\end{bmatrix}$; $t,s$ refer to real numbers
Since we have $4x_1+3x_2+2x_3=1$, this can be re-written as: $x_1=\dfrac{1}{4}-\dfrac{3}{4}x_2-\dfrac{1}{2}x_3$ Suppose: $x_2=s,x_3=t$ Then: $x_1=\dfrac{1}{4}-\dfrac{3}{4}s-\dfrac{1}{2}t$ Our desired solution set is: $\begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix}=\begin{bmatrix}\dfrac{1}{4}-\dfrac{3}{4}s-\dfrac{1}{2}t \\ s \\t\end{bmatrix}$; $t,s$ refer to real numbers