3.3 Lösning 2d

$\begin{array}{rclcl} 16 \, + \, 2\,x & = & 14 & \qquad | & - \, 16 \\ 16 \, + \, 2\,x - \, 16 \, & = & 14 \, - \, 16 \, & & \\ 2\,x & = & -2 & \qquad | & / \, 2 \\ \displaystyle \frac{2\,x}{2} & = & \displaystyle \frac{-2}{2} & & \\ x & = & -1 \end{array}$