\( \left. \begin{array} { l } { d _ { 1 } : x + a y - 4 = 0 } \\ { d _ { 2 } : 3 x - 2 y + 3 = 0 } \end{array} \right. \quad d _ { 1 } + d _ { 2 } \)

To solve for \( d_1 + d_2 \), we first need to express \( d_1 \) and \( d_2 \) in a form that makes adding them straightforward. For line \( d_1: x + a y - 4 = 0 \), we can express \( y \) in terms of \( x \): \[ y = \frac{4 - x}{a} \] For line \( d_2: 3x - 2y + 3 = 0 \), rearranging gives us: \[ y = \frac{3}{2}x + \frac{3}{2} \] Now, let’s set the two equations for \( y \) equal to each other and find the intersection, showing how to combine both lines. Setting the right sides equal gives: \[ \frac{4 - x}{a} = \frac{3}{2}x + \frac{3}{2} \] This can be simplified further based on the value of \( a \). The solution will allow us to understand how these two lines relate to each other geometrically! Now, let's move on to their properties. Adding the two lines can give us a new line, combining coefficients to see how they work together. For the result of \( d_1 + d_2 \), we'll consider the linear combination of both equations. By simply adding the two original equations, you can derive a trajectory in the coordinate plane that includes contributions to both variables. Enjoy exploring how these lines interact!