$\displaystyle -2x-5y=3, -8y+4z=-7$
How do you go about finding (x,y,z)? I can't find this in my textbook.
I would prefer Ackbeet's method. Since z does not appear in the first equation and x does not appear in the second, it is easy to solve the two equations for x and z in terms of y. Then you can use "y" as parameter. Once you have x= f(y) and z= g(y), your parametric equations for the line of intersection are x= f(t), y= t, z= g(t).
$\displaystyle \left\{\begin{array}{ccc}
x&=-\displaystyle\frac{5y+3}{2} \\ \\
z&=\displaystyle\frac{8y-7}{4}
\end{array}\right.\rightarrow
\left[\begin{array}{ccc}
x\\y\\z
\end{array}\right]=
\left[\begin{array}{ccc}
-\displaystyle\frac{5y+3}{2} \\ \\
y \\ \\
\displaystyle\frac{8y-7}{4}
\end{array}\right]=
\left[\begin{array}{ccc}
-\displaystyle\frac{5}{2}y-\frac{3}{2} \\ \\
y \\ \\
\displaystyle\frac{8}{4}y-\frac{7}{4}
\end{array}\right]=
y\left[\begin{array}{ccc}
-\displaystyle\frac{5}{2}\\ \\
1 \\ \\ 2
\end{array}\right]+\left[\begin{array}{ccc}
-\displaystyle\frac{3}{2} \\ \\
0 \\ \\
-\displaystyle\frac{7}{4}
\end{array}\right]
$
Line pass through Point $\displaystyle (-\displaystyle\frac{3}{2},0,-\frac{7}{4})$ with normal vector $\displaystyle (-\displaystyle\frac{5}{2},1,2)^T$