Question 798041
The problem here is finding the point of intersection: (a,b).
After that, the line perpendicular to the x-axis, is a vertical line, and the equation of the vertical line passing through (a,b) is {{{x=a}}} ({{{y}}} does not play a part).
Similarly, the line perpendicular to the y-axis, is a horizontal line, and the equation of the horizontal line passing through (a,b) is {{{y=b}}}

We are looking for a pair (x,y) that satisfies {{{2x+y=8}}} and {{{3x+2y=0}}}.
We need to solve {{{system(2x+y=8,3x+2y=0)}}}.
You seemed to be trying to solve the system by combining the equations, which always works, but in this case, I would prefer to solve by substitution.
Since the coefficient of {{{y}}} in {{{2x+y=8}}} is {{{1}}}, I would solve the system by substitution starting by solving {{{2x+y=8}}} for {{{y}}}.
(It seems easiest that way).
{{{2x+y=8}}}-->{{{y=8-2x}}}
Substituting into {{{3x+2y=0}}}
{{{3x+2(8-2x)=0}}}-->{{{3x+16-4x=0}}}-->{{{16-x=0}}}-->{{{highlight(x=16)}}}
 
That is the solution to part (A), {{{highlight(x=16)}}}
 
Then, returning to the solved {{{y=8-2x}}}, and plugging in {{{x=16}}}, I get
{{{y=8-2*16}}}--->{{{y=8-32}}}-->{{{highlight(y=-24)}}}
 
That is the solution to part (B), {{{highlight(y=-24)}}}
 
IN PICTURES: 
{{{drawing(300,300,-40,60,-60,40,
grid(0),
blue(line(35,-62,-20,48)),locate(2,8,blue(2x+y=8)),
green(line(-40,60,40,-60)),locate(-35,20,green(3x+2y=0)),
red(circle(16,-24,2)),locate(18,-20,red("(16,-24)")))
)}}} system solved, now the perpendicular lines {{{drawing(300,300,-40,60,-60,40,
grid(0),line(35,-62,-20,48),
line(-40,60,40,-60),
line(16,-60,16,40),locate(18,20,x=16),
line(-40,-24,60,-24),locate(-35,-24,y=-24)
)}}}
 
ANOTHER WAY TO SOLVE THE SYSTEM:
This may be what you were trying to do. Names for this procedure may vary. It may be called "by combinations", or "by elimination", or even fancier names).
The ways to show the work vary even more.
{{{system(2x+y=8,3x+2y=0)}}}-->{{{system(2x+y-8=0,3x+2y=0)}}}-->{{{2(2x+y-8)-(3x+2y)=0}}}-->{{{4x+2y-16-3x-2y=0}}}-->{{{x-16=0}}}-->{{{x=16)}}}
You can substitute {{{x=16}}} into one of the equations to find {{{y}}} or you can make a different combination.
{{{system(2x+y-8=0,3x+2y=0)}}}-->{{{3(2x+y-8)-2(3x+2y)=0}}}-->{{{6x+3y-24-6x-4y=0}}}-->{{{-y-24=0}}}-->{{{y=-24}}}