311.1.3.19 find h and k in the system

[karush]

$\tiny{311.1.3.19}$
find h and k such that the system has, no solution (parallel) ,unique solution (intersection) , many solutions (same line)
$\begin{array}{rrr}
x_1+hx_2&=2&(1)\\
4x_2+8x_2&=k&(2)
\end{array}$

Spoiler: my take

by observation if h=2 then $k=\mathbb{R}, k\ne 8$ for parallel

if $h\ne 2$ then intersection

if $h=2$ and $k=8$ many solutions

 

[HOI]

I presume that you solved the equations:
Multiply the first equation by 4 to get [math]4x_1+ 4hx_2= 8[/math].
Subtract that from the second equation, [math]4x_1+ 8x_2= k[/math] to get [math](8- 4h)x_2= k- 8[/math].

Solve for [math]x_2[/math] by dividing by 8- 4h: [math]x_2= \frac{k- 8}{8- 4h}[/math]. As long as 8- 4h is not 0, which means h is not 2, there is a unique answer. (There are two intersecting lines.)

If h= 2 and k= 8 the equation is [math]0x_2= 0[/math] which is true for all [math]x_2[/math] so there are infinitely many solutions. (There is only one line described by both equations.)

If h= 2 and k is not 8, the equation is [math]0x_2= k- 8[/math] which is NOT true for any [math]x_2[/math]. There is no solution. (There are two parallel lines.)

That is what you have. Well done!