Question 1200186
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I'll answer part (i) only. 
Part (ii) seems a bit garbled so you'll need to revise.



Part (i)


Plug k = 11 into the 2nd equation and isolate y.
2x+y = k
2x+y = 11
y = -2x+11


Plug this into the other equation.
xy = 12
x(-2x+11) = 12
-2x^2+11x = 12
-2x^2+11x-12 = 0


Turn to the quadratic formula.
We'll plug in 
a = -2
b = 11
c = -12
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-11+-sqrt((11)^2-4(-2)(-12)))/(2(-2))}}}


{{{x = (-11+-sqrt(121 - 96))/(-4)}}}


{{{x = (-11+-sqrt(25))/(-4)}}}


{{{x = (-11+-  5)/(-4)}}}


{{{x = (-11+5)/(-4)}}} or {{{x = (-11-5)/(-4)}}}


{{{x = (-6)/(-4)}}} or  {{{x = (-16)/(-4)}}}


{{{x = 3/2}}} or  {{{x = 4}}}


{{{x = 1.5}}} or  {{{x = 4}}}


If x = 1.5, then,
y = -2x+11
y = -2*1.5+11
y = -3+11
y = 8
Making (1.5, 8) one point of intersection.


If x = 4, then,
y = -2x+11
y = -2*4+11
y = -8+11
y = 3
Making (4, 3) the other point of intersection.


Graph:
{{{drawing(400,400,-3,10,-3,10,
graph(400,400,-3,10,-3,10,-100,12/x,-2x+11),
circle(1.5,8,0.02),
circle(1.5,8,0.04),
circle(1.5,8,0.06),
circle(1.5,8,0.08),
circle(1.5,8,0.10),
circle(1.5,8,0.12),
circle(1.5,8,0.14),
circle(1.5,8,0.16),
circle(1.5,8,0.18),

circle(4,3,0.02),
circle(4,3,0.04),
circle(4,3,0.06),
circle(4,3,0.08),
circle(4,3,0.10),
circle(4,3,0.12),
circle(4,3,0.14),
circle(4,3,0.16),
circle(4,3,0.18),

locate(1.5+0.5,8+0.5,"(1.5,8)"),
locate(4+0.5,3+0.5,"(4,3)")
)}}}
xy = 12 in green
2x+y = k in blue, where k = 11
1.5 = 3/2


Here's a more zoomed-out look of the graph.
{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10,-100,12/x,-2x+11)
)}}}
I recommend using either Desmos or GeoGebra as a graphing tool. 
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