Question 789115
To graph one line you only need to plot two points and connect them with a straight line. Since the problem asks for 4 points, we do the extra unnecessary work.
 
For {{{3x+4y=7}}},
{{{3x+4y=7}}}-->{{{4y=7-3x}}}-->{{{y=(7-3x)/4}}}, so chosing random values for {{{x}}} we would get lots of fractions.
However, substituting {{{x=1}}}gives us
{{y=(7-3)/4=4/4=1)}}} and we have the point (1,1).
Increasing or decreasing x by 4 gives us other integer values for y:
Substituting {{{x=5}}}gives us
{{{y=(7-3(5))/4=(7-15)/4=-8/4=-2}}} and we have the point (5,-2).
Substituting {{{x=-3}}}gives us
{{{y=(7-3(-3))/4=(7+9)/4=16/4=4}}} and we have the point (-3,4).
Substituting {{{x=-7}}}gives us
{{{y=(7-3(-7))/4=(7+21)/4=28/4=7}}} and we have the point (-7,7).
 
For {{{y=x}}}, nice points are easier to find:
We could substitute a few scattered values for x,  to get
{{{x=4}}}-->{{{y=4}}} for point (4,4),
{{{x=8}}}-->{{{y=8}}} for point (8,8),
{{{x=-4}}}-->{{{y=-4}}} for point (-4,-4), and
{{{y=-8}}}-->{{{y=-8}}} for point (-8,-8)
 
{{{drawing(300,300,-10,10,-10,10,
grid(1),blue(line(-11,10,17,-11)),
blue(circle(-7,7,0.3)),blue(circle(-3,4,0.3)),
blue(circle(1,1,0.3)),blue(circle(5,-2,0.3)),
green(circle(4,4,0.3)),green(circle(-4,-4,0.3)),
green(circle(-8,-8,0.3)),green(circle(8,8,0.3)),
green(line(-10,-10,10,10))
)}}} The lines seem to intersect at point (1,1), so {{{system(x=1,y=1)}}} seems to be the solution.
We need to verify by substituting in the original equations:
For {{{3x+4y=7}}}, {{{3*1+4*1=3+4=7}}} checks
For {{{y=x}}}, {{{1=1}}} checks.
So {{{highlight(system(x=1,y=1))}}} is the solution.