Question 869761
Linear equations, like {{{x=2}}} , or {{{y=3}}} or {{{x+2y=4}}}
graph as straight lines. (That is why they are called "linear").
Since two points determine a line, to graph a linear equation you just need to plot two points
(two (x,y) pairs) and connect them with a straight line.
One point and a slope can also be a useful cambination.
That sounds easy, but finding points that are
easy to calculate and
conveniently located is not always easy.
 
Vertical lines, like {{{x=2}}} are easy to graph,
and so are horizontal lines like {{{y=3}}}.
Slanted lines may be easy or harder to graph.
Take {{{x+2y=4}}} .
You can make {{{x=0}}} and {{{y=0}}} and get two helpful points:
For {{{x=0}}} , {{{2y=4}}}-->{{{y=4/2}}}-->{{{y=2}}} gives you point (0,2).
For {{{y=0}}} , {{{x=4}}} gives you point (4,0).
You plot those two points; connect them, and you got your line.
{{{drawing(300,300,-2,8,-2,8,
grid(0),blue(line(-2,3,8,3)),
red(line(2,-2,2,8)),locate(2.1,6,red(x=2)),
circle(0,2,0.1),circle(4,0,0.1),
circle(0,2,0.15),circle(4,0,0.15),
locate(5,3,blue(y=3)),locate(3,1,green(x+2y=4)),
green(line(-2,3,8,-2))
)}}}
 
A linear equation like {{{3x-5y=7}}} makes graphing more complicated.
You could do as above and find points (7/3,0) and (0,-7/5),
but those points are hard to locate precisely.
Sometimes points are hard to calculate;
sometimes they are hard to locate,
and sometimes the two points you get are too close together to draw the line accurately.
In that case, you try, try again.
Practice will tell you what to do next time.
For {{{3x-5y=7}}} , I would solve for {{{y}}} to get something like this:
{{{3x-5y=7}}}-->{{{3x-7=5y}}}-->{{{y=(3x-7)/5}}}<-->{{{y=(3/5)x-7/5}}} .
Leaving the equation as {{{y=(3x-7)/5}}} helps me figure out a good integer {{{x}}} value,
that will give me an integer {{{y}}} value.
For {{{x=4}}} , {{{3x-7=3*4-7=12-7=5}}} and {{{y=(3*4-7)/5=5/5=1}}} .
That gives me point (4,1).
For {{{x=-1}}} , {{{3x-7=3(-1)-7=-3-7=-1}}} and {{{y=(3(-1)-7)/5=-10/5=-2}}} .
That gives me point (-1,-2).
With those two points, I can draw the line.
However, with just one of those points, and the {{{3/5}}} slope, I can get other points without calculating.
A slope of {{{3/5}}} means that as {{{x}}} increases by {{{5}}} , {{{Y]]] increase by {{{3}}} .
That means that from any point, adding {{{5}}} to the {{{x}}} and {{{3}}} to the {{{y}}} I get another point.
From (4,1), I get (4+5,1+3) = (9,4),
and adding again I get (9+5,4+3) = (14,7).
I could subtract instead of adding, and get (4-5,1-3) = (-1,-2).
{{{drawing(300,250,-2,10,-3,7,
grid(1),
red(circle(-1,-2,0.2)),
red(circle(4,1,0.2)),
red(line(-6,-5,14,7)),
green(arrow(4,1,9,1)),green(arrow(9,1,9,3.8)),
green(line(4,1,9,1)),green(line(9,1,9,3.8)),
circle(9,4,0.2)
)}}}