Question 690089
General NOTES:
The equations {{{y=1x+2}}} and {{{y=3x+0}}} have straight lines for graphs.
They can be called linear functions or linear equations.
(When the equations start with "y=" as the ones you have, the number multiplied times the {{{x}}} is called the slope, and the number added to that product is called the y-intercept. The equation {{{y=1x+2}}} has a slope of {{{1}}} and a y-intercept of {{{2}}}).
 
If you have those two equations in one problem, it would be what we call a system of linear equations, and we would indicate it like this.
{{{system(y=1x+2,y=3x+0)}}}
The solution to that system would be a pair of values,
with a value for x, and a value for y.
You could give the solution as {{{x=1}}}, {{{y=3}}},
or you could show the pair of values as the ordered pair (1,3).
I'll show you how to graph lines and solve systems, baby step by baby step.
 
HOW TO GRAPH LINEAR EQUATIONS:
Grid paper helps.
You can buy it, and there were some sheets that I would download and print when my youngest child was in high school.
I would start with something like this:
{{{drawing(300,300,-5,5.5,-5,5.5,
grid(1)
)}}}
My children's teachers (in the USA) insisted on putting arrowheads and x and y labels on both sides of the horizontal x-axis and the vertical y-axis. They marked down homework that did not have that. (I was not taught like that, and the website does not label axes that way, sorry).
Because 2 points determine a line, you just need to know 2 points on a line to graph the line.
 
To get each point you chose a value for {{{x}}} and calculate the corresponding {{{y}}}.
I suggest choosing points that are easy to calculate, easy to plot, and give you a nice looking graph.
Choosing {{{x=0}}} often works well.
 
For {{{y=3x+0}}} (which I would write as simply {{{y=3x}}}),
I could choose {{{x=0}}} to get
{{{y=3*0}}} --> {{{y=0}}}, to get the point (0,0),
and {{{x=1}}} to get {{{y=3-1}}} --> {{{y=3}}}  to get the point (1,3).
Plotting those 2 points (circled) would get me a half done graph looking like this:
{{{drawing(300,300,-5,5.5,-5,5.5,
grid(1),
blue(circle(0,0,0.2)),blue(circle(1,3,0.2))
)}}}
The next step is to connect the two points with a line to get.
{{{drawing(300,300,-5,5.5,-5,5.5,
grid(1),
blue(circle(0,0,0.2)),blue(circle(1,3,0.2)),
blue(line(-2,-6,2,6))
)}}}
Sometimes, if your two points are too close together, it is hard to draw the accurately. In this case, it was not too bad, but it may have been better to use {{{x=-1}}} and {{{x=1}}} for the points.
The teacher may  specify details for the graph, such as how to draw/indicate the points. (Circling may not be the teacher's style).
When graphing is required, the teacher may expect a table of values for x and y.
The teacher may expect a certain style of table, and  may recommend a certain number of points.
My table would look like this:
{{{drawing(150,100,-2,4,-2,2,
line(-2,0,4,0),line(0,-2,0,2),line(2,-2,2,2),
locate(-1.2,-0.6,y),locate(-1.2,1.2,x),
locate(0.8,-0.6,0),locate(0.8,1.2,0),
locate(2.8,-0.6,3),locate(2.8,1.2,1)
)}}} with a row for the x's and a row for the y's,
but vertical tables, with columns for the x's and y's may be more popular.
 
For {{{y=1x+2}}} (which I would write as simply {{{y=x+2}}}),
I will use {{{x=-4}}} to get {{{y=-4+2}}} --> {{{y=-2}}},
for point (-4,-2), and
{{{x=3}}} to get {{{y=3+2}}} --> {{{y=5}}},
for point (3,5).
My graph looks like this
{{{drawing(300,300,-5,5.5,-5,5.5,
grid(1),
green(circle(-4,-2,0.2)),green(circle(3,5,0.2)),
green(line(-5,-3,4,6))
)}}}
 
TO SOLVE SYSTEMS OF EQUATIONS:
There are many ways.
BY GRAPHING:
One way to solve the system {{{system(y=1x+2,y=3x+0)}}}
would be to graph both lines and see where they cross.
You would get
{{{drawing(300,300,-5,5.5,-5,5.5,
grid(1),
blue(line(-2,-6,2,6)),green(line(-5,-3,4,6))
)}}} and figure out that the solution seems to be the point (1,3),
with {{{x=1}}} and {{{y=3}}}.
Because thing are not always the way they look, you are required to verify that in both equations, setting {{{x=1}}} results in {{{y=3}}}.
BY SUBSTITUTION:
If one of the equations in a sytem has "y= .." or "x= ...", you can just substitute (in the other equation) the variable for the expression that's equal to it.
The system {{{system(y=x+2,y=3x)}}} is very easy to solve by susbstitution.
(I would say it's the best method for this sytem).
You could substitute {{{3x}}} for {{{y}}} in {{{y=x+2}}}
to get {{{3x=x+2}}}.
Subtracting {{{x}}} from both sides of the equal sign, we get an equivalent equation, that simplifies further
{{{3x=x+2}}} --> {{{3x-x=x+2-x}}} --> {{{2x=2}}}
Dividing both sides of the equal sign by {{{2}}} we get an equivalent equation that simplifies into the solution for {{{x}}}.
{{{2x=2}}} --> {{{2x/2=2/2}}} --> {{{highlight(x=1)}}}
Then, substituting that value for {{{x}}} in {{{y=3x}}}, we find the value for {{{y}}}
{{{y=3*1}}} --> {{{highlight(y=3)}}}
BY ELIMINATION and USING MATRICES are other ways to solve systems of equations, but that is more than you want to know, and more than you need right now.