You are looking for the solution to the given equations by graphing them.
First you need to create a table of values. Using these values the given lines can be graphed. From the graph, the solution can be determined.
First equation is:

Using different values of x, we can find corresponding y values which can be used to graph the line.
For x=0,

For x=2,

For x=4,

For x=6,
Second equation is:

Using different values of x, we can find corresponding y values which can be used to graph the line.
For x = 0,

For x = 2,

For x = 3,

For x = 1,

We can see that the value of y is same for both the lines at x = 2. So
(2,4) is the solution of the equations. This can also be observed from the graph by plotting these lines as shown in the image below