You could try and find matching #x#s and #y#s by trial and error, but it is much more convenient to rearrange the equation so that one variable, say #y#, is isolated on the left side. This way, we can pick any value for #x# we like and compute the matching #y#.
You can add, subtract, multiply and divide algebraic equations without changing their solution if you follow two simple rules: First, always do the same thing both on the left and the right side of the equation. Second, never divide by zero.
In this case, let's try and isolate the #y#. Basically, we want to remove everything that's not #y# from the left side. We start by subtracting the #2x#. Remember that we need to do it on both sides, so we get:
#2x+3y color(red)(-2x) = -3 color(red)(-2x)#
#cancel(2x)+3y cancel(-2x) = -3 - 2x#
#3y = -3 -2x#
By convention, we write the terms with #x# first:
#3y = -2x - 3#
To remove the #3# in front of the #y#, we could try and subtract #2y#, but since we need to do it on both sides, we would reintroduce the #y# to the right side, and we want it on the left side only. So instead, we divide both sides by #3#. Remember if you divide a sum, you need to divide all terms. Thus, we get
#(3y)/3 = (-2x - 3)/3#
#y = (-2x)/3 - 3/3#
#y = -2/3x - 1#
Now that we have our #y# isolated, we can just pick a few #x# values. For #x=0#, we get
#y = -2/3*0-1 = -1#
Thus, #(0,-1)# is a solution. You can test it with the original equation if you don't trust the math yet.
For #x=3#, we get
#y = -2/3*3 -1 = -2 - 1 = -3#
Thus, #(3,-3)# is another solution for our equation.
Now, what we have here is a linear equation, which means that all solutions lie on an infinitely long straight line. We could continue calculating more points, but it is much easier to mark the two points we have, and then use a ruler to draw a straight line that passes through these points.
All solutions for the linear equations lie on this line, and conversely, every point on the line is a solution to the equation.
graph{-2/3x - 1 [-10, 10, -5, 5]}
