Let's graph the equation
[tex]12x+3y=-3[/tex]
Notice that this equation corresponds to a line.
1. First, let's clear y, so we can get the equation of the line in the slope-intercept form:
[tex]\begin{gathered} 12x+3y=-3 \\ \rightarrow3y=-12x-3 \\ \rightarrow y=-4x-1 \end{gathered}[/tex]
This way, he equation of the line in the slope-intercept form is
[tex]y=-4x-1[/tex]
2. Now, we have to expose two points contained in the line. To do so, we give x any value we want and we calculate the corresponding y value.
Let's do this with x = 0 and x = 1
[tex]\begin{gathered} y=-4x-1 \\ \\ x=0\Rightarrow y=-4(0)-1\rightarrow y=-1 \\ \text{This way, the point} \\ (0,-1) \\ \text{Belongs to the line} \\ \\ x=1\Rightarrow y=-4(1)-1\rightarrow y=-5 \\ \text{This way, the point } \\ (1,-5) \\ \text{belongs to the line} \end{gathered}[/tex]
3. Put those points in the cartesian plane:
4. Draw a straight line that passes through both points. This will be the graph of the line
