To graph lineal inequalities:
1. Solve in the inequality the variable y:
![3x+3y>12](https://img.qammunity.org/2023/formulas/mathematics/college/8tmlylxzw2yoy1l6g1lqpa40jb7wfffgjr.png)
- Substract 3x in both sides of the inequation:
![\begin{gathered} 3x-3x+3y>-3x+12 \\ 3x>-3x+12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lzqnxmm64gxj0jui3xaqfh4dergov5dr8b.png)
- Divide both sides of the inequation into 3:
![\begin{gathered} (3)/(3)x>((-3x+12))/(3) \\ \\ x>-(3)/(3)x+(12)/(3) \\ \\ x>-x+4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ttipy6a2875ipw7e41ercxmkl3gmxsr1uy.png)
2. Find two points in the line that limits the inequality
The line is y=-x+4
Find y when x=0
![\begin{gathered} y=-0+4 \\ y=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cj320ayygmtvyus0wk6cy59sjoxyzj5jae.png)
Find x when y=0:
![\begin{gathered} 0=-x+4 \\ 0+x=-x+x+4 \\ x=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/j56jtracvnnr6mzw077rw18j9dxzpb187q.png)
You have the next two points in the line: (0,4) and (4,0)
3. Put the points in the plane and draw the line that passes trought those points:
When the inequality is < or > you draw a dotted line (the line is not part of the solution)
When the inequality is ≥ or ≤ you draw a line (it is part of the solution)
4. Shadow the area that correspond to the inequality:
When the inequality is < or ≤ you shadow the area under the line
When the inequality is > or ≥ you shadow the area over the line.
In this case as the inequality symbol is > you draw a dotted line and shadow the area over the line. You get the next graph: