1) Graph the equation: \( 3 y+12=-2 x \). Plot a \( \frac{10}{p} \) minimum of 3 points.

Graph the solution set of the following system of inequalities. \[ \begin{array}{l}x+y \leq 4 \\ y \geq x^{2}-3\end{array} \]

\( \left. \begin{array} { r l r l } { 5 x + 3 x + 2 x } & { = } & { 7 y + 5 y - 9 y } & { = } \\ { 2 a + 4 a + 6 a } & { = } & { 2 a + 3 a - 5 a } & { = } \\ { 3 x + 5 x - 6 x } & { = } & { 10 x - 5 x + 12 x } & { = } \\ { - 5 y - 8 y } & { = } & { - 11 b - 7 b } & { = } \end{array} \right. \)

Resolver \( \left\{\begin{aligned} x+2 y+z & =4 \\ 3 x+z & =2 \\ x-y+z & =7\end{aligned}\right. \)

Simplify. \[ \frac{x^{2} y^{6}}{x^{3} y^{6}} \]

Simplify the expression \( 9 j-j+8 j \)