Asystem of inequalities is given. $$ \left\{\begin{array}{l}y \leq x+4 \ y3 x-1\end{array} ight. $$ Which graph shows the solution set for the system of inequ

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Asystem of inequalities is given. $$ \left\{\begin{array}{l}y \leq x+4 \ y>3 x-1\end{array}ight. $$ Which graph shows the solution set for the system of inequ

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1. Write down the inequalities and identify the boundary lines:
   - $$ y \leq x+4 $$ has the boundary line $$ y=x+4 $$ (solid line, because the inequality includes equality).
   - $$ y > 3x-1 $$ has the boundary line $$ y=3x-1 $$ (dashed line, because the inequality is strict).

2. Decide which side of each line is included:
   - For $$ y \leq x+4 $$: choose a test point (for example, $$(0,0)$$). Since $$ 0 \leq 0+4 $$ is true, the region containing $$(0,0)$$ is part of the solution (i.e. below or on $$ y=x+4 $$).
   - For $$ y > 3x-1 $$: test the point $$(0,0)$$ again. Since $$ 0 > 3(0)-1 $$ simplifies to $$ 0 > -1 $$ (true), the region containing $$(0,0)$$ is included (i.e. above the line $$ y=3x-1 $$).

3. Graph the two boundary lines:
   - Draw $$ y=x+4 $$ as a solid line.
   - Draw $$ y=3x-1 $$ as a dashed line.

4. Shade the appropriate regions:
   - Shade the area below or on the line $$ y=x+4 $$.
   - Shade the area above the line $$ y=3x-1 $$.

5. The solution set is the intersection of these shaded regions, consisting of all points that satisfy both $$ y \leq x+4 $$ and $$ y > 3x-1 $$.

Thus, the correct graph is the one showing the region that lies between (and includes) the solid line $$ y=x+4 $$ (shaded below it) and the dashed line $$ y=3x-1 $$ (shaded above it).
The solution set is the area below the solid line $$ y = x + 4 $$ and above the dashed line $$ y = 3x - 1 $$.

Asystem of inequalities is given. \( \left\{\begin{array}{l}y \leq x+4 \\ y>3 x-1\end{array}\right. \) Which graph shows the solution set for the system of inequ

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