W. Wraph the system $$ y3 x+1 $$ and $$ y

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W. Wraph the system $$ y>3 x+1 $$ and $$ y<\frac{1}{2} x-4 $$

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1. We are given the system of inequalities $$ y>3x+1 $$ and $$ y<\frac{1}{2}x-4. $$ 2. The inequality $$ y > 3x+1 $$ represents the region above the line $$ y=3x+1, $$ while $$ y < \frac{1}{2}x-4 $$ represents the region below the line $$ y=\frac{1}{2}x-4. $$ (Since both inequalities are strict, the boundary lines are not included in the solution.) 3. To have any point $$(x,y)$$ satisfying both inequalities simultaneously, the two boundaries must satisfy $$ 3x+1<\frac{1}{2}x-4. $$ 4. Solve the inequality $$ 3x+1<\frac{1}{2}x-4 $$: - Multiply both sides by 2 to eliminate the fraction: $$ 2(3x+1) < 2\left(\frac{1}{2}x-4\right) \quad \Rightarrow \quad 6x+2 < x-8. $$ - Subtract $$ x $$ from both sides: $$ 6x-x+2 < -8 \quad \Rightarrow \quad 5x+2 < -8. $$ - Subtract 2 from both sides: $$ 5x < -10. $$ - Divide both sides by 5: $$ x < -2. $$ 5. For any $$ x < -2 $$, the corresponding $$ y $$ must satisfy $$ 3x+1 < y < \frac{1}{2}x-4. $$ Note that when $$ x=-2 $$, both lines give $$ y=-5 $$ (since $$ 3(-2)+1=-5 $$ and $$\frac{1}{2}(-2)-4=-5$$); however, $$ x=-2 $$ is not in the solution set because the inequality $$ 3x+1 < \frac{1}{2}x-4 $$ must hold. 6. Therefore, the solution region of the system is $$ \{(x,y) \mid x < -2 \text{ and } 3x+1 < y < \frac{1}{2}x-4\}. $$ 7. On a graph, you would: - Draw the line $$ y=3x+1 $$ as a dashed line and shade the region above it. - Draw the line $$ y=\frac{1}{2}x-4 $$ as a dashed line and shade the region below it. - The intersection of these two regions for $$ x < -2 $$ is the solution of the system. The solution to the system is all points $$(x, y)$$ where $$ x < -2 $$ and $$ 3x + 1 < y < \frac{1}{2}x - 4 $$.

W. Wraph the system \( y>3 x+1 \) and \( y<\frac{1}{2} x-4 \)

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