Find the graph of the solution set of the following system of linear inequalities. Drag the points into the correct positions to determine the lines and then drag the shading points to select the correct shading. \[ \begin{array}{r}-x+2 y>-2 \\ 2 x+y>-3\end{array} \]

1. Start with the two inequalities: \[ -x+2y > -2 \quad \text{and} \quad 2x+y > -3. \] 2. Rewrite the first inequality by isolating \( y \). Add \( x \) to both sides: \[ 2y > x - 2. \] Then divide by 2: \[ y > \frac{1}{2}x - 1. \] The line corresponding to the equality is \[ y = \frac{1}{2}x - 1. \] Since the inequality is strict (\( > \)), the boundary line is dashed and the solution is the set of points \emph{above} this line. 3. Rewrite the second inequality by isolating \( y \). Subtract \( 2x \) from both sides: \[ y > -3 - 2x. \] The corresponding boundary line is \[ y = -3 - 2x, \] again drawn as a dashed line since the inequality is strict. The solution region is the set of points \emph{above} this line. 4. Find the intersection of the two boundary lines by solving: \[ \frac{1}{2}x - 1 = -3 - 2x. \] Multiply both sides by 2 to eliminate the fraction: \[ x - 2 = -6 - 4x. \] Add \( 4x \) to both sides: \[ 5x - 2 = -6. \] Now add 2 to both sides: \[ 5x = -4, \] so \[ x = -\frac{4}{5}. \] Substitute \( x = -\frac{4}{5} \) back into one of the line equations (using \( y = \frac{1}{2}x - 1 \)): \[ y = \frac{1}{2}\left(-\frac{4}{5}\right) - 1 = -\frac{2}{5} - 1 = -\frac{7}{5}. \] Thus, the lines intersect at \[ \left(-\frac{4}{5}, -\frac{7}{5}\right). \] 5. The final graph is constructed as follows: - Draw the line \( y = \frac{1}{2}x - 1 \) as a dashed line. - Draw the line \( y = -3 - 2x \) as a dashed line. - Shade the region that is above both of these lines. This is the intersection of the half-planes: \[ y > \frac{1}{2}x - 1 \quad \text{and} \quad y > -3 - 2x. \] Any point in the shaded region satisfies both \( -x+2y > -2 \) and \( 2x+y > -3 \).