The given system of inequalities is:
\[
\begin{array}{l}
y<-\frac{2}{3} x+5 \\
y \geq \frac{4}{3} x-1
\end{array}
\]
Find the boundary lines
The boundary lines are \(y = -\frac{2}{3}x + 5\) and \(y = \frac{4}{3}x - 1\). The first inequality is a strict inequality, so the boundary line is dashed. The second inequality has "greater than or equal to", so the boundary line is solid.
Graph the boundary lines
The first line, \(y = -\frac{2}{3}x + 5\), has a \(y\)-intercept of 5 and a slope of \(-\frac{2}{3}\). The second line, \(y = \frac{4}{3}x - 1\), has a \(y\)-intercept of -1 and a slope of \(\frac{4}{3}\).
Shade the regions
For the first inequality, \(y<-\frac{2}{3} x+5\), shade the region below the line \(y = -\frac{2}{3}x + 5\).
For the second inequality, \(y \geq \frac{4}{3} x-1\), shade the region above the line \(y = \frac{4}{3}x - 1\).
Find the point of intersection.
To find the intersection point, set the equations equal to each other:
\[ -\frac{2}{3}x + 5 = \frac{4}{3}x - 1 \]
\[ 6 = \frac{6}{3}x \]
\[ 6 = 2x \]
\[ x = 3 \]
Substitute \(x=3\) into either equation. Using the second one:
\[ y = \frac{4}{3}(3) - 1 = 4 - 1 = 3 \]
So the intersection point is \((3,3)\).
\(\boxed{\text{The solution is the region where the shaded areas for both inequalities overlap.}}\)
\(\boxed{\text{The solution is the region where the shaded areas for both inequalities overlap.}}\)
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