Question

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$$\begin{cases} y < 2x + 2 \\\\ y \\geq -x + 1 \\end{cases}$$

Explanation:

Step1: Analyze the first inequality \( y < 2x + 2 \)

This is a linear inequality. The boundary line is \( y = 2x + 2 \), which has a slope of \( 2 \) and a y - intercept of \( 2 \). Since the inequality is \( y<2x + 2 \), the line should be dashed (because the inequality is strict, \( y\) is not equal to \( 2x + 2 \)) and we shade the region below the line.

Step2: Analyze the second inequality \( y\geq -x + 1 \)

The boundary line is \( y=-x + 1 \), with a slope of \( - 1 \) and a y - intercept of \( 1 \). Since the inequality is \( y\geq -x + 1 \), the line should be solid (because \( y\) can be equal to \( -x + 1 \)) and we shade the region above the line.

Step3: Find the solution region

The solution to the system of inequalities is the region that is shaded by both inequalities. So we need to find the area that is below the dashed line \( y = 2x+2 \) and above the solid line \( y=-x + 1 \). To graph this, first plot the line \( y = 2x + 2 \) as a dashed line and shade below it. Then plot the line \( y=-x + 1 \) as a solid line and shade above it. The overlapping region of these two shadings is the solution to the system.

(If the question was to graph the system, this is the process. If it was to find a test point, for example, let's take a point like \( (0,1.5) \). For \( y<2x + 2 \): \( 1.5<2(0)+2=2 \), which is true. For \( y\geq -x + 1 \): \( 1.5\geq - 0+1 = 1 \), which is also true. So \( (0,1.5) \) is in the solution region.)

Answer:

The solution to the system \(

$$\begin{cases}y < 2x + 2\\y\geq -x + 1\end{cases}$$

\) is the region below the dashed line \(y = 2x+2\) and above the solid line \(y=-x + 1\). (If a specific answer like a test point or graph description was needed, adjust accordingly. For example, if testing the point \((0,1.5)\), it satisfies both inequalities and is in the solution region.)