Question

part 3. solve & graph each system of inequalities.

1. \\(\

$$\begin{cases} 2x + 3y \\leq 6 \\\\ x - y > -2 \\end{cases}$$

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1. \\(\

$$\begin{cases} y \\geq -3 \\\\ \\frac{1}{2}x - y \\geq -1 \\end{cases}$$

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1. \\(\

$$\begin{cases} 4x - y \\geq -1 \\\\ x - 2y < 6 \\end{cases}$$

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1. \\(\

$$\begin{cases} x \\leq 2 \\\\ -2x + y > -4 \\end{cases}$$

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Explanation:

Step1: Solve \(2x + 3y \leq 6\) for \(y\)

Subtract \(2x\) from both sides: \(3y \leq -2x + 6\). Divide by 3: \(y \leq -\frac{2}{3}x + 2\). This is a line with slope \(-\frac{2}{3}\), y - intercept 2, solid line (since \(\leq\)), shade below.

Step2: Solve \(x - y > -2\) for \(y\)

Subtract \(x\): \(-y > -x - 2\). Multiply by -1 (reverse inequality): \(y < x + 2\). Line with slope 1, y - intercept 2, dashed line (since \(>\)), shade below.

Step3: Graph the lines

• For \(y \leq -\frac{2}{3}x + 2\): Plot (0,2) and (3,0) (since \(y = -\frac{2}{3}x + 2\), when \(x = 0\), \(y = 2\); \(x = 3\), \(y = 0\)), draw solid line, shade below.
• For \(y < x + 2\): Plot (0,2) and (-2,0) (when \(x = 0\), \(y = 2\); \(x = -2\), \(y = 0\)), draw dashed line, shade below.

Step4: Find the intersection

The solution is the region where both shadings overlap.

Answer:

The solution region is where \(y \leq -\frac{2}{3}x + 2\) (solid line, below) and \(y < x + 2\) (dashed line, below) overlap. To graph, plot the lines as described and shade the overlapping area.