Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
y ≤ x + 6
y ≥ -\frac{1}{2}x - 3
graph of coordinate plane with grid, x-axis from -10 to 10, y-axis from -10 to 10
answer attempt 1 out of 2
point: (\square, \square) submit answer

Explanation:

Step1: Analyze the first inequality \( y \leq x + 6 \)

The boundary line is \( y = x + 6 \), which has a slope of 1 and a y - intercept of 6. Since the inequality is \( \leq \), we draw a solid line and shade below the line.

Step2: Analyze the second inequality \( y \geq -\frac{1}{2}x - 3 \)

The boundary line is \( y = -\frac{1}{2}x - 3 \), with a slope of \( -\frac{1}{2} \) and a y - intercept of - 3. Since the inequality is \( \geq \), we draw a solid line and shade above the line.

Step3: Find the intersection region

The solution set is the region where the two shaded regions overlap. To find a point in the solution set, we can pick a point that satisfies both inequalities. Let's try \( x = 0 \).
For the first inequality: When \( x = 0 \), \( y \leq 0 + 6=6 \).
For the second inequality: When \( x = 0 \), \( y \geq-\frac{1}{2}(0)-3=-3 \).
Let's choose \( y = 0 \). Check if \( (0,0) \) satisfies both inequalities:

• For \( y \leq x + 6 \): \( 0\leq0 + 6 \), which is true.
• For \( y \geq-\frac{1}{2}x - 3 \): \( 0\geq-\frac{1}{2}(0)-3=-3 \), which is true.

Answer:

\((0,0)\) (Other valid points like \((2,2)\) (check: \(2\leq2 + 6 = 8\) and \(2\geq-\frac{1}{2}(2)-3=-1 - 3=-4\)) are also correct)