Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y \geq \frac{1}{2}x + 3$
$y < -x + 6$

Explanation:

Step1: Analyze the first inequality

The first inequality is \( y \geq \frac{1}{2}x + 3 \). This is a linear inequality with a slope of \( \frac{1}{2} \) and a y - intercept of 3. The line \( y=\frac{1}{2}x + 3 \) should be drawn as a solid line (because of the "greater than or equal to" symbol) and the region above the line should be shaded.

Step2: Analyze the second inequality

The second inequality is \( y < -x + 6 \). This is a linear inequality with a slope of - 1 and a y - intercept of 6. The line \( y=-x + 6 \) should be drawn as a dashed line (because of the "less than" symbol) and the region below the line should be shaded.

Step3: Find the intersection region

The solution set of the system of inequalities is the region that is shaded for both inequalities. To find a point in the solution set, we can pick a point that satisfies both inequalities. Let's test the point \( (2,4) \):

• For \( y\geq\frac{1}{2}x + 3 \): Substitute \( x = 2 \) and \( y = 4 \). \( \frac{1}{2}(2)+3=1 + 3=4 \), and \( 4\geq4 \) which is true.
• For \( y < -x + 6 \): Substitute \( x = 2 \) and \( y = 4 \). \( -2 + 6=4 \), and \( 4<4 \) is false. Let's try \( (0,4) \):
• For \( y\geq\frac{1}{2}x + 3 \): \( \frac{1}{2}(0)+3 = 3 \), and \( 4\geq3 \) is true.
• For \( y < -x + 6 \): \( -0+6 = 6 \), and \( 4<6 \) is true. Another point: Let's solve the system of equations \( y=\frac{1}{2}x + 3 \) and \( y=-x + 6 \) to find the intersection of the two lines. Set \( \frac{1}{2}x+3=-x + 6 \). Add \( x \) to both sides: \( \frac{1}{2}x+x+3=-x+x + 6\Rightarrow\frac{3}{2}x+3 = 6 \). Subtract 3 from both sides: \( \frac{3}{2}x=3 \). Multiply both sides by \( \frac{2}{3} \): \( x = 2 \). Then \( y=-2 + 6=4 \). The intersection point of the two lines is \( (2,4) \), but since the second inequality is strict (\( y < -x + 6 \)), the point \( (2,4) \) is not included. Let's take \( (0,3) \):
• For \( y\geq\frac{1}{2}x + 3 \): \( \frac{1}{2}(0)+3=3 \), \( 3\geq3 \) is true.
• For \( y < -x + 6 \): \( -0 + 6=6 \), \( 3<6 \) is true. Or \( (2,3) \):
• For \( y\geq\frac{1}{2}x + 3 \): \( \frac{1}{2}(2)+3=1 + 3=4 \), \( 3\geq4 \) is false. Let's take \( (4,4) \):
• For \( y\geq\frac{1}{2}x + 3 \): \( \frac{1}{2}(4)+3=2 + 3=5 \), \( 4\geq5 \) is false. Let's take \( (0,5) \):
• For \( y\geq\frac{1}{2}x + 3 \): \( 0 + 3=3 \), \( 5\geq3 \) is true.
• For \( y < -x + 6 \): \( 0+6 = 6 \), \( 5<6 \) is true.

Answer:

A point in the solution set is \( (0,4) \) (or other valid points like \( (0,5) \), \( (2,3) \) is invalid, \( (0,3) \) is valid etc. Here we choose \( (0,4) \) as an example).