We are given a graph and asked to find and correct the mistake. Let's start by looking at the given graph.

Let's take a look at the different inequalities individually.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x \geq - 2 y > 2 \frac{1}{2} x + y \geq 0 (I) (II) (III)

In order to find the error the student has made, we will graph each inequality. Let's start with Inequality (I).

Inequality (I)

To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign.

\underline{Inequality} x \geq - 2 \underline{Boundary Line} x = - 2

Notice that the equation for this boundary line only has one variable. This equation tells us that each and every point that lies on the line will have an

x -

coordinate equal to

- 2 .

This gives us a vertical line. Also, the inequality is non-strict, so the points on the boundary line are included in the solution set. We show this by drawing a solid line.

The inequality $x \geq - 2$ describes all values of $x$ that are greater than or equal to $- 2 .$ This means that every possible $(x, y)$ coordinate pair with an $x -$ value that is greater than or equal to $- 2$ needs to be included in the shading.

Inequality (II)

To graph the inequality, we will first draw the boundary line. As before, we will write the equation of the boundary line by replacing the inequality symbol from the inequality with an equals sign.

\underline{Inequality} y > 2 \underline{Boundary Line} y = 2

Notice that the equation for this boundary line only has one variable. This equation tells us that each and every point that lies on the line will have a

y -

coordinate equal to

2 .

This gives us a horizontal line. Also, the inequality is strict, so the points on the boundary line are not included in the solution set. We show this by drawing a dashed line.

The inequality $y > 2$ describes all values of $y$ that are greater than $2 .$ This means that every possible $(x, y)$ coordinate pair with a $y -$ value that is greater than $2$ needs to be included in the shading.

As we can see, we have shaded the opposite region to the one the student has shaded in their solution. It means the student shaded the solution set of this inequality incorrectly.

Inequality (III)

In order to graph the Inequality (III), we will first isolate the

y -

variable to rewrite the inequality in slope-intercept form.

\frac{1}{2} x + y \geq 0

SubIneq

$LHS - \frac{1}{2} x \geq RHS - \frac{1}{2} x$

y \geq - \frac{1}{2} x

Let's write the equation of the boundary line of this inequality, as we have done for the two previous inequalities.

\underline{Inequality} y \geq - \frac{1}{2} x \underline{Boundary Line} y = - \frac{1}{2} x

Now that the boundary line is in slope-intercept form, we can determine its slope

m

and

y -

intercept

b .

Slope - Intercept Form : Boundary Line : y = m x + b y = - \frac{1}{2} x + 0

We can use the slope and the

y -

intercept to draw the boundary line. Note that the boundary line will be solid, because the inequality is not strict.

Note that this graph does not match the student's, which means the student has drawn the graph of this inequality's boundary line incorrectly. By testing a point that is not on the boundary line we can determine which region we should shade. Let's test the point

(1, 0) .

If it satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the opposite region.

y \geq - \frac{1}{2} x

SubstituteII

$x = 1$ , $y = 0$

0 \geq ? - \frac{1}{2} (1)

IdPropMult

Identity Property of Multiplication

0 \geq - \frac{1}{2} ✓

Since the test point satisfies the inequality, we will shade the region that contains the point.

Conclusion

Let's draw all the graphs and shaded regions in the same coordinate plane.

The correct solution set is the intersection of all the shaded regions.

Exercise

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