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Check the equations to graph the lines and use test points to check the shading.
Error: Incorrect graphing and shading.
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≥-2 & (I) y>2 & (II) 12x+y≥0 & (III) In order to find the error the student has made, we will graph each inequality. Let's start with 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. cc Inequality & Boundary Line [0.5em] x ≥ - 2 & 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.
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. cc Inequality & Boundary Line [0.5em] y > 2 & 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.
x= 1, y= 0
Identity Property of Multiplication
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.