We will begin by isolating y on the left-hand side.
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& &Inequality [0.5em]
&I: &y≥ -4x-16 ⇔ y≥ -4x-16 [0.5em]
&II: & 4y≤ 26-x ⇔ y≤ -1/4x+6.5 [0.5em]
&III: &3y+6x≤30 ⇔ y ≤ -2x+10 [0.5em]
&IV: & 4y-2x≥ -10 ⇔ y ≥1/2x-2.5
Next, the boundary lines of the inequalities should be determined to graph them. To do that we will replace the inequality symbols with the equals sign.
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& &Inequality &Boundary Line [0.5em]
&I: & y ≥ -4x-16 &y = -4x-16 [0.5em]
&II: & y ≤ -1/4x+6.5 &y = -1/4x+6.5 [0.5em]
&III: & y ≤ -2x+10 & y = -2x+10 [0.5em]
&IV: & y ≥ 1/2x-2.5 & y = 1/2x-2.5
Notice that the boundary lines are in slope-intercept form which means that we can immediately determine their slope and y-intercept.
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& &Boundary Line [0.5em]
&I: &y= -4x -16 [0.5em]
&II: &y= -1/4x+ 6.5 [0.5em]
&III: &y= -2x+ 10 [0.5em]
&IV: &y= 1/2x -2.5
Now that we know the slope and y-intercept of the inequalities, we can begin by drawing the first inequality. Let's plot its y-intercept and use its slope to find a second point. Thus, we can draw a line.
Now that we have two points, we can draw the boundary line that passes through these points. Because the inequality is non-strict, the line will be solid.
Next, we will decide which region we should shade. In this case, the inequality states that the points with y-coordinates greater than or equal to -4x-16 where x is the x-coordinate are included in the solution set. Thus, we will shade above the boundary line.
The first inequality has been graphed. The other inequalities are in the same form. Thus, they can be graphed by proceeding in the same way. Let's start!
The overlapping section of the graph above is the solution set to the system.
Finally, we have a quadrilateral. Looking at the graph, we can determine the vertices of it. We should also label the vertices for the upcoming operations.
When we check the slopes of Boundary Line III and IV, we see that they are negative reciprocals of each other.
-2* 1/2=-1
Therefore, they are perpendicular to each other at the point C. To use this angle effectively, let's draw a line segment that passes through B and D. With this, we will have two triangles.
To determine the height of △ ABD, let's draw a line that passes through the point B(2,6) and perpendicular to Boundary Line I. The slope of perpendicular lines are negative reciprocal of each other.
&Boundary Line I &&Height
&y= -4x -16 &&y= 1/4x+ b
Next, we will find it y-intercept by substituting the point (2,6) into the equation.
Now that we know the slope and y-intercept of the line, we can graph it.
To find the length of the height, we should determine where the height intersect Boundary Line I. We will solve together the equation of Boundary Line I and the equation of height to find the point of intersection.
