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| 10 Theory slides |
| 15 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Each of the following graphs represents the solution set of a certain inequality.
Pair each graph with its corresponding inequality.
As seen earlier, when two or more inequalities are graphed on the same coordinate plane, their solution sets may overlap. In these cases, the set of all inequalities being solved simultaneously forms a system of inequalities.
Notice that there is a region where the solution sets of the inequalities overlap. All the points in this region satisfy both inequalities simultaneously. Therefore, the overlapping region is the solution set of the system. In the next graph, only the common region is shaded.
Since the boundary lines in their entirety are not part of the solution set, they can be cropped to show only the edges of the overlapping region, or the exceeding parts can be drawn with lower opacity.
Jordan, feeling jolly, is thinking about giving gifts to her teammates — there are 30 players. Shopping at a stationery store, she decides it is best to buy some fancy ballpoint and fountain pens. She wants to spend less than $240 and is now unsure whether to give all or only some of her teammates a gift.
Let x and y be the number of ballpoint and fountain pens Jordan will buy, respectively.
With the first option, 19 teammates will receive a gift, while 20 teammates will get a gift with the second option. Therefore, if Jordan buys the maximum number of fountain pens she can, the maximum number of teammates that could get a gift is 20. Of those 20 teammates, only one would get a ballpoint pen.
Minimum: 101
Finally, Jordan was able to save enough money to buy a used car! She now plans to take a mini-road trip to visit a relative across-state. First, she will pick up her sister Ramsha, who lives in a different city. The travel distance depends on the path she chooses, but the entire route is no less than 990 kilometers. Jordan would like to drive for a maximum of 8 hours.
Jordan plans to drive at 70 kilometers per hour from her house to Ramsha's, and from there, she plans to increase the speed to 110 kilometers per hour until reaching her relative's house.
For each given system of linear inequalities, select the region corresponding to its solution set, if any.
To determine which of the options is the solution set, let's graph the system of inequalities. y ≤ 2x-2 & (I) y > - x -4 & (II) Note that both inequalities are written in slope-intercept form. Let's plot the first inequality. To do this, we first identify the corresponding boundary line. ccc Inequality & & Boundary Line y ≤ 2x-2 & & y = 2x-2 To determine which region to shade, we will test (0,0) in the inequality.
We got a false statement. Therefore, the region to be shaded is the one not containing the origin. Remember, since the inequality is non-strict we will draw the boundary line solid. We are now ready to plot our first inequality.
Next, let's plot the second inequality. Again, we start by writing its boundary line. ccc Inequality & & Boundary Line y > - x -4 & & y = - x -4 As before, to determine which region to shade let's test (0,0) in the inequality.
We got a true statement, which implies that the we have to shade the region containing the origin. Since the second inequality is strict, the boundary line is dashed. Let's graph this inequality on the same coordinate plane we graphed the first inequality.
Finally, let's graph only the solution set of the system of inequalities — the overlapping region.
Comparing the solution set we got with the four given ones, we conclude that the correct choice is option C.
To determine which of the regions describes the solution set, let's graph the system of inequalities. y > 2x + 3 & (I) y > 2x - 1 & (II) The first thing we note is that both inequalities are already written in slope-intercept form. Let's continue by writing the equations of the boundary lines. ccc Inequality & & Boundary Line [0.5em] y > 2x + 3 & & y = 2x + 3 [0.25em] y > 2x - 1 & & y = 2x - 1 A second thing to note is that both boundary lines have the same slope, which means that the lines are parallel. Also, both inequalities are strict, so we will draw both boundary lines dashed.
Let's focus for a moment on the first inequality. To determine which region to shade, we will test (0,0) in the first inequality.
We got a false statement. Therefore, we have to shade the region not containing the origin — the region above the line y=2x+3.
In a similar fashion, let's test (0,0) in the second inequality to determine which region to shade.
We got a true statement. Therefore, we have to shade the region containing the origin — the region above the line y=2x-1.
Finally, let's graph only the solution set — the overlapping region.
Consequently, option C describes the solution set of the system of inequalities. Option C The region above the liney=2x+3 without including it.
To determine which of the regions describes the solution set, let's graph the system of inequalities. y < 12x + 4 [0.1cm] y ≤ 12x -2 The first thing we note is that both inequalities are already written in slope-intercept form. Let's continue by writing the equations of the boundary lines. ccc Inequality & & Boundary Line [0.5em] y < 12x + 4 & & y = 12x + 4 [0.15cm] y ≤ 12x -2 & & y = 12x -2 A second thing to note is that both boundary lines have the same slope, which means that the lines are parallel. Since the first inequality is strict, we will draw the first boundary line dashed. In contrast, the second inequality is non-strict and therefore, we will draw the second boundary line solid.
Let's focus for a moment on the first inequality. To determine which region to shade, we will test (0,0) in the first inequality.
We got a true statement. Therefore, we have to shade the region containing the origin — the region below the line y= 12x+4.
In a similar fashion, let's test (0,0) in the second inequality to determine which region to shade.
We got a false statement. Therefore, we have to shade the region not containing the origin — the region below the line y= 12x-2.
Finally, let's graph only the solution set — the overlapping region.
Consequently, option B describes the solution set of the system of inequalities. Option B The region below the liney= 12x-2, including it.
To determine which of the regions describes the solution set, let's graph the system of inequalities. y ≥ 13x - 4 [0.1cm] y ≤ 13x + 3 The first thing we note is that both inequalities are already written in slope-intercept form. Let's continue by writing the equations of the boundary lines. ccc Inequality & & Boundary Line [0.5em] y ≥ 13x - 4 & & y = 13x - 4 [0.15cm] y ≤ 13x + 3 & & y = 13x + 3 A second thing to note is that both boundary lines have the same slope, which means that the lines are parallel. Also, both inequalities are non-strict which means that we have to draw both boundary lines solid.
Let's focus for a moment on the first inequality. To determine which region to shade, we will test (0,0) in the first inequality.
We got a true statement. Therefore, we have to shade the region containing the origin — the region above the line y= 13x-4.
In a similar fashion, let's test (0,0) in the second inequality to determine which region to shade.
We got a true statement. Therefore, we have to shade the region containing the origin — the region below the line y= 13x+3.
Finally, let's graph only the solution set — the overlapping region.
Comparing this region to the four given descriptions, we conclude that option A describes the solution set of the system of inequalities. Option A The region between the linesy= 13x-4 andy= 13x+3, including both lines.
To determine which of the claims is true about the system of inequalities, let's solve it graphically. y ≤ x - 1 y > x + 1 The first thing we note is that both inequalities are already written in slope-intercept form. Let's continue by writing the equations of the boundary lines. ccc Inequality & & Boundary Line [0.5em] y ≤ x - 1 & & y = x - 1 y > x + 1 & & y = x + 1 A second thing to note is that both boundary lines have the same slope, which means that the lines are parallel. Since the first inequality is non-strict, we will draw the first boundary line solid. In contrast, the second inequality is strict and therefore, we will draw the second boundary line dashed.
Let's focus for a moment on the first inequality. To determine which region to shade, we will test (0,0) in the first inequality.
We got a false statement. Therefore, we have to shade the region not containing the origin — the region below the line y=x-1.
In a similar fashion, let's test (0,0) in the second inequality to determine which region to shade.
We got a false statement. Therefore, we have to shade the region not containing the origin — the region above the line y=x+1.
As we can see, the individual solution sets do not overlap each other. This means that the system of equations has no solutions. Consequently, option D is the true statement about the system. Option D The system has no solution.
Determine if the point (-2,-2) is a solution to the system whose solution is graphed.
Let's plot the point (- 2, - 2) in the given coordinate plane. If it lies in the shaded area, then the ordered pair is a solution of the system of inequalities.
Since the point lies within the shaded area, the ordered pair (- 2,- 2) is a solution to the system of linear inequalities.
We will plot the point (-2,-2) in the same coordinate plane as the system.
The point is not in the shaded region, and thus it is not a solution to the system of inequalities.
When determining if a given point satisfies an equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation.
Example Equation: y=x+1 | ||
---|---|---|
Point | Substitution | Solution? |
( 1, 0) | 0 &? = 1+1 0 &= 2 * | No |
( 0, 1) | 1 &? = 0+1 1 &= 1 ✓ | Yes |
For systems of inequalities, we can use the same method. However, substituting the point must create true statements in every inequality of the system. Let's test (3,11) to see if it is a solution to the given system.
Since 11 is greater than 7 and less than 17, both statements are true. Therefore, the point (3,11) is a solution to the given system.
Let's graph each of the inequalities on the same coordinate plane. To start, note that each of the inequalities is already written in slope-intercept form. y ≥ 2 y ≤ x +5 y ≤ - x +5
In order to graph an inequality, we start by graphing the boundary line. A boundary line can be written by replacing the inequality symbol with an equals sign. ccc Inequality & & Boundary Line y ≥ 2 & & y = 2 The boundary line of the inequality is y=2, which is a horizontal line whose y-intercept is 2. Notice that the line is solid because the inequality is not strict.
To know which region to shade, let's test (0,0) in the inequality.
Since we got a false statement, (0,0) is not in the solution set of the first inequality. Therefore, let's shade the region above the line.
Let's start by determining the boundary line. ccc Inequality & & Boundary Line y ≤ x+5 & & y = x+5 Because the inequality is not strict, we will draw the boundary line solid.
As before, let's test (0,0) in the inequality to determine which region to shade.
Since the origin satisfies the inequality, we will shade the region containing the point (0,0). We will graph this inequality on the same coordinate plane where we graphed the first inequality.
In the same manner as before, let's determine the boundary line for the third inequality. ccc Inequality & & Boundary Line y ≤ - x+5 & & y = - x+5 Once again, let's test the point (0,0) in this third inequality.
Since (0,0) satisfies the inequality, we will shade the region containing the origin. This boundary will be solid as well because the inequality is not strict. Let's add it to the same coordinate plane as the first two inequalities.
Next, let's graph only the solution set to the given system of inequalities — that is, let's remove the parts that are not common for the three regions.
The solution set is a triangle. To calculate its area, we need to determine the base and height. Then, we can use the formula for the area of a triangle.
As we can see from the graph, the base is b=6 and the height is h=3. Let's find the area.
In the following diagram, a system of linear inequalities has been graphed.
Let's plot the given points on the same coordinate plane as the system of inequalities.
When looking at the four points, the first thing we can notice is that R(0,-4) and S(-1,-6) lie squarely within the solution set, and the points Q(1,-2) and P(2,-4) rest on the boundary lines. Let's remember what it means when a line is dashed and when it is solid.
With the above in mind, we can conclude that P(2,-4) is a solution to the system because it lies on a solid line. However, the point Q(1,-2), which lies on both a solid and a dashed line, is a solution to one of the inequalities but not a solution to the other. Therefore, Q(1,-2) does not belong with the others, as it is not in the solution set.