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| | 10 Theory slides |
| | 15 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Each of the following graphs represents the solution set of a certain inequality.
Pair each graph with its corresponding inequality.
In the following applet, each of the solution sets of the inequalities presented previously can be viewed on the same coordinate plane. Plane regions are formed once two inequalities are drawn. Only two can be drawn at a time.
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.
A system of inequalities is a set of two or more inequalities involving the same variables. For example, the set formed by the following two inequalities is a system. y ≤ - 0.5x+3 y > x The solution set of a system of inequalities is the set of all ordered pairs that satisfy all the inequalities in the system simultaneously. The ordered pair ( 0, 1), for example, is a solution to the above system. 1 ≤ - 0.5( 0)+3 & ✓ 1 > 0 & ✓ Usually, systems of inequalities are solved by graphing each inequality on the same coordinate plane. When the inequalities in a system are graphed, the coordinate plane is divided into different regions. These regions provide an insight into the determination of the solution set. To see what these regions represent for the aforementioned system, move point P.
A system of linear inequalities can be solved by graphing all inequalities on the same coordinate plane and then finding the region of intersection, if any. For example, consider the following system. x+y<7 & (I) x+2y ≤ 10 & (II) To solve the system graphically, these three steps can be followed.
(I): LHS-x<RHS-x
(II): LHS-x≤RHS-x
(II): .LHS /2.≤.RHS /2.
Similarly, the boundary line of the second inequality is y=-0.5x+5. Inequality II:& y< - 0.5x + 5 Boundary Line II:& y = -0.5 x + 5 Since the inequality is non-strict, its boundary line is solid. In addition, the region to be shaded is the one below the line. This inequality will be graphed on the same coordinate plane.
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.
x + y ≤ 30 Jordan does not want to go over budget and wants to spend less than $240. Therefore, the total cost of the pens has to be less than 240. Total Cost < 240 The total cost equals the number of ballpoint pens bought multiplied by its price plus the number of fountain pens bought multiplied by its price. Using the fact that each ballpoint pen costs $6.00 and each fountain pen costs $12.00, a second inequality can be set. 6x + 12y < 240 In consequence, the following system of inequalities models Jordan's situation. x + y ≤ 30 & (I) 6x+12y < 240 & (II)
x + y ≤ 30 & (I) 6x+12y < 240 & (II) Start by rewriting each inequality in slope-intercept form. To do so, isolate the y-variable in the left-hand side of each inequality.
To graph inequality (I), first, draw the boundary line whose equation is y=- x+30. Since the inequality is non-strict, the line is solid. In this case, the region to be shaded is the one below the line.
Next, graph inequality (II) on the same coordinate plane. To do so, draw the boundary line whose equation is y=- 12x+20. Since the inequality is strict, the line will be dashed. As before, the region to be shaded is the one below the line.
The solution to the system of inequalities is the overlapping region. However, since buying a negative number of pens does not make sense, the negative values should be discarded. Therefore, the shaded region will include but also be limited by the coordinate axes.
Additionally, buying 3.5 ballpoint pens is nonsensical. Therefore, only points with integer coordinates are part of the solution. Consequently, the solution set of the system looks as follows.
As seen, in the shaded region, there are two points with the greatest integer y-coordinate — (0,19) and (1,19). Therefore, Jordan has two options, buying no ballpoint pens and 19 fountain pens or buying one ballpoint pen and 19 fountain pens.
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.
Jordan enjoyed being able to buy gifts for her friends. Now she wants to save up to buy a family members car. To achieve this goal, she dares to work two jobs this summer — one in a mechanic shop and another in a bakery. She can make $9 per hour in the mechanic shop and $8 per hour in the bakery.
Bakery & [-0.15cm] ^(↑) y &≥ ↓x + 4 [-0.15cm] & Mechanic Shop On the other hand, it is said that Jordan plans to earn no more than $108. That is, her daily income is at most $108. Daily Income ≤ 108 The daily income equals the hourly rate multiplied by the number of hours worked. Since Jordan has two jobs, her daily income equals 9 multiplied by the number of hours worked in the mechanic shop added to 8 multiplied by the number of hours worked in the bakery. 9x + 8y ≤ 108 Consequently, the following system of inequalities models Jordan's situation. y ≥ x+4 & (I) 9x + 8y ≤ 108 & (II)
y ≥ x+4 & (I) 9x + 8y ≤ 108 & (II) The first inequality is already in slope-intercept form. However, the other inequality needs to be written in slope-intercept form.
To graph inequality (I), its boundary line y=x+4 will be drawn first. Since the inequality is non-strict, the line is solid. Since (0,0) does not satisfy the inequality, the region to be shaded is the one not containing the origin.
Next, inequality (II) will be graphed on the same coordinate plane. To do so, draw the boundary line, whose equation is y=-1.125x+13.5. Since the inequality is non-strict, this line will also be solid. In this case, the region to be shaded is the one below the line.
The solution to the system of inequalities is the overlapping region. Notice that negative values do not make sense in this context. Therefore, the shaded region will include but also be limited by the y-axis. Consequently, the system's solution looks as follows.
Because the second line drawn represents Jordan's maximum possible daily income, the further a point is from this line, the lower the income that the point represents. Of the points, the one that is farthest from the line is (1,5) which represents the case when Jordan worked 1 hour in the mechanic shop and 5 hours in the bakery. ( 1, 5) ⇒ 👨🏼🔧 * 1 hour 👨🏼🍳 * 5 hour With this combination of worked hours, the minimum income that Jordan could have earned last Friday can be found. To do so, substitute x= 1 and y= 5 into the expression representing the daily income.
x= 1, y= 5
Multiply
Add terms
Consequently, the minimum amount of money that Jordan could have earned last Friday is $49.
Jordan now has a bit of extra cash to spend. She is excited for a new movie release. Ticket prices vary depending on the customer's age — $6 for children and $9 for adults. The cinema expects more than 250 people to attend the premiere but they do not expect a revenue of more than $2700.
Minimum: 101
x + y > 250 Also, it is said that the estimated revenue will be no more than $2700. Therefore, the revenue is less than or equal to 2700. Revenue ≤ 2700 The cinema's revenue equals the number of children who attended the premiere multiplied by 6 added to the number of adults who attended the premiere multiplied by 9. 6x + 9y ≤ 2700 Consequently, the following system of inequalities models the described situation. x + y > 250 6x + 9y ≤ 2700
x + y > 250 & (I) 6x + 9y ≤ 2700 & (II) First, rewrite each inequality in slope-intercept form.
To graph inequality (I), draw the boundary line whose equation is y=- x+250. Since the inequality is strict, the line is dashed. In this case, the region to be shaded is the one above the line.
Next, graph inequality (II) on the same coordinate plane. To do so, draw the boundary line whose equation is y=- 23x+300. Since the inequality is not strict, this line will be drawn solid. In this case, the region to be shaded is the one below the line.
The solution to the system of inequalities is the overlapping region. However, because of the context, only non-negative and integer values make sense. Therefore, the region where both inequalities overlap will also be limited by the axes, including them.
Instead of showing each point separately on the graph, because it is difficult to do so, the region where these points lie is shown. However, remember that only the points with integer coordinates are meaningful.
Recall that these two conditions were written as a system of inequalities and solved graphically. x + y > 250 6x + 9y ≤ 2700 The solution set of the system shows all the possible cases satisfying both conditions. Now, knowing that there were 150 children in the premiere eliminates most of the possible cases.
The number of adults who could have attended the premiere can be any point whose x-coordinate is 150 and lie inside the shaded region. Of those points, ( 150,200) has the the maximum integer y-coordinate and ( 150,101) has the minimum y-coordinate. ( 150,200) ← Maximumy-coordinate ( 150,101) ← Minimumy-coordinate Therefore, the number of adults who could have attended the premiere is any number between 101 and 200, inclusive.
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.
x + y ≤ 8 On the other hand, it is said that the entire route is no less than 990 kilometers, which means that the distance to drive is greater than or equal to 990 kilometers. Distance to Drive ≥ 990 This distance can be written in terms of x and y by using the speed formula d=r* t. According to Jordan's plan, she will drive to Ramsha's house at a rate of 70 kilometers per hour. Multiplying this rate by the time elapsed to arrive at Ramsha's house gives the distance traveled from her house to Ramsha's. Distance from Jordan's house to Ramsha's ⇓ 70x Similarly, multiplying the second rate, 110 kilometers per hour, by the time it takes for the rest of the route, the distance from Ramsha's house to their relative's will be obtained. Distance from Ramsha's house to their relative's ⇓ 110y The sum of these distances equals the distance needed to drive. Thus, a second inequality can be set. 70x + 110y ≥ 990 These two inequalities need to be satisfied simultaneously. Therefore, the two inequalities together form a system of inequalities that models Jordan's travel plan. x + y ≤ 8 70x + 110y ≥ 990
x + y ≤ 8 & (I) 70x + 110y ≥ 990 & (II) First, rewrite the inequalities in slope-intercept form.
To graph inequality (I), draw the boundary line y=- x+8. Since the inequality is not strict, the line is dashed. In this case, the region to be shaded is the one below the line.
Next, graph inequality (II) on the same coordinate plane. To do so, first draw the boundary line y=- 711x+9. Since the inequality is not strict, this line will also be drawn solid. However, the region to be shaded is the one above the line.
The solution to the system of inequalities is the overlapping region. However, in the context of the given situation, x and y cannot take negative values as they represent time. Therefore, only solutions in the first quadrant, if any, should be considered.
As seen, the system of inequalities has no realistic solution, which means Jordan's plan is not feasible. She will not be able to meet her travel expectations. For a more realistic plan, she should either increase the traveling speed or increase the maximum time she plans to drive. The second option is the safest, though.
For each given system of linear inequalities, select the region corresponding to its solution set, if any.
When working with a system of inequalities, there is no upper limit to the number of inequalities that can be analyzed. Consider, for example, the following system with three inequalities. -3x + 4y ≥ -8 & (I) 2x + y > -6 & (II) x + 4y ≤ 4 & (III) The steps to solve this system are the same as those used to solve a system with only two inequalities.
The inequalities in the given system can be rewritten as follows. y ≥ 34x-2 & (I) y > -2x-6 & (II) y ≤ - 14x+1 & (III) The second step can be performed with the help of the applet.
Before starting the trip, the friends decided that Ramsha will drive more hours than Tearrik every day.
From the given information, the system we need to write will have three inequalities.
For all three inequalities, let $x$ and $y$ be the number of hours that Tearrik and Ramsha will drive each day, respectively.
It is given that the guys plan to drive less than $13$ hours. We can write an inequality to represent this situation in the following way. \begin{gathered} x+y < 13 \end{gathered}
Tearrik will drive about $75$ miles per hour and Ramsha about $60$ miles per hour. Using the fact that the guys will drive at least $420$ miles a day, we can write our second inequality. \begin{gathered} 75x +60y \geq 420 \end{gathered}
The fact that Ramsha will drive more hours than Tearrik allows us to write the last inequality as follows. \begin{gathered} y > x \end{gathered}
Having written each of the three inequalities, we can write the following system that models the trip. \WriteSysEqnIIIb{x+y < 13}{75x +60y \geq 420}{y > x} Let's solve the system graphically. To start, we write each inequality in slope-intercept form.
Let's continue by graphing the inequalities one at a time. We will start by writing the equation of the boundary line corresponding to the first inequality. \begin{array}{ccc} \underline{\textbf{Inequality}} & & \underline{\textbf{Boundary Line}} \\[0.5em] \col{y< \N x+ 13} & & y = \N x+ 13 \end{array} Since the inequality is strict, we will draw the boundary line dashed. To determine which region to shade, let's substitute $(0,0)$ into the inequality.
Since we got a true statement we will shade the region containing the origin.
In a similar manner, let's write the boundary lines corresponding to the other two inequalities. We will also test an arbitrary point on each inequality to determine which region to shade.
| Inequality | Boundary Line | Test Point | Region to Shade |
|---|---|---|---|
| $\colII{y \geq \N\dfrac{5}{4}x + 7}$ | $y = \N\dfrac{5}{4}x + 7$ Solid |
$(0,0)$ $\Downarrow$ \begin{aligned} y &\geq \N\dfrac{5}{4}x + 7 \\ \colII{0} &\stackrel{?}{\geq} \N\dfrac{5}{4}(\col{0}) + 7 \\ 0 &\geq 7 \quad\StatementFalse \end{aligned} |
Not containing the origin |
| $\colIV{y > x}$ | $y = x$ Dashed |
$(1,0)$ $\Downarrow$ \begin{aligned} y &> x \\ \colII{0} &> \col{1} \quad\StatementFalse \end{aligned} |
Not containing the point $(1,0)$ |
Using the information in the table, let's graph the last two inequalities. We will graph them on the same coordinate plane we graphed the first inequality.
Based on the context, $x$ and $y$ cannot be negative. Therefore, we need to limit the solution set to the first quadrant. Then, the solution set looks as follows.
Comparing the shape of the solution set to the four given polygon answer choices, we conclude that the correct choice is D.
Since $y$ represents the number of hours that Ramsha drives in one day, and we are told that Ramsha drives for $5$ hours on a certain day, then $y=5.$ To determine the minimum integer number of hours that Tearrik has to drive that day, let's plot the line $y=5$ along with the solution set we found in the previous part.
Note that the minimum integer value of $x$ that is inside the solution set and on the line $y=5$ is $2.$
Consequently, the minimum integer number of hours that Tearrik has to drive to fulfill the daily goals is $2$ hours.
Since $x$ represents the number of hours that Tearrik drives in one day, and he said that the next day he will drive for $1$ hour, then $x=1.$ To determine the minimum integer number of hours that Ramsha has to drive the next day, let's plot the line $x=1$ along with the solution set we found in the Part A.
Note that the minimum integer value of $y$ that is inside the solution set and on the line $x=1$ is $6.$
Consequently, the minimum integer number of hours that Ramsha has to drive on this day to fulfill the daily goals is $6$ hours.
Solve the following system of inequalities. { l y < x+2 8x-4y ≤ 24 2x + y ≥ 2 y ≥ 0 . Which of the following polygons resembles the shape of the solution set?
To determine which of the polygons resembles the solution set, let's solve the system graphically. y < x+2 & (I) 8x-4y ≤ 24 & (II) 2x + y ≥ 2 & (III) y ≥ 0 & (IV) Note that the inequalities (I) and (IV) are already written in slope-intercept form, but inequalities (II) and (III) still need to be rewritten. Then, let's do it.
We can now proceed by graphing the inequalities starting with the first one. To do this, we identify the corresponding boundary line. ccc Inequality & & Boundary Line [0.25em] y < x+2 & & y = x+2 Since the inequality is strict, we will draw the boundary line dashed. To determine which region to shade, we will test (0,0) into the inequality.
We got a true statement, which implies that we have to shade the region containing the origin.
Let's summarize the computations for the other three inequalities in a table.
| Inequality | Boundary Line | Test Point | Region to Shade |
|---|---|---|---|
| y ≥ 2x-6 | y = 2x-6 Solid |
(0,0) ⇓ 0 ? ≥ 2( 0)-6 0 ≥ -6 ✓ |
Containing the origin |
| y ≥ -2x + 2 | y = -2x + 2 Solid |
(0,0) ⇓ 0 ? ≥ -2( 0)+2 0 ≥ 2 * |
Not containing the origin |
| y ≥ 0 | y = 0 Solid |
(0,1) ⇓ 1 ≥ 0 ✓ |
Containing the point (0,1) |
Next, let's use all the information in the table to graph the last three inequalities on the same coordinate plane we graphed the first one.
Let's remove the parts that do not belong to the solution set.
Now that we can see the solution set clearly, and after comparing its shape to the six given polygons, we conclude that the polygon that resembles the solution set is C.