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Every individual benefits from the efficient use of their resources such as time and money. To use such resources effectively, several inequalities might need to be solved simultaneously. Together, these inequalities form a system. This lesson will teach how to write systems of inequalities and how to solve them using graphs.
### Catch-Up and Review

**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.

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class=\"mrel\">><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><\/span><\/span><\/span>"},{"id":2,"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8304100000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2265<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>"},{"id":3,"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.7335400000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>"},{"id":4,"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8304100000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">x<\/span><\/span><\/span><\/span>"}],[{"id":0,"text":"Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">A<\/span><\/span><\/span><\/span>"},{"id":3,"text":"Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.02778em;\">D<\/span><\/span><\/span><\/span>"},{"id":1,"text":"Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.05017em;\">B<\/span><\/span><\/span><\/span>"},{"id":2,"text":"Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">C<\/span><\/span><\/span><\/span>"},{"id":4,"text":"Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.05764em;\">E<\/span><\/span><\/span><\/span>"}]],"lockLeft":false,"lockRight":false},"formTextBefore":"","formTextAfter":"","answer":[[0,1,2,3,4],[0,3,1,2,4]]}

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.

Think about what each of these regions — overlapping, shaded, and unshaded — represent.

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.
*simultaneously*. The ordered pair $(0,1),$ for example, is a solution to the above system.

${y≤-0.5x+3y>x $

The solution set of a system of inequalities is the set of all ordered pairs that satisfy all the inequalities in the system ${1≤-0.5(0)+31>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 $P.$
Of the regions formed, the overlapping region represents the solution set of the system.

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.
*expand_more*
*expand_more*
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Keep in mind that if there is no overlapping region, the system has no solution.

${x+y<7x+2y≤10 (I)(II) $

To solve the system graphically, these three steps can be followed.
1

Write the Inequalities in Slope-Intercept Form

To simplify the process of graphing the inequalities, start by writing them in slope-intercept form.

2

Graph the Inequalities

To graph the inequalities, begin with writing the corresponding boundary lines. The boundary line of the first inequality is $y=-x+7.$

$Inequality I:Boundary Line I: y<-x+7y=-x+7 $

Since the inequality is strict, the boundary line should be dashed and, in this case, the shaded region is the one below the line.
Similarly, the boundary line of the second inequality is $y=-0.5x+5.$

$Inequality II:Boundary Line II: y<-0.5x+5y=-0.5x+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.
3

Find the Overlapping Region

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.

a Help Jordan and write a system of inequalities that represents this situation.

b Graph the solution set of the system.

c If Jordan wants to buy the maximum number of fountain pens she can under her budget, what is the maximum number of teammates that could get a gift? How many of them will get a ballpoint pen?

a ${x+y≤306x+12y<240 $

b **Graph:**

c A maximum of $20$ teammates could get a gift. Of that number, only one would get a ballpoint pen.

a Recall that Jordan can buy *at most* $30$ gifts and that the total cost has to be *less than* $$240.$

b To graph the solution set of the system of inequalities, start by writing the inequalities in slope-intercept form. Note that only points with non-negative and integer coordinates are meaningful.

c Of the points in the solution set, which one has the greatest $y-$coordinate?

a Since Jordan has $30$ teammates, she can buy at most $30$ gifts. In other words, the sum of the number of ballpoint pens $x$ and the number of fountain pens $y$ can be *at most* $30.$ This leads to writing the first inequality.

$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 $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≤306x+12y<240 (I)(II) $

b To solve the system written in Part A graphically, both inequalities will be graphed on the same coordinate plane.

${x+y≤306x+12y<240 (I)(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.
${x+y≤306x+12y<240 (I)(II) $

${y≤-x+30y<-21 x+20 $

Next, graph inequality (II) on the same coordinate plane. To do so, draw the boundary line whose equation is $y=-21 x+20.$ 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.

c Since $y$ represents the number of fountain pens that Jordan will buy, in the solution of the system found in Part B, look for the points with the greatest $y-$coordinate.

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.
### Answer

### Hint

### Solution

*at most* $$108.$
b To solve the system written in Part A graphically, both inequalities will be graphed on the same coordinate plane.
To graph inequality (I), its boundary line $y=x+4$ will be drawn first. Since the inequality is non-strict, the line is solid. Since $(0,0)$ does not satisfy the inequality, the region to be shaded is the one not containing the origin.
Consequently, the minimum amount of money that Jordan could have earned last Friday is $$49.$

Jordan plans to work every day for at least $4$ more hours in the bakery than in the mechanic shop. However, since she does not want to overwork, she does not plan to earn more than $$108$ per day.

a Let $x$ and $y$ be the number of hours that Jordan will work in the mechanic shop and in the bakery every day, respectively. Write a system of inequalities that models Jordan's situation.

b Solve the system graphically.

c Last Friday, Jordan worked whole number of hours in both places. What is the minimum amount of money she could have earned last Friday?

a ${y≥x+49x+8y≤108 $

b **Graph:**

c $$49$

a Note that $y$ must be *greater than or equal to* $x+4.$ Also, be aware that Jordan plans to earn *at most* $$108$ per day.

b Both boundary lines have to be drawn solid, and negative values do not make sense.

c Look for the points with integer coordinates inside the region drawn in Part B. Which of these points is farthest from the line representing the maximum daily earnings? Use the coordinates of that point to determine the amount of money earned.

a Since Jordan wants to work in the bakery at least $4$ more hours than in the mechanic shop, $y$ must be *greater than or equal to* $x$ plus $4.$

$Bakeryy_{↑} ≥↓x +4≥Mechanic Shop $

On the other hand, it is said that Jordan plans to earn no more than $$108.$ That is, her daily income is $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+49x+8y≤108 (I)(II) $

${y≥x+49x+8y≤108 (I)(II) $

The first inequality is already in slope-intercept form. However, the other inequality needs to be written in slope-intercept form.
${y≥x+49x+8y≤108 (I)(II) $

${y≥x+4y≤-1.125x+13.5 $

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.

c It is given that Jordan worked an integer number of hours in both jobs last Friday. Then, look for the points with integer coordinates inside the shaded region drawn in Part B. Since the points on the $y-$axis represent the case where Jordan did not work in the mechanic shop, these points should not be considered.

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)⇒{👨🏼🔧×1hour👨🏼🍳×5hour $

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.
$Daily Income=9x+8y$

Substitute coordinates and evaluate

SubstituteII

$x=1$, $y=5$

$Daily Income=9(1)+8(5)$

Multiply

Multiply

$Daily Income=9+40$

AddTerms

Add terms

$Daily Income=49$

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.$
### Answer

### Hint

### Solution

b To solve the system written in Part A graphically, both inequalities will be graphed on the same coordinate plane.

Jordan finds herself wanting to solve this problem during her break time at the bakery. Let $x$ and $y$ be the number of children and adults who attended the movie premiere, respectively.

a Write a system of inequalities that models the given situation.

b Solve the system graphically.

c If the cinema's expectations were met and $150$ children attended the premiere, what is the maximum and minimum possible number of adults who could have attended the premiere?

a ${x+y>2506x+9y≤2700 $

b **Graph:**

c **Maximum:** $200$

**Minimum:** $101$

a The number of attendees is greater than $250.$ The cinema's revenue is less than or equal to $2700.$

b The line representing the attendance is dashed while the other line is solid. In this context, only non-negative and integer values for $x$ and $y$ make sense.

c Look for the points whose $x-$coordinate is $150$ and lie inside the shaded region drawn in Part B. Of those points, what is the maximum integer $y-$coordinate?

a Since the cinema expects an attendance of more than $250$ people, the number of children plus the number of adults must be greater than $250.$

$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>2506x+9y≤2700 $

${x+y>2506x+9y≤2700 (I)(II) $

First, rewrite each inequality in slope-intercept form.