{{ article.displayTitle }}

Kevin's class is having a work experience internship this week. He is working with a farmer to test the speed of an autonomous tractor in a field.

In the first test, the tractor started from a point $2$ miles away from the barn. After a half hour, the tractor was at least $10$ miles away from the barn. In the next test, the tractor started at a point $1.5$ miles away from the barn. When it stopped $45$ minutes later, it was less than $20$ miles from the barn.

Answer

a $16\le r<24.7$

b

Solution

a To find the possible speeds, both test needs to be considered. It is given that in the first test, the tractor started $2$ miles away from the barn and stopped at least $10$ miles away from the barn after half an hour. This means that the sum of the distance traveled and the starting distance of $2$ miles has to be greater than or equal to $10$ miles.

$$\begin{array}{c}\text{distance traveled}+2\ge 10\end{array}$$

To find the distance traveled, a modification of the speed formula will be used.

$$\begin{array}{c}r=\frac{d}{t}\end{array}$$

In this formula, $r$ is the speed or rate, $d$ is the distance traveled, and $t$ is the time it took to travel the distance. To find the distance, $d$ needs to be the isolated variable.

$$\begin{array}{c}r=\frac{d}{t}\Rightarrow d=rt\end{array}$$

Therefore, this formula can be substituted for the distance traveled in the written inequality.

$$\begin{array}{c}rt+2\ge 10\end{array}$$

Since the speed is asked in miles per hour and the tractor drove for half an hour, the given time is written as $\frac{1}{2}=0.5.$

$$\begin{array}{c}0.5r+2\ge 10\end{array}$$

Likewise in the second test, it is given that the tractor starts from $1.5$ miles away and ends less than $20$ miles away from the barn after $45$ minutes. This means that the sum of the distance traveled and $1.5$ miles has to be less than $20$ miles.

$$\begin{array}{c}\text{distance traveled}+1.5<20\end{array}$$

Then, the formula for the distance traveled can be substituted in this inequality as well.

$$\begin{array}{c}rt+1.5<20\end{array}$$

In this case $45$ minutes can be converted to hours using a conversion factor. $45$ minutes are equal to $\frac{45}{60}=0.75$ hours.

$$\begin{array}{c}0.75r+1.5<20\end{array}$$

Since the possible values for the speed are the values of $r$ that satisfy both inequalities, these inequalities can be combined as a compound inequality with and.

$$\begin{array}{c}0.5r+2\ge 10\text{ and }0.75r+1.5<20\end{array}$$

To solve this compound inequality, the properties of inequalities will be used. To solve the inequality on the left, first $2$ will be subtracted from both sides of the inequality.

$0.5r+2\ge 10$

▼

Solve for $r$

SubIneq

$\text{LHS}-2\ge \text{RHS}-2$

$0.5r\ge 8$

DivIneq

$\text{LHS}/0.5\ge \text{RHS}/0.5$

$r\ge 16$

Similarly, to solve the inequality on the right, $1.5$ will be subtracted from both sides of the inequality.

$0.75r+1.5<20$

▼

Solve for $r$

SubIneq

$\text{LHS}-1.5<\text{RHS}-1.5$

$0.75r<18.5$

DivIneq

$\text{LHS}/0.75<\text{RHS}/0.75$

$r<24.666666\dots$

RoundDec

Round to $1$ decimal place(s)

$r<24.7$

The compound inequality can be written with the found solutions.

$$\begin{array}{c}r\ge 16\text{ and }r<24.7\end{array}$$

It should be noted that some compound inequalities that are written with and can be rewritten as follows.

$$\begin{array}{c}16\le r<24.7\end{array}$$

This means that the possible speeds of the tractor are greater than or equal to $16$ miles per hour and less than $24.7$ miles per hour.

b To graph the solution set of the compound inequality, it is useful to graph the solution set of each individual inequality first. Since the inequality is non-strict, the solution set of the inequality $r\ge 16$ is made of all the numbers to the right of $16,$ including $16.$ This is shown by using a closed circle at $16.$

On this case, since the inequality is strict, the solution set of the inequality $r<24.7$ is made of the numbers to the left of $24.7.$ Since $24.7$ is not included in the solution set, an open circle is used instead.

Since the compound inequality is written with and, its solution set is made of the numbers that satisfy both inequalities.

Ignacio is interning at the local disaster preparedness center. He is using a simulator to help to secure the unsafe area near a volcano that is about to erupt. He is in charge of marking safe distances to the east and the west of the volcano. He initially marked the safe distance with flags, one $30$ miles to the east of the base of the volcano and the other $15$ miles to the west.

As time passes and data comes in, Ignacio realizes that his estimates were wrong. He notes that the flag to the west is less than half the distance away from the volcano that it should be. On the other hand, he calculates that the eastern flag covers less than two thirds of the actual necessary safe distance from the volcano.

Answer

a $d<\text{-}30\text{ or }d>45$

b

Solution

a To write the safe distance as a compound inequality, the individual inequalities must first be identified. It is given that the western flag is less than half the actual necessary safe distance. This can be written as an inequality, with $d$ representing the total safe distance.

$$\begin{array}{c}15\text{ miles west}<\frac{1}{2}d\end{array}$$

Since the distance is to the west of the volcano, it is written as a negative number. Also, since this value is negative, further distances from the volcano are less than $\text{-}15.$ Therefore, the inequality has to be rewritten as follows.

$$\begin{array}{c}\frac{1}{2}d<\text{-}15\end{array}$$

Similarly, it is given that the eastern flag is less than two thirds of the safe distance at $30$ miles from the volcano. This can also be written as an inequality. Since the distances to the east are written with positive numbers, the inequality does not need to be modified.

$$\begin{array}{c}30<\frac{2}{3}d\end{array}$$

These inequalities can be solved by using the Properties of Inequalities, a similar fashion to solving an equation. The first inequality is solved by multiplying by $2.$

$\frac{1}{2}d<\text{-}15$

▼

Simplify

MultIneq

$\text{LHS}\cdot 2<\text{RHS}\cdot 2$

$\frac{2}{2}d<2(\text{-}15)$

SimpQuot

Simplify quotient

$d<2(\text{-}15)$

Multiply

Multiply

$d<\text{-}30$

The safe distance to the west is less than $\text{-}30$ miles, or more than $30$ miles. The second inequality can be solved by multiplying both sides by the reciprocal of $\frac{2}{3},$ $\frac{3}{2}.$

$30<\frac{2}{3}d$

▼

Simplify

MultIneq

$\text{LHS}\cdot \frac{3}{2}<\text{RHS}\cdot \frac{3}{2}$

$\frac{3}{2}\cdot 30<\frac{3}{2}\cdot \frac{2}{3}d$

MultFracByInverse

$\frac{a}{b}\cdot \frac{b}{a}=1$

$\frac{3}{2}\cdot 30<d$

MoveRightFacToNum

$\frac{a}{c}\cdot b=\frac{a\cdot b}{c}$

$\frac{3\cdot 30}{2}<d$

Multiply

Multiply

$\frac{90}{2}<d$

CalcQuot

Calculate quotient

$45<d$

RearrangeIneq

Rearrange inequality

$d>45$

The safe distance to the east is more than $45$ miles. Since the safe distance can be either more than $30$ miles west or more than $45$ miles east, the compound inequality is written with or.

$$\begin{array}{c}d<\text{-}30\text{ or }d>45\end{array}$$

b To graph the solution set of a compound inequality, it is convenient to first graph each individual inequality. First, the solution set of $d<\text{-}30$ is made of all the numbers to the left of $\text{-}30,$ not including $\text{-}30$ because the inequality is strict.

Similarly, the graph of the solution set of $d>45$ is made of every number to the right of $45,$ not including $45.$

Finally, since it is written with or, the graph of the compound inequality is the combination of both solution sets and does not need to be adjusted or limited.

Tearrik is spending his week of work experience at a local bakery. On his first day, he bought $100$ cookies at a discount. He decides to eat 5 cookies a day until they are all gone. Tearrik is also allowed to bring home $3$ cookies per day, which he gives to his brother. Tearrik's brother decides that he will not eat his cookies until he has at least $90$ saved up.

Answer

a $d<20\text{ and }d\ge 30$

b

Solution

a To find on which days both brothers will eat cookies together, it is important to identify when each brother will eat cookies. Tearrik starts with $100$ cookies and he eats $5$ cookies every day until he reaches $0$ cookies. This situation can be written as an inequality.

$$\begin{array}{c}100-5d>0\end{array}$$

Since Tearrik will eat cookies throughout this time, the values of $d$ are the possible numbers of days in which he eats cookies. On the other hand, Tearrik's brother will save $3$ cookies every day until he has at least $90$ cookies. Then, Tearrik's brother will start eating his cookies.

$$\begin{array}{c}3d\ge 90\end{array}$$

For both to eat cookies, the value $d$ must satisfy both inequalities at the same time. This kind of compound inequality is written with and.

$$\begin{array}{c}100-5d>0\text{ and }3d\ge 90\end{array}$$

To solve this compound inequality, both inequalities must be solved. First, to solve the inequality on the left, $100$ will be subtracted from both sides of the inequality.

$100-5d>0$

▼

Solve for $d$

SubIneq

$\text{LHS}-100>\text{RHS}-100$

$\text{-}5d>\text{-}100$

DivNegIneq

Divide by $\text{-}5$ and flip inequality sign

$d<20$

This means that Terrik will eat his cookies over the first $20$ days. Since this inequality is strict, Tearrik will not reach the $20\text{th}$ day before he finishes his cookies. Then, the second inequality is solved by dividing both sides of the inequality by $3.$

$3d\ge 90$

DivIneq

$\text{LHS}/3\ge \text{RHS}/3$

$d\ge 30$

This means that Terrik's brother will save his cookies for $30$ days. Since the inequality is non-strict, he will start eating his cookies on the $30\text{th}$ day. Now the compound inequality can be rewritten using the solutions.

$$\begin{array}{c}d<20\text{ and }d\ge 30\end{array}$$

This means that Tearrik will be able to eat cookies for $20$ days, while his brother will save his cookies for $30$ days before eating any of them. Therefore, they will not eat any cookies on the same day.

b To graph the solution set of the compound inequality, first it is necessary to graph the solution set of each individual inequality. The solution set of the inequality $d<20$ are the numbers less than $20,$ not including $20.$ Since $20$ is not included in the solution set, it is marked with an open circle.

Similarly, the solution set of the inequality $d\ge 30$ is made of the numbers greater than or equal to $30.$ Since thi inequality is not strict, the circle is closed.

The solution set of the compound inequality is made of the numbers that satisfy both inequalities at the same time. Since there are no such numbers, the compound inequality has no solution.

This confirms that the brothers will not eat cookies together.

For his internship, Davontay is working in a research lab. He is researching what tools are used to measure really low and really high temperatures. He found out that a thermocouple thermometer can measure temperatures lower than $3272^{\circ }\text{ F},$ while a pyrometer thermometer can measure temperatures greater than or equal to $973\text{ K}.$

To help find these temperatures in degrees Celsius, the relationships between the different temperature scales are shown in the following table. It should be noted that $C$ refers to a temperature in degrees Celsius, $F$ is the temperature in degrees Fahrenheit, and $K$ refers to kelvins.

Answer

a $C<1800\text{ or }C\ge 700$

b

Solution

a To find the range of temperatures in degrees Celsius that can be covered by the thermometers, first it is needed to write the given information as inequalities. It is given that a thermocouple can measure temperatures that are lower than $3272^{\circ }\text{ F}.$ This can be written as an inequality, with $F$ as the temperature in degrees Fahrenheit.

$$\begin{array}{c}F<3272\end{array}$$

Since the temperature is needed in degrees Celsius, the conversion formula will be substituted for $F$ in the inequality above. Let $C$ be the temperature in degrees Celsius.

$$\begin{array}{c}\frac{9}{5}C+32<3272\end{array}$$

On the other hand, it is given that a pyrometer can measure temperatures as low as $973\text{ K}.$ This can be written as an inequality, using $K$ as the temperature in kelvins. It should be noted that kelvins are not degrees since it is an absolute scale.

$$\begin{array}{c}K\ge 973\end{array}$$

Substituting the conversion formula from the given table, the inequality can be rewritten in terms of degrees Celsius $C.$

$$\begin{array}{c}C+273\ge 973\end{array}$$

Since the question asks to describe the temperature range that both thermometers can measure, if either one can be used, the compound inequality is written with or.

$$\begin{array}{c}\frac{9}{5}C+32<3272\text{ or }C+273\ge 973\end{array}$$

Now, to solve this compound inequality, each inequality will be solved individually using the Properties of Inequalities. To solve the inequality on the left, first subtract $32$ from both sides of the inequality.

$\frac{9}{5}C+32<3272$

▼

Solve for $C$

SubIneq

$\text{LHS}-32<\text{RHS}-32$

$\frac{9}{5}C<3240$

MultIneq

$\text{LHS}\cdot \frac{5}{9}<\text{RHS}\cdot \frac{5}{9}$

$C<\frac{5}{9}\cdot 3240$

MoveRightFacToNum

$\frac{a}{c}\cdot b=\frac{a\cdot b}{c}$

$C<\frac{5\cdot 3240}{9}$

Multiply

Multiply

$C<\frac{16200}{9}$

CalcQuot

Calculate quotient

$C<1800$

A similar process can be done to the inequality on the right.

$C+273\ge 973$

SubIneq

$\text{LHS}-273\ge \text{RHS}-273$

$C\ge 700$

Therefore, the compound inequality can be rewritten with these solutions.

$$\begin{array}{c}C<1800\text{ or }C\ge 700\end{array}$$

b To graph the temperature range from Part A, first the solution set of each individual inequality should be graphed. The solution set of the inequality $C<1800$ is made of all the numbers to the left of $1800,$ not including $1800.$

Similarly, the solution set of the inequality $C\ge 700$ is made of every point to the right of $700,$ including $700.$

Since the inequality is written with or, the solution set of the compound inequality is made of the numbers that satisfy either inequality. By combining the graphs it can be noted that every number is a solution to the compound inequality.

This means that any temperature can be measured using either of the thermometers.

Diego is helping at a local car dealership for his work experience. After spending time at the dealership, he decides to start saving money to buy a car. His father told him that he would double the amount of money that Diego saves, starting from now. Also, Diego will receive extra $500$ dollars from his uncle to help buy the car when he finishes saving.

When looking for prices, Diego notices that most of the cars he likes range from $15$ thousand dollars to $18$ thousand dollars.

Answer

a $15000\le 2m+500\le 18000$

b Diego must save between $\$7250$ and $\$8750,$ including these values, to afford the car.

c

Solution

a To write the situation as a compound inequality, first the total amount of money Diego needs to save should be expressed. Let $m$ be the total amount of money that Diego saves. It is given that Diego's dad is going to double the amount of $m.$ This means that $m$ is multiplied by $2.$

$$\begin{array}{c}2m\end{array}$$

Also, Diego's uncle will give Diego $\$500$ dollars after finishing saving. This adds $500$ to the amount after it is multiplied by $2.$ With this information, it is possible to write Diego's total amount of money as an expression.

$$\begin{array}{c}2m+500\end{array}$$

Diego needs from $15$ thousand dollars to $18$ thousand dollars to afford a car he likes. Therefore, the total amount of money that Diego needs must lie between these values in order for him to be able to afford the car. This can be expressed as a compound inequality.

$$\begin{array}{c}2m+500\ge 15000\\ \text{ and }\\ 2m+500\le 18000\end{array}$$

Since the inequalities are written with and, they can be rewritten as follows.

$$\begin{array}{c}15000\le 2m+500\le 18000\end{array}$$

b To find the amount of money that Diego himself needs to save to be able to afford the car, the compound inequality set in Part A needs to be solved for $m.$ This can be done using the Properties of Inequalities to solving each individual inequality first.

$15000\le 2m+500$

SubIneq

$\text{LHS}-500\le \text{RHS}-500$

$14500\le 2m$

DivIneq

$\text{LHS}/2\le \text{RHS}/2$

$7250\le m$

Since the expression is the same, the second inequality is solved following the same steps.

$2m+500\le 18000$

SubIneq

$\text{LHS}-500\le \text{RHS}-500$

$2m\le 17500$

DivIneq

$\text{LHS}/2\le \text{RHS}/2$

$m\le 8750$

The resulting inequality is obtained by combining these results, similar to how it was written in Part A.

$$\begin{array}{c}7250\le m\le 8750\end{array}$$

Since the inequality is non-strict, Diego can start looking to buy the car after he saves some amount of money from $\$7250$ to $\$8750,$ including these values. After he saves enough, he can stop saving money.

c To graph the solution set of the compound inequality set in Part B, the solution set of each individual inequality will be graphed first. The solution set of $m\ge 7250$ is made of all the numbers to the right of $7250,$ including $7250.$

Similarly, the graph of the solution set of inequality $m\le 8750$ is made of all the points to the left of $8750,$ including $8750.$

Finally, the solution set of the compound inequality is made of every number that both graphs share. In this case, the overlapping space from $7250$ to $8750,$ inclusive.

At the beginning of the lesson, it was asked how far Vincenzo's house was from the beach. The following information was given.

• Vincenzo will drive at $50$ miles per hour.
• He will arrive home in $40$ to $55$ minutes.

Then, the following questions were asked.

Solution

a To find the distance from the beach to Vincenzo's house, a modification of the speed formula will be used.

$$\begin{array}{c}r=\frac{d}{t}\end{array}$$

In this formula, $r$ is the rate or speed, $d$ is the distance traveled, and $t$ is the time duration of the movement. Since two time estimations are given, the time has to be the isolated quantity in the formula.

$r=\frac{d}{t}$

▼

Solve for $t$

MultEqn

$\text{LHS}\cdot t=\text{RHS}\cdot t$

$tr=d$

DivEqn

$\text{LHS}/r=\text{RHS}/r$

$t=\frac{d}{r}$

It is given that it will take Vincenzo between $40$ and $55$ minutes to get home. Therefore, the time $t$ in the formula has to be at least $40.$ This is the same as writing that $t$ is greater than or equal to $40$ minutes.

$$\begin{array}{c}t\ge 40\text{ minutes}\end{array}$$

However, since the speed is given in miles per hour, the $40$ minute interval must be written in terms of hours. This can be done using a conversion factor.

$$\begin{array}{c}40\cancel{\text{ minutes}}\cdot \frac{1\text{ hour}}{60\cancel{\text{ minutes}}}=\frac{2}{3}\text{ hours}\end{array}$$

The inequality can be rewritten using this information.

$$\begin{array}{c}t\ge \frac{2}{3}\text{ hours}\end{array}$$

Now the formula for the time in terms of the distance and the speed can be substituted into the inequality. It is given that Vincenzo will drive at $50$ miles per hour. This means that $r$ is equal to $50.$

$$\begin{array}{c}\frac{d}{50}\ge \frac{2}{3}\text{ hours}\end{array}$$

The same can be done with the upper limit of $55$ minutes. The first to note is that the statement can be written as $t$ is less than or equal to $55$ minutes

$$\begin{array}{c}t\le 55\text{ minutes}\end{array}$$

Using the same conversion factor, the time can be rewritten in terms of hours.

$$\begin{array}{c}55\cancel{\text{minutes}}\cdot \frac{1\text{ hour}}{60\cancel{\text{minutes}}}=\frac{11}{12}\text{ hours}\end{array}$$

And using this time span, the inequality can be rewritten.

$$\begin{array}{c}t\le \frac{11}{12}\text{ hours}\end{array}$$

Finally, the expression in terms of the distance can be substituted into the inequality.

$$\begin{array}{c}\frac{d}{50}\le \frac{11}{12}\text{ hours}\end{array}$$

Since the possible values for $d$ satisfy both inequalities, the compound inequality is written with and.

$$\begin{array}{c}\frac{d}{50}\ge \frac{2}{3}\text{ and }\frac{d}{50}\le \frac{11}{12}\end{array}$$

Recall that compound inequalities written with and can be rewritten as follows.

$$\begin{array}{c}\frac{2}{3}\le \frac{d}{50}\le \frac{11}{12}\end{array}$$

Finally, to find the possible values of $d,$ each section of the compound inequality is multiplied by $50.$

$\frac{2}{3}\le \frac{d}{50}\le \frac{11}{12}$

▼

Solve for $d$

MultEqn

$\text{LHS}\cdot 50=\text{RHS}\cdot 50$

$\frac{50\cdot 2}{3}\le d\le \frac{50\cdot 11}{12}$

Multiply

Multiply

$\frac{100}{3}\le d\le \frac{550}{12}$

CalcQuot

Calculate quotient

$33.333333\dots \le d\le 45.833333\dots$

RoundDec

Round to $1$ decimal place(s)

$33.3\le d\le 45.8$

This means that Vincenzo's house is somewhere between $33.3$ and $45.8$ miles away from the beach. To graph the solution set of this compound inequality, each point between $33.3$ and $45.8$ will be included. Note that the compound inequality is non-strict, so the endpoints are denoted with closed circles.

b From Part A, it was written that the possible values for $d$ can be written using a compound inequality.

$$\begin{array}{c}33.3\le d\le 45.8\end{array}$$

This compound inequality can be written with the word and.

$$\begin{array}{c}d\ge 33.3\text{ and }d\le 45.8\end{array}$$

The values of $y$ are the values that fall outside of this compound inequality, so each individual inequality should be reversed. Note that because $33.3$ and $45.8$ are included in the original interval, these values cannot be included in the solution set for $y.$ Since $d$ is greater than or equal to $33.3$ miles, then $y$ must be less than $33.3$ miles.

$$\begin{array}{c}y<33.3\end{array}$$

Similarly, since the values of $d$ are less than or equal to $45.8$ miles, $y$ has to be greater than $45.8$ miles.

$$\begin{array}{c}y>45.8\end{array}$$

The possible values of $y$ are those that are either less than $33.3$ or greater than $45.8.$ Therefore, the compound inequality is written with or.

$$\begin{array}{c}y<33.3\text{ or }y>45.8\end{array}$$

The solution set of this compound inequality is made of every number left of $33.3$ and right of $45.8.$