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Here are a few recommended readings before getting started with this lesson.
Ignacio, has just been offered a paid internship as a junior rocket scientist for a space exploration company. The company has offered Ignacio two options for how to be paid.
Ignacio is unsure which is the best option. Help him make the best choice by answering the following questions. His salary depends on it!
Similar to equations, inequalities have some properties that allow their manipulation without changing their solution set. When these properties are applied, an equivalent inequality is obtained. The Addition and Subtraction Properties are two of them.
Adding the same number to both sides of an inequality generates an equivalent inequality. This equivalent inequality will have the same solution set and the inequality sign remains the same. Let $x,$ $y,$ and $z$ be real numbers such that $x<y.$ Then, the following conditional statement holds true.
If $x<y,$ then $x+z<y+z.$
Identity Property of Addition
Rewrite $0$ as $z−z$
Commutative Property of Addition
$-a−b=-(a+b)$
Add parentheses
If $x<y,$ then $x+z<y+z.$
Subtracting the same number from both sides of an inequality produces an equivalent inequality. The solution set and inequality sign of this equivalent inequality does not change. Let $x,$ $y,$ and $z$ be real numbers such that $x<y.$ Then, the following conditional statement holds true.
If $x<y,$ then $x−z<y−z.$
Identity Property of Addition
Rewrite $0$ as $(-z)−(-z)$
Commutative Property of Addition
Remove parentheses
$-a−b=-(a+b)$
Commutative Property of Addition
Add parentheses
If $x<y,$ then $x−z<y−z.$
The Addition and Subtraction Properties of Inequalities do just a portion of the work. That is because they do not create the ability to isolate variable terms that contain coefficients. Not to worry, the Multiplication and Division Properties can help in these cases. Together, these properties help to solve inequalities by creating equivalent inequalities.
Multiplying both sides of an inequality by a nonzero real number $z$ produces an equivalent inequality. The following conditions about $z$ need to be considered when applying this property.
Positive $z$ | If $z$ is positive, the inequality sign remains the same. |
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Negative $z$ | If $z$ is negative, the inequality sign needs to be reversed to produce an equivalent inequality. |
For example, let $x,$ $y,$ and $z$ be real numbers such that $x<y$ and $z =0.$ Then, the equivalent inequalities can be written depending on the sign of $z.$
The case when $x<y$ will be proven. The remaining cases can be proven following a similar reasoning. Before starting the proof, the following properties of real numbers need to be considered.
Using these properties, the following conditional statements can be proven.
Each conditional statement will be analyzed separately.
Distribute $(-z)$
$(-a)b=-ab$
$a−(-b)=a+b$
$LHS+zy>RHS+zy$
If $x<y$ and $z<0,$ then $zx>zy.$
Dividing both sides of an inequality by a nonzero real number $z$ produces an equivalent inequality. However, the following conditions need to be considered.
Positive $z$ | If $z$ is positive, the inequality sign remains the same. |
---|---|
Negative $z$ | If $z$ is negative, the inequality sign needs to be reversed to produce an equivalent inequality. |
For example, let $x,$ $y,$ and $z$ be real numbers such that $x<y$ and $z =0.$ Then, the equivalent inequalities can be written depending on the sign of $z.$
The case when $x<y$ will be proven. The remaining cases can be proven following a similar reasoning. Before starting the proof, the following properties of real numbers need to be considered.
Using these properties, the following conditional statements can be proven.
Each case will be analyzed separately.
If $x<y$ and $z>0,$ then $zx <zy .$
Put minus sign in numerator
$-(b−a)=a−b$
Write as a difference of fractions
$LHS+zy >RHS+zy $
If $x<y$ and $z<0,$ then $zx >zy .$
Knowing which property to use when solving an inequality is of importance because this can minimize mistakes. In the applet, select the property used to produce each equivalent inequality.
Applying the Properties of Inequalities to one inequality will produce equivalent inequalities. These equivalent inequalities can have a simpler form, making their solutions more straightforward to identify. Since the equivalent inequalities have the same solutions, the solution set of the original inequality can be determined.
Here, a circle representing the boundary point is drawn on the number line. If the inequality is strict, the circle is open. If the inequality is non-strict, the circle is closed. For this example, the inequality is strict, and the boundary point is $6.$ Then, an open circle will be drawn on the number line on the number $6.$
Finally, the rest of the solution set will be shaded by drawing an arrow that goes along the solution set and starts on the boundary point. For this situation, the solution set corresponds to all numbers less than $6,$ which means the arrow will be along the left of the boundary point.
It is worth mentioning that the graph of inequalities whose solution sets are all the real numbers are represented with bidirectional arrows that cover all the number line.
An object must travel at a speed of at least $11.2$ kilometers per second to escape Earth's gravitational field. At Gravitasi Z, engineers built a rocket and were tasked with the mission of exploring celestial objects faraway from Earth. However, there is a big problem: The rocket was made to travel at a speed of only $8$ kilometers per second!
Gravitasi Z's engineers plan to improve the rocket in order to accomplish its mission. Solve the following predicaments to help them succeed.
at leastmeans greater than or equal to. Therefore, the inequality is non-strict and the symbol must be $≥.$
$s+8≥11.2$ | |
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Solution Set | Boundary Point |
$s≥3.2$ | $3.2$ |
Since the inequality is non-strict, a closed circle will be drawn on the number line on its boundary point $3.2.$
Now, the rest of the solution set needs to be shaded. Because the speed added needs to be greater than or equal to $3.2,$ the region on the right of the boundary point will be shaded.
This corresponds to Graph II.
Ignacio is an employee at Gravitasi Z. He feels unsure about the success of their space exploration, so he decides to diversify his money making opportunities by investing in cryptocurrency. He has found a crypto exchange company that does all of the actual investing work for him.
The exchange company charges a subscription of $$299$ plus a commission of $$5$ for every cryptocurrency bought. Also, for an additional fee of $$35,$ each cryptocurrency can be sold again later. Ignacio wants to determine the number of cryptocurrencies that will make this investment profitable. Fractions of cryptocurrencies can be bought.
$LHS−5c>RHS−5c$
Subtract term
$LHS/30>RHS/30$
Use a calculator
Round to $2$ decimal place(s)
$35c>299+5c$ | |
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Solution Set | Boundary Point |
$c>9.97$ | $9.97$ |
Now, the rest of the solution set needs to be shaded. Because the number of cryptocurrencies bought needs to be greater than $9.97,$ the region on the right side of the boundary point will be shaded.
This corresponds Graph I.
Gravitasi Z wants to sale coffee mugs to the families of their astronauts — at a low price, of course. A company that produces mugs wants to charge Gravitasi Z a fixed amount of $$185$ for production plus a tax of $$5$ per mug. Gravitasi Z plans to sell each mug for $$18.$
The space exploration company, busy with making rockets, decides to hire a financial advisor to make these calculations. They hope to make at least a small profit. Find the answers to the following situations to become Gravitasi Z's trusted financial advisor.
$185+5m<18m$ | |
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Solution Set | Boundary Point |
${x∣xis an integer greater than14}$ | $14$ |
Since the inequality is strict, an open circle will be drawn on the number line on its boundary point $14.$
Now the solution set needs to be shaded on the number line. Since integers represent the number of mugs, only the points representing integers should be plotted on the number line instead of shading the region greater than $14.$ The points will be closed as they are the part of the solution set.
This corresponds to Graph IV. Congratulations! Gravitasi Z was impressed by this data, and it looks like someone has a new job as financial advisor.
Ignacio is feeling stress from doing so much rocket science, so he goes with his friend Tearik to an escape room adventure! After some tough challenges, they have reached the final level where only one can win! They stand before two giant doors. Each can be unlocked only by solving a riddle.
The riddles are as follows.
Ignacio's Riddle |
Two less than the product of four and a number is less than one-half of the sum of eight times the number and six. |
Tearrik's Riddle |
Three less than the product of five and a number is greater than or equal to twice this number plus three times the sum of this number and two. |
It seems like each riddle is an inequality! All they have to do now is give the correct number. Answer the following questions to decipher how Ignacio and Tearrik can unlock their respective doors to get out of the escape room.
Graph:
Graph:
$LHS⋅2<$