[fs-toc-h2]Basics

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Solving and writing linear inequalities is almost exactly the same as solving and writing linear equations; the only thing that's different is the equal sign.

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[fs-toc-h3]Less than: \(<\)

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Whatever is left of this inequality is less than whatever is right of it.

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For example:

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\[3 < 22\]

\[x < 12 \quad \]

\[ \text{(this tells us that } x \text{ must be a value less than 12)}\]

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[fs-toc-h3]Greater than: \(>\)

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Whatever is left of this inequality is greater than whatever is right of it.

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[fs-toc-h3]Less than or equal to: \( \leq \)

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Whatever is left of this is less than OR equal to whatever is right to it.

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\[x \leq 12 \quad \]

\[ \text{(this is saying that } x \text{ could either be less than 12 or equal 12)}\]

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[fs-toc-h3]Greater than or equal to \(\geq\)

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Whatever is left of this inequality is greater than or equal to whatever is right of it.

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Just remember, the alligator eats the bigger number. Whichever way the open side of the inequality is facing is the greater side.

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[fs-toc-h2]Solving inequalities

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If you know how to solve a linear equation, you can solve a linear inequality. The process is almost identical other than one step. Refer to the inequality below:

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\[-5x + 13 > 43\]

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Just like any equation, we can begin by subtracting 13 from both sides:

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\[-5x > 30\]

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Now it's time to divide by -5, and this is where inequalities differ from equations. When multiplying or dividing by any negative number, you must switch the way the inequality is facing. So when you divide by -5, our final answer becomes:

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\[x < 6\]

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\(x\) is going to be less than 6.

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[fs-toc-h4] Example 1

Let's start off by simply solving for \(x\). Subtract \(3x\) from both sides and add 3 to both sides to end up with:

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\[-2 \geq x\]

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We now know that -2 is greater than or equal to \(x\).

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-5 and -3 are both less than -2, and -2 is equal to -2. This means that the only number which is NOT a solution is -1, making A the correct answer choice.

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If you have extra time, a good way to check is to simply plug in the answer:

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\[3(-1) - 5 \geq 4(-1) - 3\]

\[-8 \geq -7\]

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-8 is NOT greater than or equal to -7, meaning we're right.

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[fs-toc-h2]Multivariable inequalities

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Just like equations which can have two variables, for example: \(y = 3x + 4\), inequalities can too. Let's take our linear equation and turn it into an inequality:

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\[y \leq 3x + 4\]

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To find the solution sets of this equation, the easiest way is to go into Desmos and graph it.

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In this inequality, any coordinate that lies on the line OR in the shaded region is a solution, anything outside of it is not.

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The first coordinate shown above, (-3, 7), is NOT a solution to the inequality, as it lies outside of the boundaries.

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Our next two though, since they lay either on the line or within the shaded area, are solutions.

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If we switch the less than or equal to sign into a regular less than sign, notice what happens.

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The line has now become dotted. If it is a regular inequality and not an “and equal to” inequality, anything on the line is no longer a solution. So in this case, (2, 8) isn't a solution anymore.

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We can also check if a coordinate set is a solution or not by plugging in the respective \(x\) and \(y\) values and seeing if the inequality is true or not.

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Let's look at the inequality \(y > 3x + 7\) and see if the following points are solutions or not without graphing.

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Try (2, 9):

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\[y > 3x + 7\]

\[9 > 3(2) + 7\]

\[9 > 12\]

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Since this inequality is simply untrue, (2, 9) is NOT a solution.

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Try (-3, 8):

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\[y > 3x + 7\]

\[8 > 3(-3) + 7\]

\[8 > -2\]

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Since 8 is i fact greater than 2, this inequality is true, deeming (-3, 8) as a solution.

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[fs-toc-h2]Writing inequalities

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Writing inequalities, alongside systems of inequalities, is the most tested subtopic within inequalities. Let's look at some examples. Try to write out an inequality before looking at the answer.

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Dylan sells lemonade for $3 a cup. By the end of the day, he made more than $51.

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Let's assign \(c\) to the amount of cups he sells. If he sells \(c\) cups for $3 each, the total money he would make is \(3c\). It's given that this amount is more than 51, so our final inequality is:

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\[3c > 51\]

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William sells burgers for $6 each and fries for $3 each. By the end of the day, he made less than $200.

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Using \(b\) to represent burgers and \(f\) to represent fries, the amount of money made for \(b\) burgers can be written as \(6b\), and fries can be written as \(3f\). Since he made less than $200, we’ll use the less than sign, so our final inequality will be:

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\[6b + 3f < 200\]

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[fs-toc-h4] Example 1

The first piece of information given is that there is a $25 service fee, so 25 will have no variable attached to it in the equation (this number is known mathematically as the initial value or y-intercept).

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We’re also told that it costs $10 an hour, with the variable \(t\) representing hours. So the total cost after \(t\) hours will be \(10t\) plus the initial $25 service fee, we can write this out as:

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\[25 + 10t\]

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Finally, we’re given that the person intends to spend a maximum of $75, meaning the total price has to be equal to or less than $75, this gives us a final inequality of:

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\[25 + 10t \leq 75\]

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Or answer choice D.

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However, most inequality questions will show up as a system (two equations). This doesn't change any of the methodology though. The only difference is that you have to write two equations rather than one.

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[fs-toc-h4] Example 2

First, let's take care of one equation at a time. At the end of the question, we’re told that \(x\) is the possible number of $50,000 policies and \(y\) is the possible number of $100,000 policies. We’re also told that he did NOT meet his goal of selling 57 policies, meaning he sold less than 57. So, the sum of \(x\) and \(y\) (both types of insurance policies) is less than 57, giving us a first inequality of:

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\[x + y < 57\]

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Looking at the answer choices, neither B nor D include this inequality, meaning you can cross them out.

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Next, we have to look at the price. It’s given that he sold more than $3,000,000 worth of policies. So the sum of $50,000\(x\) (the total price of \(x\) policies he sold), and $100,000\(y\) (total price of \(y\) policies he sold) will be more than $3,000,000, giving us an inequality of:

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\[50,000x + 100,000y > 3,000,000\]

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Our two equations side by side are:

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\[x + y < 57\]

\[50,000x + 100,000y > 3,000,000\]

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This corresponds to answer choice C.

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[fs-toc-h4] Example 3

This question requires us to write and solve an inequality. Let's start by writing an expression. We can assign the number of hours to the variable \(h\).

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Since the rate is $60 per hour, the total amount she spends is \(60h\) plus the initial fee, $10, giving us:

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\[60h + 10\]

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She wants to spend no more than $280, meaning that 280 is greater than or equal to the total cost. Making our final inequality:

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\[280 \geq 60h + 10\]

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All we have to do now is solve for \(h\):

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\[280 \geq 60h + 10\]

\[270 \geq 60h\]

\[4.5 \geq h\]

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However, the boat rental is only available for a whole number of hours. You may be tempted to round up to 5 hours but this would be incorrect. In a real-world situation, to rent out the boat for 5 hours, you would need enough money (or more) for 5 hours. In this case, she only has enough for 4.5 hours.

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Since she can't rent out half an hour, we have to round down to 4 hours, making 4 the final answer.