Absolute Value Inequalities

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From the course by University of California, Irvine

Pre-Calculus: Functions

298 ratings

University of California, Irvine

Pre-Calculus: Functions

298 ratings

This course covers mathematical topics in college algebra, with an emphasis on functions. The course is designed to help prepare students to enroll for a first semester course in single variable calculus. Upon completing this course, you will be able to: 1. Solve linear and quadratic equations 2. Solve some classes of rational and radical equations 3. Graph polynomial, rational, piece-wise, exponential and logarithmic functions 4. Find integer roots of polynomial equations 5. Solve exponential and logarithm equations 6. Understand the inverse relations between exponential and logarithm equations 7. Compute values of exponential and logarithm expressions using basic properties

From the lesson

Linear Equations and Inequalities. Absolute Value Equations and Inequalities.

In this module, we will learn to solve linear and absolute value equations and inequalities. We will also explore a range of story problems which can be solved utilizing linear equations.

Absolute Value Equations 6:35

Absolute Value Inequalities 6:32

Inequalities Involving Radicals 5:49

Meet the Instructors

Dr. Sarah Eichhorn
Associate Dean, Distance Learning & Tenured Lecturer
Dr. Rachel Cohen Lehman

[MUSIC] Let's learn how to solve Absolute Value Inequalities.

Let's solve the following inequality for X and put our answer in interval

notation. If the absolute value is < or = 9, that

means what's inside the absolute value is between -9 and 9.

In other words, we are going to use the following factor to help us.

If the absolute value of Y is < or = to a Means that -a <= y <= a.

Lets apply that here. Namely, -9 <= 5 minus 2 X is less than or

equal to 9. Now remember, when solving this type of

compound inequality, we can work with all the parts at the same time.

So let's start by subtracting 5 everywhere.

Which gives us negative 9 minus 5. <= 5 - 5 - 2x <= 9 - 5,

or -14 <= -2x <= 4. And now let's divide everything by

negative two, however remember when we divide by a negative, we better flip or

reverse the direction of both of these inequalities.

Which give us 7 >= x >= -2. Which we usually write in the other

direction, -2 <= x <= 7. Let's look at this on the number line.

Let's say this is -2 and this is 7. x is greater than or equal To negative 2,

so x can equal negative 2. And then we go to the right, however x

has to be less than or equal to 7, so we stop here.

And we're asked to put our answer in interval notation.

Therefore our answer would be closed bracket.

So we want to include -2, up to 7. Again, closed bracket, because we want to

include 7. Alright lets look at another example.

[SOUND] Lets solve this inequality, and again put our answer in interval

notation. Done.

Now the difference here is that we have greater than rather than less than like

the last example. Now if the absolute value is greater than

1, then what's inside the absolute value is either larger than 1 or smaller than

-1. In other words, we're going to use the

following fact to help us solve this problem.

That the absolute value of y greater than a means that y is greater than a or y is

smaller than negative a. And this is what we're going to apply

here. And this gives us that 3 + 2x is either >

1 or 3 + 2x < -1. And now, when solving this type of

compound inequality, we work with each side separately.

So let's start with this 1st inequality here.

We'll bring the 3 to the right or subtract 3 on both sides, which gives us

2x is greater than 1 minus 3 or 2x is greater than negative 2.

And now dividing both sides by 2. We get x is greater than -1.

And what about this 2nd inequality? Again we'll bring the 3 to the right-hand side,

which means that 2x is less than -1 minus 3.

Or 2x is less than -4. Again, dividing by 2, gives us x is less

than -2. And since these inequalities are joined

by the word or, then the solution is all x values that make at least one of these

True. Let's look at this on the number line.

Let's say that this is -2 and this is -1. Greater than -1, we do not want x to

equal -1, so we put an open circle. Greater than we go to the right.

X less than -2. Again, we put an open circle because X

can not equal -2. Less than, we go to the left.

But again we're asked to put our answer in interval notation.

And doing this gives us our answer of negative infinity up to -2 open

parentheses because we do not want x to equal -2.

union, again open parentheses -1 up to infinity.

And this is how we solve Absolute Value Inequalities.

Thank you, and we'll see you next time. [MUSIC]

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