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In each of the following exercises, solve the given inequality.$$\frac{(x-1)(x+2)}{x-3} \leq 1$$

$$x<3$$

Algebra

Chapter 0

Reviewing the Basics

Section 5

Solving Non-Linear Inequalities

Equations and Inequalities

Missouri State University

Campbell University

Oregon State University

Lectures

05:39

In each of the following e…

03:28

03:06

02:30

03:40

03:41

01:20

For the following exercise…

01:50

02:31

03:23

05:55

04:15

02:07

Solve each inequality alge…

All right, we've got a question here. X minus one X plus two X minus three is greater than one. Sorry. Less than or equal to one. Okay, so to solve for this inequality, what you gonna do is you're gonna take each of your factors here and just set them equal to zero. And you will be able to solve your values of X. Alright, So X minus one was set to equal was set equal to zero. Then you know your ex would have to be one if exports to was set to equal zero. You know, your X would be equal to negative two X minus three was set equal to zero. And you know your X would be equal to three and take a number line you're gonna draw. Excuse me. I'm gonna draw a number. Line is gonna label your ex values negative to one and then three. And here you have ah, less than an equal to symbol. So for your values that you obtained your excited that you've seen from the numerator, you'll put solid symbols there. You've got a one on the negative too. But since your X minus three, where X is equal to three came from the denominator. We would have to put a hollow circle there. And the reason why is because if you actually plug in a three here three months, there would be zero. And if you have a zero on the denominator, you would have an undefined function. Therefore, we could not make that into a solid point there. All right, now you're gonna pick values within these regions and determine whether or not those regions of X values would satisfy the inequality. So let's start from a value between three and infinity, and we'll choose a random value and I'll go ahead and pick four. So we'll do four minus one times, four plus two over four, minus three. And we're just plugging that four and for X. So you get three times six divided by one, which would just be a positive 18. And we know a positive 18 is not gonna be less than or equal to one, and therefore this region is not satisfied. Pick a number between one and three. You could pick too minus one, two plus two tu minus three, and you would get a negative four over negative one. You get negative. One negative. Four were negative. One is four. That's also not gonna be less than able to. Um, excuse me. That's also not going to be less than one. I might be doing something wrong here. Hold on to minus one. It's 11 times four. It's 44 Divided by too much would actually give us a negative for. Okay, that's right. All right, so we have a negative four and negative four is actually less than one. Therefore, this region is actually satisfied. All right, let's plug in a value between one and negative to we could plug in zero. So it just be negative. One multiplied by two, divided by a negative three. That will be negative. Two divided by negative three, which is equal to two thirds. And then, um, you have two thirds is less than negative one, So we can confirm that exists in this region as well. Okay, now, let's pick a value between negative two and negative infinity. We could choose negative three negative three minus one negative three plus two, divided by a negative three minus three. So you would have a negative four multiplied by negative one, which is 44 divided by a negative six. So four over negative six is certainly less than one. Therefore, this region is also satisfied. So basically all the values less than three satisfied this inequality. And so you would just write that when excess less than three all of the values of X less than three would satisfy this inequality. All right, well, I hope that clarifies the question. Thank you so much for watching.

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