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In $3-14,$ solve and check each inequality.$$\frac{2}{x}+\frac{3}{x}<10$$

$\left\{x | x<0 \text { or } x>\frac{1}{2}\right\}$

Algebra

Chapter 2

THE RATIONAL NUMBERS

Section 8

Solving Rational Inequalities

Fractions and Mixed Numbers

Decimals

Equations and Inequalities

McMaster University

Baylor University

Idaho State University

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Okay. Hey, guys. So in this problem, we're giving inequality to over X plus three over eggs left in 10. As in the problems that we have just been dealing with are variable access in the denominators. So it's going to make this problem little slightly tricky. So let's get on with the problem. First step, let's combine all of our rational functions toe one rational function on put it on one side of the inequality. We're gonna do that. Oh, and luckily, oh, we've been given two rational functions that have the same common denominator. So we consume easily. Cem easily. Just simplify this to overexpose three over X is just five over X and five over X is less than 10. Now we're going to subtract are 10 to the left hand side of the inequality and basically subtracting time. Both sides gives us five over X minus 10 less than zero. Now, we're gonna find a common denominator to combine these two numbers. So our common denominator is is ex. So this is gonna simplify to fuck. Basically, expand in a sense. Five over X minus 10 x over X less than zero. Let's bring this up a little bit, so it's gonna be easier for us to solve more space. So if you simplify five over X minus 10 X over X, you're gonna get five minus 10 X over X is less than zero. Now we're gonna factor out if I just to make our lives easier in the future. And so when we factor the five you're left with one minus two x over X. Next, we're going to get rid of this five by multiplying by 1/2 on both sides. Really? These these two steps that I just did here just for to make our life easier in the next step will explain what the next step is again. So we have one over to X over X less than zero. Now, as I go into the next whiteboard, Paige, what we're gonna do is we're gonna solve for when the numerator is zero with the denominator zero. Put it on a number line and check on check on the regions that the numbers create and check which regions satisfy this inequality. So let's get that started. Oh, our inequality basically simplifies to this Inequality one over one minus two x over X is less than zero. So we saw for when one minus two X is equal to zero and that is actually equal to one have and we solve. When X is equal to zero, that's just X equals zero. So on our number line, let's Well, it's a bad number line. So honor, number line let's make a little bigger shovel is plot the 0.0 and 1/2. So we have zero right here, 1/2 right here, and we split this into regions one, two and three. We already know that these are gonna be open circles here because we're doing less than and that we're not doing less than are equal to or anything. So there's no closed circles, only open circles. So we're gonna choose a value from each region, plug it into this inequality and see if the inequality holds. So in region one, let's choose the value negative one if you choose negative one in region one. So what's gonna happen is you're gonna plug in one minus two times negative one over negative one, which is equal to one plus two over negative one, which is equal to three over negative one which is equal to negative three and is negative. Three less than zero. Yes, it is. So Region one is in the solution set. Now, let's take a look at region to we're gonna plug in the value 1/4 when you plug in the value 1/4 you're gonna do one minus two over 14 times, 1/4 over. 1/4 one minus toot. All times 1/4 is the same thing is one minus 1/2 over 1/4. And that's gonna simplify too. 1/2 over 1/4. And that's gonna be equivalent to two and is too less than zero. No, it is not. So we know that region to is not in the solution set. Now let's take a look at Region three. Region three is going to be. What value should we choose? We're gonna choose the value one. Choose simple values, guys, just to make it easier for you yourselves. So when you call you in one minus 21 times one over one, you're gonna get one minus two. Divided by one. I'm gonna give you one minus two is minus one over one, which is equivalent to minus one and his minus one less than zero. Yes. So the third region is in the subset as well. And if we want to graft the solution set completely, this is what it's gonna look like. It's gonna look like all the values left of zero while the values right of 1/2. Now, if we want to write this out, it's gonna look like it's gonna It's gonna look like ex less than zero or X greater than 1/2. And this is our answer to the inequality problem. This is the answer. Explicit answer. This is the This is the answer. This is the graphical representation of the answer. And this entire process is the verification process to show why Our answer, our solution said is correct. Thanks for listening, guys. I hope this helped and yeah.

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