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In $13-24,$ divide and express each quotient in simplest form. In each case, list any values of the variables for which the fractions are not defined.$$\frac{c^{2}-6 c+9}{5 c-15} \div \frac{c-3}{5}$$

$1 ; c \neq 3$

Algebra

Chapter 2

THE RATIONAL NUMBERS

Section 3

Multiplying and Dividing Rational Expressions

Fractions and Mixed Numbers

Decimals

Equations and Inequalities

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In mathematics, the absolu…

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Now, before we actually start solving this problem, I see that I've got a quadratic try. No meal in the C squared, minus 60 plus nine. So I want a real quick Just do real quick refresher on how to factor those just in case anybody needed it. Um, that is a quarterback. Try no meal. Meaning it is written in the form a X squared plus B x plus C. Meaning you should look at it as the three terms that it is. You have, ah, squared term. You have the variable to the first power, and then you have a constant that has no variable with it at all. Okay, there are many ways to teach this. The way I teach is I call it the ex meth that there's a bunch different ways to show It doesn't really matter. I have it as an ex. The idea is, what you're doing is you want to take your a times your C, and then you want to find two numbers that multiply to the same value as a time See, but add to your be value you want to side numbers that will multiply to a time. See but we'll add to be okay. Meaning, for example, if you look up at this problem here, we have an A value of one, an invisible one. In front of that C squared. We have a B value of negative six that negative six in front of the sea, and then we have a see value in nine. So if I was gonna factor C squared minus six C plus nine, then I'm looking for two numbers that multiply to nine because one times nine is nine and add to negative 682 numbers that multiply the positive nine. But add to negative six. In this case, that would be negative. Three and negative three right negative times a negative is a positive and negative. Three times negative three would give me a positive nine negative three plus negative three would give me negative six in this case because our A value in this case, because our A value is one. We already have our factor. We can just go straight to C minus three and C minus three. If are a value was not one, we'd have extra steps, but we don't have to deal with that in this problems case. All right, There's a quick review on Factoring in general. Um, hopefully that helped. If not, you may to go back and review that stuff because you have learned factoring before this section already. Now, as far as solving this actual problem, Okay, as far as solving this actual problems will get rid of all of this stuff. We do have division going on here. We know we aren't supposed to leave it as division. We need to keep our first fraction the same. So we'll just rewrite it as c squared. Minus 60 plus nine over five C minus 15. We need to change that division to multiplication and flip or find the reciprocal of our second fraction. Meaning now is five over C minus three. Now we can work on factoring each piece separately and see what cancels out. We already talked about what C squared minus six C plus nine factors to that would be C minus three and C minus three, which, yes, you could also write that as C minus three squared if you wanted to. Okay, five C minus 15 week in G C f. Factor that because five C and 15. Both have five in common. So five c, divided by five would be see minus 15 divided by five would be three. And then we've got times five over C minus three. Everything is factored. So if anything matches up between the numerator and the denominator, we can cancel it out. For example, we have a C minus three in my numerator and a C minus three. In my denominator, those can cancel. We have a C minus three in my numerator and a C minus three and my denominator again. So those canceled. There is a five in the numerator, and there is a five in the denominator, So it was also cancelled, so it looks like literally everything cancelled here. Meeting Our answer is simply one. We do, however, need to still determine any values that could cause this to be undefined. Remember, meaning that if we had zero in the denominator zero on the bottom, that would be an issue. So I have to look at all of my denominators both pre and post reciprocal to make sure I account for all the terms. So here, I've got five C minus 15. Here I have C minus three. So five C minus 15 is the denominator, and we do not want it to equal zero. Hence the slash through the equal sign C minus three is a denominator. And again, we do not want a T equals zero. Well, we saw what happened when we had five C minus 15. We actually factored that five out. So actually, we only have to solve this out once. But we can do it both just to make sure we are confusing anybody. So add 15 over. You've got five. C does not equal 15 divide by five and you've got that see does not equal three. Which is the same thing you would get on the second equation. You would also get See does not equal three meaning. Our final answer is this equation simplifies to one as long as C does not equal positive three

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