in-3-20-perform-the-indicated-additions-or-subtractions-and-write-the-result-in-simplest-form-in--16

Question

In $3-20$ , perform the indicated additions or subtractions and write the result in simplest form. In each case, list any values of the variables for which the fractions are not defined.
$$
\frac{1}{a^{2}-a-6}-\frac{1}{2 a^{2}-7 a+3}
$$

Georgiann Andersen · Accepted Answer

it was simplified while wearing expression. This is 18. We have one over a squared minus a minus six. We&#x27;re going to subtract one over to a square in a seven A. That&#x27;s three. We&#x27;re gonna first factor out each denominator. So this one turns out to be a minus. Hey, plus two, 2nd 1 it is going to be a minus three, thanks to a I mean this morning. Now we can create adopts nominators. Um, you&#x27;d have to be the same. This is going to change, too, to a minus one from the top to a minus. One times a minus. Three times a plus two. We&#x27;re going to subtract. We&#x27;re gonna multiply the second run by a plus two. And that&#x27;s gonna end up being now to a minus one minus a minus two all over. April. It&#x27;s too. You minus three to a minus one. We&#x27;re gonna have a minus three on the top, and we&#x27;re gonna have a plus two a minus three and to a minus one. We can eliminate that. A minus three. That&#x27;s leaving us with one on the top, A plus two to a minus one on the bottom. That&#x27;s a ce faras. We can simplify that now, the, uh, numbers that would make this undefined meaning they cannot be a negative too three or ever.

Derrick Hanson · Answer

This is an interesting question because not only do we not have a common denominator, but this is the first problem out of this section where we can actually factor the denominators. Also, if you don&#x27;t have a common denominator and you are able to factor them, you should, because you might end up finding that it&#x27;s not as hard to find a common denominator as it looks. Right now. It may end up still being Justus hard, but you should factor just in case. So, looking at what we have here for my first fraction, we have a denominator of a squared minus four. There is that is a quadratic because we have in a squared, we&#x27;ve got two terms, we&#x27;ve got subtraction, and both of those terms are capable of being squared it. So this would be a difference of squares, meaning the square root of a squared is a. The square of four is, too. So are factors would be a plus two and a minus two. If you have for gotten difference of squares, you should go back and review that before continuing on here. Looking at my second fraction, we also have a quadratic, but this one is not a difference of squares because we have addition. So it&#x27;s not a difference, and we can&#x27;t square both these. However, this one can be factored using G CF because a squared into a both heaven a in common so we can pull that a out a square divided by a would be a plus to a divided by a would be to. So now, when we look at these denominators, it&#x27;s still not going to be super easy. Not gonna act like this is not a tougher problems some other ones, but notice that we have an A plus to in my first denominator and in a plus two in my second denominator. So because of that, I don&#x27;t have to account for those when finding a common one. They&#x27;re already in common, meaning. The only things that are uncommon is my first denominator has an A minus two, and my second denominator has an A. Neither of those have anything in common with each other, so the easiest thing to do is going to be the multiplying my first fraction by what it&#x27;s missing from the second fraction, which would be a and all multiply my second fraction. By what the first fraction has the dozen, which is a minus two. Meaning we&#x27;re just gonna multiple age fraction by the other fractions. Denominator. Okay, so for example, first fraction of two over a plus two a minus two. We&#x27;re going to be multiplying that by a But of course, we have to do that on both top and bottom denominator and numerator. For this to stay the same value. So that would give me to a on top two times A is to a on the bottom member. We don&#x27;t actually have to multiply these out. We can keep them separate. So we have a We have a plus two and we have a minus two. Looking at my second fraction, we have one over a times a plus two. So that&#x27;s a great, but it&#x27;s missing the A minus two. So we&#x27;re gonna have to multiply by the quantity of a minus two both on the numerator and the denominator top and bottom time. One times anything is just that. So it&#x27;s a minus two in the numerator. We really don&#x27;t need princes there. And the denominator we have a we have a plus two. And now we have the A minus two that we&#x27;ve multiplied in there. So recognized now, weaken. Take both of these two new fractions that we found. Not really new value. Is that the same value? They&#x27;re just They look different and we can rewrite it. So what we&#x27;re doing here is we&#x27;re taking to a over a times a plus, two times a minus two, and we&#x27;re subtracting that from a minus two over the exact same denominator. A times a plus, two times a minus two. All right, so do make sure you remember that this minus right here goes to both of these. So that means it&#x27;s a negative. A and a positive to essentially. Okay, two a minus. A would give me just a There is no numbers there. No constant in my first fraction. So you just have to remember that this negative does distribute to the two. So I guess you could think of this as a zero. If you wanted to write zero minus a negative, to would be positive, too. And then all of that denominator carries down. So this is what we&#x27;ve got. Notice we have a plus to the quantity in my numerator, and we have a plus to the quality in my denominator. That means we can actually simplify this further, which would give me just a one on top. Because remember, if everything cancels out when we&#x27;re dealing with division, it&#x27;s just the one. It&#x27;s not zero. And then we&#x27;d have a and a minus two, both left over in the denominator. That would be our answer for this, Certainly probably the longest, most intensive problem that we&#x27;ve done on this section so far. We&#x27;re not done yet, because remember, we still have to account for any values that could cause us to get zero in the denominator meeting, causing the whole question to be undefined. We actually have three different denominators. We have a plus two that we have to set not equal to zero. We have a minus two that we have to set not equal to zero, and we have a that we have to set not equal to zero. All of those have to be solved out. Subtract two over. We get a does not equal negative, too. Add to over. We get a does not equal positive, too. This one&#x27;s already solved so we get a does not equal zero. Remember, the reason we&#x27;re doing this is we are using the literal algebraic representation of saying we don&#x27;t want zero in the denominator. So any variable in the denominator, we simply set not equal to zero and see what we can&#x27;t have. So our final answer is one over a times a minus two, assuming that a does not equal negative too zero or positive, too.

Georgiann Andersen · Answer

we&#x27;re gonna do that. We&#x27;re gonna simplify this expression. This is a problem Number 16 and our expression is be over. You minus one minus one over to minus two B. Okay. First thing we&#x27;re gonna do is we&#x27;re going to put the second traction the second for action into its simplest form. We&#x27;re gonna take one over to minus two B and we&#x27;re gonna turn that into one over two times one minus b. Now we&#x27;re gonna make it so that it&#x27;s a positive variable. First, in order to do that, we&#x27;re going to change this to a negative one over two times being minus one. So now we can change our expression to be over being minus on, attracting a negative one over two times B minus one. Now you can see that main difference between the first fraction of the second production is that too, so we can multiply the first fraction by two. What to do at the top, you have to do the bottom. So then we can subtract Negative one over. Now we can go ahead and, uh, combine these so it&#x27;s gonna be to be subtract a negative one, which is gonna meet at one all over too. You mind us want, and that is this Father ist We can take that. That&#x27;s a simplest form. Now, in order for this to, um, not be undefined, be cannot be won, cause if you do the one it&#x27;s gonna make B minus one, that&#x27;s gonna be zero, and so that would be an undefined.

Derrick Hanson · Answer

first thing I noticed with this problem is that one term is written as a fraction and one term is not. I do not like trying to solve that math problems where I have different pieces being expressed in different forms. Just I think it just opens you up to accidentally making mistakes that you&#x27;re better than so first thing I want to do is rewrite this and so that a can be represented as a fraction. I&#x27;m simply just going to put it over one. Right? A over one is certainly the same thing as a cause. Anything divided by one is itself. So that&#x27;s that&#x27;s fair. So I&#x27;m really doing a over one minus 3/2 A. All right now, I can see I do not have a common denominator here, so I&#x27;m gonna have to find one. Since one can be multiplied by anything, I&#x27;m not gonna need to multiply both fractions by something. I can already tell that I&#x27;m going to be able to just rewrite my second fraction because I should easily be able to get my fraction of a over one to change into a fraction that has to a for its denominator because I know that if I take one times to A, I will get to a now the catches just like with everything else we&#x27;ve been doing. If you want to change a rewrite a fraction and not change its value, whatever you multiply the denominator by you also have to multiply the numerator by meaning we have to take a times to a well a times to a would be to a squared right. Eight times a is a squared to is along for the ride Essentially. So my first fraction instead of a minus one could be expressed as to a squared over two A. I now have a common denominator, which means I am ready to combine these. Remember, when we combine fractions by adding or subtracting, we do not combine the denominators. We just rewrite it. That minus has nothing to do with the denominator, right. If you have a negative fraction, remember for a negative fraction, you can choose whether that negative is attributed to the numerator or the denominator. We don&#x27;t want to put that negative to the denominator because then they wouldn&#x27;t be in common, so instead just attributed to the numerator and will take to a squared minus three. Well, to a squared minus three would give me to a squared minus three right to a squared and three or not common terms. Not like terms. So I can&#x27;t actually combine those. This just has to stay as a quantity. So this would be my answer. You can&#x27;t reduce anything because everything in the numerator is together as a quantity. So unless you have exactly to a squared times three, then to a start here. So unless you have exactly to a squared minus three in the denominator, you would not be able to cancel it out. Okay, Now the last part of this, we do have to account for any undefined values for a So I looked back in my original problem. This a is a numerator, as we showed right here, so I don&#x27;t need to worry about that. But I do have a denominator of to a So the question is, if I have to a what value of a would go would cause this equation to be zero, because that&#x27;s what we don&#x27;t want. Well, to time, zero would be zero. So the only problem for this equation would be if a happens to equal zero that would cause our answer to be undefined, which is basically just math word for an air. Right? So our answers to a squared minus three over to a as long as a does not equal zero.

Georgiann Andersen · Answer

we&#x27;re gonna simplify the following expression says number chat our Expressionists a plus one all over three A. We&#x27;re gonna add rehouse to this. In order to do this, we have to have the denominators the same. So we&#x27;re gonna take two times if it&#x27;s one all over two times three A. Remember what you did to the bottom. You have to do the top. And then the next one, we&#x27;re gonna take the three times the three a and the two times the three that&#x27;s gonna leave us with two es plus two all over six A was nine a over six A. So we&#x27;re gonna have to a plus two. That&#x27;s nine a all over six A that&#x27;s gonna leave us. We&#x27;re gonna combine their like terms. That&#x27;s gonna be 11 a plus two all over six a. And that is our are this week. And simplify that, too. Um, the factors that would make this undefined would end up being zero, so they cannot be zero

Alisa Lu · Answer

Okay, so this question were asked to perform subtraction on these two fractions, we have a minus one over five bias, a plus one over four. So these two don&#x27;t have any hate limiting factors for a as a Zen, the numerator and can equal anything, and you&#x27;re real number. But what we do need to do is make the denominators a common denominator. And to do this, the greatest common factor for the least common multiple for these two would be 20. So for the first fraction to equal have a denominator 20 money to multiply by four before this gives us for a minus for over 20 and for the second fraction, to be equal to have a denominator of 20 minutes need to multiply by five over five, which gives us five a plus five. Okay, so now we can just go ahead and do more. Most attraction, I guess, on the numerator. So here will have for a minus 5,000,000 which gets you negative A and the negative four minus phi, which gives your negative not so final answer is negative. A minus nine over 20

Problem

In $3-20$ , perform the indicated additions or su…

Problem 18 Medium Difficulty

In $3-20$ , perform the indicated additions or subtractions and write the result in simplest form. In each case, list any values of the variables for which the fractions are not defined.
$$
\frac{1}{a^{2}-a-6}-\frac{1}{2 a^{2}-7 a+3}
$$

Answer

$\frac{1}{(a+2)(2 a-1)}, a=3, a=-2, a=\frac{1}{2}$

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$\frac{1}{(a+2)(2 a-1)}, a=3, a=-2, a=\frac{1}{2}$

Algebra 2 and Trigonometry

Catherine R.

Heather Z.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

Ann Xavier GantertAlgebra 2 and Trigonometry

Catherine R.

Heather Z.

Caleb E.

Michael J.

Ann Xavier Gantert

Algebra 2 and Trigonometry