in-3-20-perform-the-indicated-additions-or-subtractions-and-write-the-result-in-simplest-form-in--10

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.
$$
5-\frac{1}{2 y}
$$

Alisa Lu · Accepted Answer

So this question were asked to do this attraction and then also list any values for the variable X, which the fractions are not defined. So right off the bat, we can say that wife cannot be equal to zero, as that would make the denominator here equal to zero. And we can&#x27;t have that. So to do the subtraction, we need Thio find a common denominator. So in this case, the common denominator would be too I So I would have to multiply five by two. I over to I, which gives me 10. Why? Over to y minus one of the two I. And so then this would combine into 10. Why minus one over to I, um c could, I guess, simplify this a little bit more after splitting the fraction. But that would just give you get you back to where we started. So we can just leave her answer like this. 10 Y minus one over to us.

Derrick Hanson · Answer

we&#x27;re being asked to add these two fractions, but we have a problem because in order to add or subtract fractions, we have to have a common denominator. And these do not is the first thing that we would need to do is try and find some number that both two and three can multiply up, too, so they&#x27;re denominators would be in common. In this case, I would say the least common down to the lowest number of the two and three can both multiply up to would be six. So our goal right now is we want to rewrite both of these fractions so that they are over six. Then once we&#x27;ve done that, we can actually combine them together. All right, so let&#x27;s start with why. Plus 2/2. If I want to get y plus 2/2 to be something over six, I would definitely need to take two times three. But if I&#x27;m going to take that denominator Times three, I would also need to multiply that numerator by the exact same thing, which means I&#x27;m going to take the quantity of Y plus two times three is well, that means we need to distribute that three, right? I know that&#x27;s not written the way you&#x27;re used to seeing it. Normally, you guys, we&#x27;re used to seeing it written like this, but we would still want to. This is still a distributive problem, and so we still want to be using that property. So we&#x27;d be taking three times. Why gives us three why? And three times to gives us six. So we would have three y plus 6/6. So there&#x27;s one fraction. Now I need to get the second fraction. That is to why minus 3/3. We also need to change that to something over six. Well, to get from 3 to 6, we would need to multiply by two. But if I&#x27;m gonna multiply the denominator by two that I would have to multiply the numerator by the exact same thing. This is a quantity of two y minus three. So that would mean we have a distributive property again. Two times two. Why would be four? Why, two times negative three would be negative. Six. Now that we have a common denominator, I can actually combine these. Remember, when we add or subtract fractions, we do not combine the denominators, we just rewrite it. Meaning it is still going to be over six for the numerator. We want to combine like terms. So I have a three y and a four y three y plus for why would be seven? Why? I have a six and I have a negative 66 plus negative six would be zero. Meaning we don&#x27;t really need to write that. We just have seven. Why over six? There isn&#x27;t anything that 76 have in common that I am aware of, which means we aren&#x27;t able to reduce this fraction, so that would be an answer.

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

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 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.

Alisa Lu · Answer

So for this question, were asked to combine these fractions and then also define any values of the variable X, which the fraction is not defined. So right off the bat and consider ex cannot equal to zero because that would make this denominator zero, and we can&#x27;t have that. So you combine these fractions, we need to find a common denominator. So one can be re written as one over one and still get a common denominator. We just need to multiply one by X over X to spend becomes one over X plus X over fix, which then becomes one plus X over X and you can&#x27;t reduce this anymore. So that&#x27;s your final answer.

Problem

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

Problem 12 Easy 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.
$$
5-\frac{1}{2 y}
$$

Answer

$\frac{10 y-1}{y \neq 0}$

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Ann Xavier Gantert

Algebra 2 and Trigonometry

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Absolute Value - Example 1

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$\frac{10 y-1}{y \neq 0}$

Algebra 2 and Trigonometry

Catherine R.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

Ann Xavier GantertAlgebra 2 and Trigonometry

Catherine R.

Caleb E.

Maria G.

Michael J.

Ann Xavier Gantert

Algebra 2 and Trigonometry