in-3-14-solve-and-check-each-inequality-fraca4fraca26

Question

In $3-14,$ solve and check each inequality.
$$
\frac{a}{4}>\frac{a}{2}+6
$$

Dale Sanford · Accepted Answer

we&#x27;re going to be solving the inequality. A over four is greater than ever to plus six. Let&#x27;s get started. So first off, what I want to do here is get the A&#x27;s all in one place. So I&#x27;m going to be subtracting and a over to from both sides, giving us on the left, a over four minus a over to eyes greater than six. And now we&#x27;re going to get some common denominators on the left that we can combine these two fractions. So we&#x27;re going to be multiplying the second term. That is that a over two by 2/2, giving us a over four minus two a over four eyes greater than six so we can subtract that giving us negative a over four is greater than six. Then we can multiply across by this four. Giving us negative A is greater than 24. Ah, and finally, we can divide over by this negative. Now the important thing to remember here is that divided by a negative will flip the sign or flip the direction of the inequality around. Giving us A is less than on negative 24 so the answer&#x27;s here is the set of all a such that a is less than negative. 24

Dale Sanford · Answer

we are going to be solving the inequality as pictured here on the street, without further ado, Let&#x27;s get started. So when solving inequalities, the first thing you want to do is get the numbers and the letters all by themselves. That is the variables, not letters. So we&#x27;re going to be adding a to to both sides as well as an A over six to both sides. There for the left side will become a plus 1/4. Plus a over six is greater than and then, since we added the two, we just get 11 plus two or 13. All right now on the left side, it&#x27;s We&#x27;re going to have to make some common denominators here, but we can make things a little bit easier for ourselves by first multiplying both sides by two, giving us a plus 1/2 because the two will cancel. Plus a over three for the same reason is greater than 13 times two or 26. Okay, now let&#x27;s start finding that common denominator on the left, so we&#x27;ll be multiplying the first term by 3/3 and the second term by 2/2. All right, let&#x27;s get to it. So we&#x27;re going to have three times a plus one plus to a all divided by three times two or six. And that is greater than 26. So we can multiply the six across as though we don&#x27;t have to do with it. Giving us three a plus three. I just did distribution on that three right there, plus to a is greater than six times 26 which is just 156. Now we can combine like terms giving us five A plus three is prettier than 1 56 Subtract the three, giving us five a is greater than 1 53 and finally divide by five. Giving us what we wanted a by itself or a is greater than 153 over five. Thus, the answer to this inequality is the set of all. A such that a is greater than 153 divided by five

Dilip Paruchuri · Answer

in this problem, writers were given the inequality three minus two over equals one is less than five and were asked to solve the inequality and give a solution set for the expert variable A. That makes the inequality true. And also to verify the solutions that as well. So let&#x27;s start by bringing everything over to one side of the equation. Since we have our variable in the denominator, we&#x27;re gonna have to do an extra step in the process like the previous problem. So what we&#x27;re gonna do first use, we&#x27;re gonna combine everything on left side like I said. So first step is gonna be to bring the night the five over in the left hand side. We do that by subtracting five on both sides. Subtracting pile on the left hand side just gives us zero on the right hand side. We&#x27;re left with minus two minus two over. We want to combine everything into a woman rat of one left. Remember here. So what we&#x27;re going to do is we&#x27;re gonna buy the common denominator of a plus. What? So when you do that, you get minus to a minus two over eight plus moderne minus two over a plus one. Just a rewrite with re written form of this stuff right here. Now we combine it and we get minus to a minus for over a plus One is less than zero. I&#x27;m gonna bring my work up to hear from more space. Next up we can do is we can factor out of minus two just to make our life a little bit easier. Uh, it&#x27;s nicer to work with cleaner functions. So you get minus two times a plus two over a plus one is equal to up, not equal to is less than zero. And then if you multiply by negative 1/2 on both sides So let&#x27;s multiply by negative. So let&#x27;s do this differently. So what they were It&#x27;s gonna look kind of weird kind of crammed. So what I&#x27;m gonna do is I&#x27;m gonna write. This is a secret second step. So if we do negative 1/2 times negative to over times a plus two over a plus one less than zero times negative one have we get rid of the negative too in this multiplication stuff and you&#x27;re you&#x27;re left with a plus two over a plus, with one is greater than zero now because we must live by a negative number. Like the previous problem, we&#x27;re gonna set our new major denominator equal to zero, and we&#x27;re gonna solve for the other coins. And then we&#x27;re just gonna plug in values from each region to check whether the inequality holds. So let&#x27;s go on to a separate board right here. Did you find our pivot points? So are better points we&#x27;re gonna be. So this is our inequality. What are Inequality has become to a plus. Two is greater than zero. So we&#x27;re going to solve the equations. A plus two equals zero and a close one equals zero. This&#x27;d zero. And the values that we get is a who&#x27;s minus to an a equal minus one. So on our number likes to run till didn&#x27;t work. So we have three regions one, two on three on. Let&#x27;s pick a value from each region, see which one works. So in Region one, let&#x27;s choose negative three. So if you choose 93 it this is gonna become night of three plus two over negative three plus one. Uh, this is gonna simplify to let&#x27;s write that again at one does not look it. So this is gonna be called negative one on the numerator. Negative two on the denominator, which is 1/2. And that&#x27;s greater than zero. No, we know that this works. Region one is a check that is a part of the solutions. Now let&#x27;s take a look at region too. And in Region two, we&#x27;re gonna use the value negative through. Perhaps so when we do negative three halves plus two over negative three halves plus one, we get minus 1/2. I mean, uh, apologies. 1/2 in the numerator and minus 1/2 of the denominator that simplifies the negative one. And that if we set the inequality, it&#x27;s supposed to be graded on zero. We know that&#x27;s not true. So we know that region to does not work now. Finally, if we take a look at Region three, Region three, let&#x27;s choose value zero. When we plug in zero, we played in zero plus two over zero plus one. That gives us two over one, which gives us two, which is greater than zero. So that is a part of the solution. Is that and solutions that three works. So let&#x27;s draw this on the ground. We use open circles here because this is a greater than sign not a greater than or equal to on our solution set on a graph graft. Looks like this. All values to the left of minus two. All values to the right of minus one. So if we want to write out our solution gonna look like this, it&#x27;s gonna be a less than minus two or a greater than minus one. This is the answer to the inequality problem. And this all this step right here is our verification process. Thanks for listening, guys. And I hope this helps solve this inequality.

Dale Sanford · Answer

we were given this inequality on screen, and we&#x27;re going to be solving it without further ado. Let&#x27;s get started now. The first step to any inequality with the X or the whatever variable on the bottom is to multiply by that denominator turn that is the variable on the bottom. So we&#x27;re gonna be multiplying both sides by x minus for giving US X minus four times four x over X minus four plus two is less than two over X minus four times X minus four. Now this is pretty easy to multiply through. Giving us four X plus two x minus eight. That&#x27;s just disturbing out. The to is less than just to those are going to cancel, and now we have to split this up into two different paths. That is, since the X Plus four or X minus four Rather was on the bottom. We have to split this up into two different cases. One where X minus four is greater than zero and one where X minus four is less than zero. Let&#x27;s start with the greater than case in this one. Nothing changes besides the fact that ah you now have this initial condition that X minus four is greater than zero. So let&#x27;s plug that into our equation or are in equality rather in salt. So we have four X plus two X minus eight is greater than two. Okay, so we can combine like terms on the left. Giving us six X minus eight is less than two. And then we have, uh, adding this eight across will give us six X is less than 10 and then we can divide over. Giving us X is less than 10/6, which simplifies down to 5/3. Um, so now we can take our initial condition of X minus four is greater than zero, and rearing is that to get that X is greater than four. Well, this is a bit of a problem. Our initial condition supposed that actual is greater than four. However, we just solved and found the axe must be less than 5/3. You can&#x27;t have a number that&#x27;s less than 5/3 and greater than four. So we therefore must disregard this side entirely, as it has given us an impossible results. Hopefully, the right side will give us a better result I&#x27;m going to solve the initial condition right off the bat, getting us X less than four. All right, so let&#x27;s hope our answer holds true to that. So here, since this is less than zero, we have to flip the sign or rather, the the direction of our inequality. Giving us four x plus two X minus eight is greater than two. Okay, once again, we can use the same processes before that will give us six. X minus eight is greater than two or six X is greater than 10 and then dividing through well, give us X is greater than five or three. So now we have X is greater than 5/3 and X is less than four. Well, this is possible. There are plenty of numbers which are greater than five or three and less than four like 6/3, Also known as to for example. All right, so we have therefore founder answer on the right side of this equation. Thus, our answer is the set of all X such that uh, 5/3 is less than X, which itself is less than four

Brandon Fox · Answer

to solve this inequality, we&#x27;re gonna take four and multiply times X for X in four times a minus two, which is minus eight. Then we&#x27;re gonna take the minus sign and go and distribute that into the second set of parentheses on minus four X and a minus minus three, which would be a plus. Tary. Now we can take the extremes and combine them. Ah, for X minus four x cancel. Cancel each other out there. Gone. Make zero in negative. Eight plus tree is a negative five. Now, this is true. Negative. Five is less than negative. Six. And so what this means, is it all X values? All exes are solutions, all ex values.

Kimberly Konzack · Answer

question 23 four acts plus three is less than three X plus six. What we want to do our strategy is, is that we wouldn&#x27;t try to get. The X is not a one side, so it typically move the want smaller core, efficient over to the other one side but using the additive inverse property. So we subtract three acceptable sides and we get that X plus three is less than so, at which point we subtract three from both sides and we get that axe is less than three.

Problem

In $3-14,$ solve and check each inequality. $$ …

Problem 3 Easy Difficulty

In $3-14,$ solve and check each inequality.
$$
\frac{a}{4}>\frac{a}{2}+6
$$

Answer

$\{a | a<-24\}$

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Ann Xavier Gantert

Algebra 2 and Trigonometry

Grace H.

Anna Marie V.

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Absolute Value - Example 1

Absolute Value - Example 2

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$\{a | a<-24\}$

Algebra 2 and Trigonometry

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Absolute Value - Example 1

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Ann Xavier GantertAlgebra 2 and Trigonometry

Grace H.

Anna Marie V.

Maria G.

Kristen K.

Ann Xavier Gantert

Algebra 2 and Trigonometry