in-3-20-solve-each-equation-and-check-fracx25fracx-23

Question

In $3-20,$ solve each equation and check.
$$
\frac{x+2}{5}=\frac{x-2}{3}
$$

Derrick Hanson · Accepted Answer

When you have a problem like this, you actually have two different ways that you can choose to solve this out. You can go the normal route that we would use for any rational equation that we want to solve, which would be to find a common denominator and then multiplied by that common denominator to get everything. Um, all our denominators gone, all of them cancelled out, right. However, in the case of this problem, there is another option. Now, this Onley works if you have exactly one fraction on each side of the equation. But if you have only one fraction on each side of the equal sign, this would be something that you can treat like a proportion meaning. If you remember probably a long time ago, when you&#x27;ve done proportions, you are able to cross multiply here, meaning we can set this diagonal of three terrible three of three times X plus two equal to this diagonal of five times X minus two. Again, this is the same thing you would end up getting if you multiplied by a common denominator, right? If you look for a common denominator here, the common denominator between five and three would be to multiply by 15. So if we multiplied, everything here by 15 will think about it. 15 over five out of you put 15 here. 15/5 would cancel 23 right, 15 divided by three for the second fraction would cancel to five. So it works the same way. Either way, you look at it. It really doesn&#x27;t matter. And I don&#x27;t know that either one saves you a whole bunch of time. But I didn&#x27;t want to point out that since there&#x27;s only one Fraction, one rational expression on each side of the equation, you can just cross multiply. It only works if you have one each side that either way, whichever method you choose to go about, you would end up with where we&#x27;re here. Three times the quantity of explosive two equals five times the quantity of X minus two. So we would distribute the 33 times X is three x three times to it&#x27;s six. Then we distribute the 55 times X is five x five times Negative too, is negative. 10. So now our goal is to get all the exes on one side of the equation. Doesn&#x27;t matter which and get all our constants on one side of the equation. I am going to choose to subtract three X to the other side because I like to keep my ex positive, if possible. And five x minus three X would give me a positive two X. All right. And so then because of that, if I&#x27;m gonna put on LAX is on the right side. I need to move that negative 10 over to the left side so that all my constance air on the opposite side. So the opposites attraction is addition. So I would need to add 10 over here, which would give me 16 equals two X. I need to then divide by two to solve out for X. And that would give me that eight is equal to X or X equals eight. Whichever way you prefer to read it.

Dilip Paruchuri · Answer

all right. Hey, guys. So in this problem, we have the rational equation X over experts. Five equals 2/3. Now, the first step in this problem is to make it easier for us for solve this by a kn fraction ing everything so we don&#x27;t want we don&#x27;t really want to deal with the fraction. So what we&#x27;re gonna do is we&#x27;re gonna first multiply by X plus five on both sides of the equation. So first step, we&#x27;re gonna multiply by expose five on the left hand side, and we&#x27;re gonna most by by experts five on the right hand side. So on the left hand side of the to expose fives here, cancel out and we&#x27;re left with X on the left hand side. When the rights hand side, we&#x27;re left with 2/3 times X plus five. And now, in order to make things a little bit more easier for ourselves, what we can do is we can cross multiply, not cross month by, but in a sense, divide out the 2/3 from the right hand side. So we&#x27;re just left with the X plus five. So what we&#x27;re gonna do is we&#x27;re going to multiply by three halves on the right hand side and three halves on the left hand side. The goal of this basically this step right here is just so we could make this equation just little bit easier to solve. Make out, make the algebra nicer for us. And so what happens on the right hand side is the 2/3 cancels out with it that we have so on the left hand side, you&#x27;re just left with experts five, which is nice, which is really nice to work with. And on the left hand side, you&#x27;re left with the rehabs times X, which is also a nice basically a nice function to work with. And so we want to get all our variables to one side now. So let&#x27;s subtract X on both sides, tracking acts on both sides. We get 1/2 X equals five and in order to get her ex by itself, we&#x27;ve multiplied by two on the left hand side and two on the right hand side. Then we get that X is equal to 10. Now we plug in, shrug and see whether the left hand side equals the right hand side. We have our left hand side is equal to X over explode. Oh, that&#x27;s right. That better is equal to X over X plus five. We plug in our value of 10. So we have 10 over templates. Five at 10 over. Tempus five is the same thing as 10 over 15 which is the same thing as 2/3. And as you noticed, our right hand side is also equal to 2/3. So we know that our left inside right to inside are equal. So value X equals 10 is the correct solution for this equation, and we have verified it. So thanks for listening, guys. And I hope this help solve this rational.

Dilip Paruchuri · Answer

All right. Hey, guys, showing this problem, We have the equation X over two equals three x three over two X plus one trying to solve for acts and verifying if our exes correct. So we&#x27;re going to do first is we&#x27;re going to multiply by two X plus one on both sides of the equation. And we&#x27;re doing this so we could get our ex out of the denominator on the right hand side. So on the right hand side of the two X plus ones, cancel out we&#x27;re left with the re on the right hand side on the left. Inside we&#x27;re left with X times two x, less one all over, too. And we&#x27;re gonna do something similar, like the first step. We&#x27;re gonna try to get rid of the denominator of the left hand side. Not did not. The denominator on the left hand side is just a constant, but it&#x27;s just nicer when the denominators, when it&#x27;s just normal members and we don&#x27;t have any fractions, it&#x27;s easier to deal with the equation. So we&#x27;re gonna multiply by two on both sides on the left hand side, the twos kids allowed, and you&#x27;re left with X plus two x plus one x times the two whole quantity of to us This one. My bed on the right hand side, you&#x27;re left with six. You see that? This is a quadratic equation. We&#x27;re gonna We&#x27;re gonna get a quadratic equation. So on the left hand side, if we distribute, we get to explore close X. Well, six. If you subtract six on both sides, you get Q X squared plus X minus six. Now we just factor. You could use the quadratic formula to solve this. Just plug in Shug on is gonna factor this out. We know are two X squared plus X minus six is gonna be equal to two X squared plus four X minus three X minus six. Uh, this is just a simple factoring trick. I&#x27;m just gonna factor really quickly. And in order to do this, we&#x27;re gonna group these two terms together and these two terms together, we&#x27;re gonna have to x times x plus two x plus two minus three times X plus two and we&#x27;re gonna factor that out again. Oh, let&#x27;s ah, rewrite this really quickly. Let&#x27;s go back a sec. So we have this equation will go back to this step because I think I wrote something incorrectly. So we have our two x squared plus four X minus three expected sticks. We said That&#x27;s equal to zero. So let&#x27;s go up here to solve the equation. If we factor these two terms that I just grouped up together, these two sets of terms will get to X explodes too. Minus three X plus two equals zero. And on the left, basically on the left hand side. If we distribute if you we factor out once again if you factor out the Expos to we get to X minus three, I&#x27;m expose to equal zero and solving this. We know either the two X minus three equals zero or the experts to equal zero. So the solution says gonna be X equal to three halves and X is equal to minus two. And so let&#x27;s verify. Ah, this equation. So let&#x27;s go to another screen right here. So the equation we&#x27;re trying to verify his X over two is equal to three over to X plus one. Let me see if I copy that. Correct. And I did. And so the first hole that we&#x27;re trying to verify his X equals the re half. So on the left hand side, we have to re have divided by two because we just it&#x27;s x over to our X. We&#x27;re saying is three halves we have three has divided by two so three halves times 1/2 is equal to three Fords. And on the right hand side, we have three over two times, three over two plus one, just plugging in our value of three halves into to expose one. And we have three over the twos. Cancel out. Still three plus one that&#x27;s equal to three over four A. CZ. You see the left hand side on the right hand side of the same. So X equals three have is a correct solution. Now let&#x27;s try. X equals minus two. X equals minus two on the left hand side. If you plug that into X over two, you&#x27;re gonna have negative to over two, which is equal to negative one. And on the right hand side, you&#x27;re gonna have the re over two times. Negative two plus one two times in 32 is negative force. You have three over negative four plus one negative four plus one is equal to 93. So you simplify this to three overnight of three, which is equal to negative one, and negative one is equal to negative one. So X equals minus two is also a valid solution to this equation. Let&#x27;s get a check mark and then let&#x27;s go back to our first screen. And we can easily and definitively say that X equals three halves and and X equals minus to our solution set to the equation X over two equals three over to expose one. Thanks for listening eyes, and I hope this hope.

Derrick Hanson · Answer

we&#x27;re being asked to solve this equation out for X. First thing we need to do is we want to try and get, um, all of our denominators simple, cancelled out and get everything moved up to the numerator so that we can actually work towards trying to solve this problem out. Okay, in order to do that, we need to find ourselves a common denominator so that we can multiply the entire equation by that Well, I have a denominator of X plus three, and I have a denominator of X plus two and X plus three. Don&#x27;t try and make this harder than necessary. If we&#x27;re trying to find a common denominator, it&#x27;s going to have to be both X Plus two and X plus three that we multiply by because that&#x27;s the only way that will cancel out all of our denominators. All right, so if we&#x27;re gonna multiply with that, we have to multiply every single term. That includes this one. So one Times X plus two and X plus three would give us X Plus two and X plus three on the left side of our equation. Now, if we multiply this to our first fraction our first rational expression of five over X plus three. The X plus threes would cancel, leaving us with five times X plus two. All right, this X plus three right here and this X plus three right here would cancel. And then finally, if we multiply X Plus two and X plus three times the last fraction of five over X plus two x plus three. Well, I have an X Plus two in my numerator and in my denominator, and I have an X plus three of my numerator and in my gin on near. So all of that would cancel leaving on Lee that five. Okay, now that I&#x27;ve gotten rid of all the denominators, I have to actually try and solve this out. Well, I&#x27;ve got two different multiplication problems that I have to do here. I have to foil these two on the left here, so x Times X would be X squared. X Times three would be three x two times X would be two x two times three would be six on the right side of the equation. I need to distribute this 55 times X would be five x five times to would be 10 and then this plus five can&#x27;t be for gotten as well. Okay, cleaning up the left side three X plus two X would give me five X over there and then on the right side. Cleaning up 10 plus five is 15 and that would be what I have on that side. So now I can see that I&#x27;ve got myself a quadratic writes, I do have an X squared as well as excess, which means I&#x27;m probably more than likely gonna have to be either using the quadratic formula or factoring to end up solving this out. But either of those options does require that I first get the equation equal to zero. So I want to move everything to one side again. I don&#x27;t want to move just the exes to one side and the constants to the other side because I haven&#x27;t X squared. So that changes how we look towards solving things. Since I have an X squared, my goal is to move everything. The ones I&#x27;d get the equation equals zero, and then we can try to either factor quadratic formula, whatever we end up deciding. OK, so I already have X squared on the left side, so I&#x27;ll try and move everything to the left side. I will subtract this five x over, and I will subtract this 15 over five X minus five x just straight up. Cancels six minus 15 would be negative nine and that is now equal to zero. It is up to you whether you would prefer to factor this or used the quadratic formula. Quadratic formula will work. Always factoring doesn&#x27;t necessarily always work, But in this case, factoring is a possibility. So I&#x27;m going to choose to factor this just because if it can, if it if it&#x27;s able to be factored, I usually do. In this case, we have two terms in X squared and a nine. There&#x27;s subtraction between them and both X squared and nine conspired perfectly, meaning this would be a difference of squares. Remember, a difference of two squares is a formula that allows us to say the square root of X squared is X. The square of nine is three, so my factors are X plus three and X minus three. After we&#x27;ve taken care of that, which, if you don&#x27;t remember difference of two squares you should go back and review that. Obviously, don&#x27;t just take my word for it after we&#x27;ve factored that way. Now we can use the zero product property, which says, since these air equal to zero weakens that each factor equal to zero separately and solve out meaning we&#x27;re gonna be getting two answers here. On my 1st 1 I would subtract three to the other side. Giving me X equals negative three. On my second equation, I would add three to their side. Giving me X equals positive. Three. So right now we have two answers to this question, which are X equals negative three and X equals positive three. Before I say that&#x27;s my final answer, though, we do have to go back to their original problem and plug these back in to make sure that they don&#x27;t accidentally cause this equation to become undefined. Remember, the way we get undefined is if you have zero in the denominator. So the question is, if I plug either of these X values back into the equation, will one of them or both of them cause me to get zero in the denominator? So let&#x27;s start with negative three. If I plug in negative three for X. That means we would have won equals five over negative three plus three plus five over negative three plus two and times negative three plus three. Well, what is negative? Three plus three? It zero that would cause this problem to be undefined. So because of that, we didn&#x27;t do anything wrong. Algebraic Lee To make sure you understand that we did not do anything wrong. Algebraic Lee. It&#x27;s just negative. Three doesn&#x27;t actually work when we plug it back in. Let&#x27;s try positive. 35 Plug in positive three. Then I would have one equals five over positive three plus three plus five over positive three plus two times positive. Three plus three. Three plus three is six that checks out. Three plus two is 53 plus three or six. None of those are zero. So this one actually does work out All right. Meaning our final answer would be the three and the three alone. Not with the negative three on Lee. The positive three

Derrick Hanson · Answer

So we&#x27;re trying to solve this problem for X first thing notice here is that obviously we do have some rational expressions here. Some fractions. I don&#x27;t want to leave those those fractions. So the first thing I want to do is I want to find a common denominator so that I can multiply the entire problem by that number to get rid of all of my denominators. Right now, I see a denominator of three and denominator of four. So the question is, what is the least what is the lowest number that both three and four can multiply up to? In this case, I would say that is 12. Okay, so we&#x27;re going to multiply this entire problem then by 12. Understand? This is 12/1, meaning 12 is being multiplied to the numerator. Okay, I have to multiply every single part of the equation by 12 so that I&#x27;m not actually changing the value of this equation. Meaning I&#x27;m going to take 12 times two x over three. Well, 12 times two x would be 24 x over three. Then I have to take 12 times one pretty straightforward. 12 times one is just 12 right, And then finally, I have to take 12 times three x over 4 12 times three x would be 36 x over four. So now, assuming that we chose something that a number that truly was a common denominator, all of my fraction should be able to reduce well. 24 x over three could reduce down to eight. X 12 Doesn&#x27;t need to reduce its on a fraction anyways. And 36 x over four should reduce to nine x 36 divided by four is nine so good? Looks like we did do the right thing here We did pick a common denominator cause all my fractions are gone. So now I simply just need to solve this equation out to get X by itself and figure out what happens equal. Ah, I would say they use this way to do that would be dis attract eight x to the right side because that goes ahead and isolates the Constance as well as the exes at the same time. So that would leave me with just 12 on the left side of the equation. Nine X minus eight X would be one x or just x so looks like our answers. X equals 12. That&#x27;s it.

Christopher Stanley · Answer

I know this problem can be a little deceiving because it is already factored, but it&#x27;s not equal to zero on that right hand side, so I haven&#x27;t factored really doesn&#x27;t help him. So the first thing you need to do is foil out this left hand side. So the X squared plus two acts plus three acts plus six is equal to 20 whenever we foil out that left hand side. Then let&#x27;s combine like terms. Here is we have X squared then plus five X. Since its we&#x27;re solving a quadratic here, I want to get it equal to zero anyway. And so I&#x27;m going. Take that 20 over So we give me minus 14 is zero. We have experience plus five. X minus 14 is equal to zero, right? So now we can factor this Left inside, it&#x27;s an X squared, and so I need an expert in the next. These last terms need to multiply to give me negative 14. And since I want that plus five accident middle, what&#x27;s gonna work is X plus seven and X minus two. So now we have this factor using a zero product property. This tells us that X plus seven zero where X minus 20 if X plus seven zero in fact, Paul 70 and then tells us accident. You have seven. The X minus 20 Then we had to. And that tells us access to we should have two answers here. Negative seven on positive, too.

Problem

In $3-20,$ solve each equation and check. $$ \f…

Problem 5 Easy Difficulty

In $3-20,$ solve each equation and check.
$$
\frac{x+2}{5}=\frac{x-2}{3}
$$

Answer

$x=8$

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Absolute Value - Example 1

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$x=8$

Algebra 2 and Trigonometry

Grace H.

Catherine R.

Alayna H.

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Absolute Value - Example 1

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Ann Xavier GantertAlgebra 2 and Trigonometry

Grace H.

Catherine R.

Alayna H.

Maria G.

Ann Xavier Gantert

Algebra 2 and Trigonometry