08 - Learn to Solve Fractional Equations in Algebra, Part 1

well welcome back to algebra the title of this lesson is solving fractional equations this is part one and there are so many different types and flavors of these things we call fractional equations that we're gonna have part one then part two then part three and part four and so on to give you lots and lots of practice with different types of problems but the basic idea is we're learning how to solve more complex equations if you remember we learned how to simplify these rational expressions we learned how to simplify what we called complex fractions in algebra which have weird numerators and denominators and then in the last few sections we learned how to simplify equations or solve equations that have fractional coefficients that sounds very similar to this this is something different though it's called solving a fractional equation and so the easiest way to explain the difference between what we've done before and what we're doing now is literally to go ahead and just put a couple of them on the board and show you but let me warn you up front we're gonna have three problems the first two were not difficult at all but the third one is pretty lengthy I really really want you to make it all the way to the end of this lesson though because and the third problem I need to get some ground work under us but in the third problem we're going to introduce the concept of an extraneous root of the equation an extraneous solution of the equation what we're gonna find out is sometimes we're gonna solve these equations and we're gonna get some answers and when we check the answer sometimes we will find out that some of these roots some of these answers aren't actually valid solutions so it's really important that you get to the end so that you can see an example of that because when solving these kinds of things we're gonna basically have to check and make sure that all of the answers are valid and so I need to show you an example of when you're not valid so let's work through some easier ones and get to a different more difficult ones here in just a minute now before we show you a fractional equation let's look at something that we learned how to solve in the last couple of lessons that would have been something like x over 3 plus 1/2 equals you know 1/10 or something this is what we called a equation with a fractional coefficient why did we call it that because even though it's written as x over 3 you can kind of think of it and the way I think of it really is one-third which is the coefficient times X plus 1/2 equals 1/10 all right so you can see that the coefficient which is just the number in front of the X is a fraction and we learned how to solve those we had three or four or five lessons solving these equations what we call fractional coefficients and the main solution technique was we said we need to cancel and clear out all of these denominator numbers we have to multiply the left in the right hand side by something some number to kill the three to also kill the two and also kill the ten and you have freedom right but I can tell you right now there's there's choices you can all have your own choice you can have your own number in your mind but I know that I can multiply left and right hand side by 30 for instance in this case I know that 30 will work and that will clear all of the numbers we did lots of problems like this so if you haven't already if you don't understand that then go back and look at those equations with fractional coefficients if we multiply the left by 30 then the 30 will multiply into both terms and the 30 divided by the three will give you 10 that's a whole number the 30 divided by the 2 will give you the 15 that's a whole number and the 30 divided by the 10 will give you 3 that's a whole number so by multiplying left and right by 30 you will eliminate all of the fractions and the fractions are what make these things a pain to solve because adding fractions means you have to get common denominators and so on so that's a trip down memory lane that's what we did last time learning how to solve these things involve multiplying left and right hand side by something to kill the denominators to make it easy all right now this is different this is the kind of problem that we're going to do here and we call these fractional equations fractional equations whoops equal kind of spell right all right and so here's an example our first example of a fractional equation 3 over t minus 1 over 3 times T equals 2/3 now take a look at this equation and compare to the one above we've solved these kinds of problems before this is new see in a equation with what we call a fractional coefficient the variable is basically still on the top right so all of the denominators of all the fractions are just numbers so in order to clear away the denominators we had to do is pick a number in this case we said we'd choose 30 but you could choose another number that also would clear them and then by multiplying by that number you would clear all of the denominators because they would divide away okay but notice in this one this is not a fractional coefficient this is a fractional equation that means that the variable is also allowed to be in the denominator of the fraction so notice that the variable is in the top here but the variables is in the bottom here it's also in the bottom down here now we could have rolled both of these concepts together the fractional coefficients and the fractional equations a lot of students say well they're two different skills right well they kind of are two different skills but really what it boils down to more than anything is when you have a fractional coefficient it's just the easier case because you just have denominators that are numbers so all you have to do is pick a number to multiply to clear everything out but now we have something different that looks slightly more complicated but it's in the same family it's a fractional equation the denominators of these fractions can also have variables that's why we say they're fractional equations so it's a little bit more challenging but the same process is going to apply what we need to do is get rid of these denominators just like we were getting rid of these denominators so let me write the equation one more time we're going to decide what we need to multiply by three t equals two-thirds what we need to multiply by in order to to get rid of these denominators now I can multiply anything I'd like left and right hand side let's say I multiply the left and the right by just the number three if I multiply by three and I multiply by 3 the 3s would cancel here that would get rid of that the 3s would be distributed in and cancel this three but it wouldn't affected the T at all the T's would still be in the bottom which would be make it very difficult to isolate T get it and divide or multiply it would just make it a little more difficult to to actually isolate T because in the denominators of fractions it just makes it hard to collect them all to one side so what we really want to do is we want to figure out what can we multiply this equation by left and right hand side to clear the threes but also at the same time clear the t's so you guessed it we want to multiply by 3 times T if we do it here we also have to multiply by 3 times T here why are we choosing this because we know that the three will cancel with this three we know the three once we multiply it all in we'll cancel this three but once we multiply this whole thing in the T will cancel with this and the T will cancel with this yes of course this T is not going to cancel with anything it'll still be in the top but we don't care about introducing things in the top all we care about is eliminating all denominators because then we don't have any fractions to deal with and by now you should know that dealing with fractions is just a pain you can solve this thing by subtraction of course but it's just not as easy so let's let's move ahead through it what we have then we're going to do exactly the same technique as before 3t is written as 3t over 1 multiply distributing it into the first fraction 3 over T then we have the minus sine 3t will be distributed in as a 1 over 3 T so it's distributed into both terms and then on the right hand side will have the 2/3 multiplied by the 3t over 1 so now we have some glorious cancellations right because we have a 3 cancelling with a 3 notice I did not strike through the T it's not gone this 3 cancels with this 3 this T also cancels with this T this T cancels with this team now by the way when I say it cancels of course it's in the talk on the bottom but what's really happening is in any fraction like 6 tenths or something you can divide the top and the bottom of that fraction by anything you want to simplify like 6 tenths I might say 6 divided by 2 on the top 10 divided by 2 in the bottom I divide top and bottom by 2 right so when I cancel these T's what I'm really doing is I'm saying divide the top by T that gives me 1 also divide the bottom by T that also gives me 1 that's why they poof disappear because I'm dividing top and bottom by the same thing and gives me 1 in both cases so now we can cruise along all I have left I have nothing on the denominator except ones so the denominators just disappear they're really there but they're just they're dividing it by 1 so they're gone 3 times the 3 is gonna give you 9 here's the minus sign this is a 1 times a 1 so don't forget it doesn't totally disappear you still have 1 times 1 so you still have one here all right okay and on the right hand side all you're gonna have is to tie the tea over here and it's divided by one as well so notice you've eliminated all of the variable T's on the left hand side the only variable that you have left over of T is on the right now you have nine minus one it's just eight 2t and then you can of course divide so I'm gonna rewrite it and say T is equal to 8 over 2 so then you know that T is equal to 4 T is equal to 4 and that is the correct answer so here's the here's the gotcha in this very simple problem it's not going to be an issue all right but I need you to kind of go through the rest of the problems with me because what's gonna end up happening is sometimes when you multiply the left and the right hand side by variables in addition to numbers sometimes some of the solutions you get at the end of the problem are actually not valid but I can't show you that it's not valid because in this case everything's fine but work with me through we have one more problem and then we have the last problem the last problem will have two solutions we get but one of them isn't actually valid and I can't get there until we get there so for now you say well T is equal to the for so the thing you should look at is you should go and plug it back in and see if it causes any problems see if it causes any problem so let's do that really quickly we're gonna go back and plug it into this equation we're gonna plug it in to what we started with and just verify that it's right so it's 3 over 4 okay that's what uh yeah because 4 is T so goes in here then we have minus 1 over 3 T 1 over 3 times 4 because T is now 4 equals 2/3 if everything is correct and this should actually work out fine so we have three-fourths minus 1/12 is equal question mark 2/3 2/3 like this now we have to change this from 3/4 we have to multiply 3 over 3 to give it a common denominator of 12 like this so we have 9 twelfths minus 1/12 so what we have here when we get down to this point is we have a common denominator so 9 minus 1 is 8 12 right and if you divide the top by 4 you're gonna get 2 and if you divide the bottom by 4 you're gonna get 3 so you should all know that that's equal also 2 Thirds so it's exactly equal to what we had in the right hand side so we get a solution we check a solution and we see the solutions fine so then at that point we can circle it and say no problem but what's going to happen in the third example here is we're going to get down to the end we're going to find out that sometimes it doesn't quite work for these simple problems it does so you have to be on the lookout for it should what we call extraneous solution solutions which fall out of the math but aren't valid so for now just understand that the concept is the same we clear the denominators by multiplying by anything we want oftentimes we'll have to multiply by variables in addition to numbers and then we proceed with the rest of the solution as we have done all right next problem if we have let's say 1 over x equals 2 over X minus 3 like this X minus 3 then we want to solve that equation what do we multiply by we want to get rid of the denominators that means we want to get rid of the X and we also want to get rid of this entire denominator X minus 3 we don't want to have any denominator here we don't want to have a 3 here and we also don't want to have an X here so let's rewrite it and let's decide what to multiply by what do we have to multiply by actually I didn't quite leave myself enough room okay sorry about that we'll do it like this 1 over X what do we have to multiply by on the left and on the right to clear everything well we know that if we multiply by X it'll kill this but if we multiply by X it's not gonna kill the whole thing it's not gonna cancel with the whole X minus 3 alternatively we can multiply by X minus 3 the quantity X minus 3 which will cancel this but if we multiply by X minus 3 by itself it's not going to cancel that so the solution is we have to multiply by both denominators we have to multiply by x times X minus 3 which looks crazy weird but remember the only thing you're trying to do is cancel those denominators you don't care how ugly it looks in the top you just want to clear everything it's kind of like the nuclear option you go to multiply by the whole enchilada to clear everything in the bottom so what we have then is X X minus 3 over 1 multiplied by the 1 over x equals 2 over - three let me wrap parentheses around this just to group it times the X X minus three now on the on the left hand side we have an X here and an X here those cancel notice that this thing we had two x didn't cancel anything on the left so it's still there but this X minus three cancels with this X minus three and notice that this X that was just sitting along here didn't cancel anything on the right but that's okay all we wanted to do is clear all of the denominators which we have now accomplished so on the Left we have X minus three times the number one which gives you X minus three on the right we have two times the X that's all that's left now look how simple this equation is to solve you can of course solve that very easily let's take the two X and move it over here by subtraction so I have one minus two so we give you a negative one X minus three is equal to zero we'll take the three and move it over here so I have negative X is equal to positive three we'll add three to both sides and then we'll say X is equal to three over negative one right which means X is equal to negative three X is equal to negative three now we think this is the answer so what we what we showed in the last lesson and the last problem is we take this and we try to plug it in and verify but really what you're I'm kind of inching along here to you what you're really on the lookout for to see if you have an extraneous solution which means a solution that's not even valid at all is you just want to really make sure you don't have to check the problem every single time to verify it's right but what you really have to do is make sure that the answer you got doesn't cause your is your equation to blow up to infinity or do something really crazy or not something that's not well behaved so what you do is you notice that over here you have a negative three for the answer so you go and you say okay if I want to put negative three in here one over negative three is that well behaved yes one over negative three is just negative one third here if I put negative three here it'll be negative three minus three which will be negative six so two over negative 6 that's also a well-behaved number it's not it's not over zero in other words what you're looking for is if any of these solutions you get when you stick them back in there you get a zero on the bottom that will cause the solution to go blowing up to infinity or the equation to blow up to which means that what you got wasn't really valid at all it's not allowable right so in this case we take this number we stick it into both plate we don't even do the math we just make sure the bottom doesn't go to infinity and it doesn't so this is totally fine well-behaved okay we'll go back to this I showed you the answer was t is equal to 4 we actually plugged in there and verified it was true but really now I'm telling you that you don't really have to verify it just take the 4 and stick it back in here you have a t7 equal to 4 does this go to infinity no it's fine if you put a 4 here 3 times 4 is 12 right 112 is very well-behaved it doesn't go to infinity it's fine but if in some alternate universe let's say for this answer I got T is equal to 4 but somehow through the math have also got T is equal to 0 that was another solution let's say that it fell out of the math if I take 0 up here 3 over 0 instantly kills the solution it goes to infinity 0 here as well makes makes another 1 over 0 basically if you ever get a solution such that you plug it back in and you get a 0 in the denominator then that means that that value of x or T or whatever isn't even allowable to be a solution of your equation because it makes your equation explode it can't be a solution if it does that so use the technique we're doing here but just keep in mind that when you multiply both sides by variable sometimes some of the answers you get aren't gonna work and if they don't work you say that you have an extraneous solution which means it's not valid and you circle the ones that are not extraneous so in this last problem we're finally going to get to the to the part of the matter where I'm gonna show you that and I've kind of given away the punchline but that's ok because I don't care about surprising you I care about you learning so we have 3 over x squared of course this is going to be a more complicated problem but it's solved the same way so we have this guy plus the number 2 equals X minus 4 on the top over here and then we have X minus 5 all right so this is a significantly harder looking equation to solve but you have to remember that the same thing happens this is a fractional equation because the variable exists in the denominator of these fractions so what we want to do is multiply by anything to kill this and your term and also to kill this entire turn so let me rewrite it again three over now I have a polynomial of course I can multiply by that but we always like to factor polynomials so what we're gonna do in the right and the bottom here is I'm gonna factor it as I go now I'm gonna factor it in a second you still have the plus two and you still have the X minus four over five I haven't really changed anything X minus five on the bottom here what is this factor two you have an x squared so you have an X and you have an X and then you have ten of course you can do one times ten we try two times five and then of course we know two plus 5 is 7 so the only way it really works is minus minus check yourself x squared here you have negative 2x negative 5x they add to the negative 7x and then you have this times this gives you a positive 10 so all we've done is rewrite the problem but we factored the polynomial which you almost always want to do and now we have to decide and it looks like I didn't do a great job with keeping myself giving myself room sorry about that but anyway we want to multiply the left hand side and the right hand side by anything by something in order to cancel this fraction and also cancel this one so notice we have an X minus 5 and we have an X minus 5 so the rule of thumb is you want to multiply both sides by everything you see on the bottoms everything because the only way that you can cancel all of it is to multiply by the product of everything that you see so here we have an X minus 5 we know we got to multiply by that here we have an X minus 5 that's going to cancel that one but we also have an X minus 2 so we have to also multiply by that so the way you write it is we're going to multiply this by X minus 2 and also X minus 5 multiplied together now unfortunately I kind of wrote this big and I'm sorry about that but we're gonna multiply the left by X minus 2 and also X minus 5 and this is unfortunate I'm sorry about that but it gets multiplied by the entire left-hand side so let me rewrite it in a cleaner way in the next line will be a lot clearer I'll I'll do a better job of writing a little bit smaller so what we're gonna have then this entire term is multiplied by this entire left-hand side so it gets multiplied by this and then distribute it in also to the two so you'll have as X minus two X minus five this can be written as a fraction of over 1 then you're going to multiply it by this first term which is 3 over X minus 2 X minus 5 like this then you're gonna have this same thing multiplied by 2 so you'll have X minus 2 X minus 5 this is gonna be over 1 you're going to multiply by 2 right and then I'm gonna have this set equal to I can't fit it all over here comfortably so I'm gonna have this which will be X minus 4 over X minus 5 multiplied by X minus 2 X minus 5 so make sure you understand all I'm doing is take this thing multiplied by this this thing multiplied by the two equals and then everything on the right hand side so now you can finally do what you want to do the X minus 2 cancels the X minus 2 X minus 5 cancels the X minus 5 and then over here when you're multiplying by 2 nothing at all cancels but then what you have on the right-hand side of the equal sign is X minus 5 cancelling with the X minus 5 so now we can tidy everything up this is a giant one implied ones everywhere this is one so all you have left is really a 3 that's all you have left there but then you have that Plus this term which is all of this stuff the numerators multiplied and then the denominator is multiplied but you just have one so it's 2 times this stuff 2 times X minus 2 X minus 5 don't try to do all the multiplication in your head you just multiply by 2 divided by 1 so it's there's nothing on the bottom and then on the right hand side you have X minus 4 times X minus 2 now this is ugly I'm not gonna lie to you but it's solvable you know how to multiply these binomials that's gonna give you lots of terms and then you draw everything to one side you solve it so let's first multiply these guys here so you have 3 plus 2 x here we have x times X is x squared the inside terms give me negative 2x the outside terms give me negative 5x the last term's negative two times negative five give me the positive ten on the right hand side x times X gives me x squared inside terms give me negative 4x outside terms give me negative 2x last term's negative times a negative gives me positive eight so now I have a lot of things to do but before I can collect terms I have this notice I had the two out here so I multiplied this but I still have the two so let's go ahead and and and and distribute that 2 in so it'd be three plus here we're gonna have 2x squared now let's be careful we have the negative 2x and the negative 5 we can add those that's gonna give us negative 7x in here but then that 2 will be times the negative 7x so it'd be negative 14x 2 times when you add these together and then the 2 times the 10 just becomes a 20 all right on the right hand side you'll have an x squared negative 4 and negative 2 gives you a negative 6x and you have a positive 8 now let me just sync up with myself and make sure I didn't make any mistakes now so what I have this 3 plus 2x squared minus 14x plus 20 x squared minus 6x plus 8 okay I'm correct so far so at this point all you want to do is is collect terms essentially so let's go ahead and add these together so you're gonna have two x squared minus 14x plus 20 whoops not plus 20 plus 20 plus 3 is 23 and you have the x squared minus 6x Plus 8 all right so you have a polynomial on the left and a polynomial on the right so all you do is you grab all these terms you move them over to the left by subtraction so you'll have a polynomial equals 0 so let's do them one at a time we'll switch colors one thing at a time we're gonna smooth this x squared over by subtracting it so we have minus 1 x squared so 2 minus 1 gives me 1 x squared minus 14x plus 23 equals now it's gone from the right so we have a negative 6x Plus 8 okay now in the next step I'll switch colors again we're gonna move the 6x over by addition we're gonna add it over here so what we're going to have here when we add it is we'll have the x squared now we have negative +6 when you have negative 14 plus 6 you'll get a positive 8x you'll have the 23 along for the ride on the right-hand side all you have is the 8 and then the only thing that remains is we have to move this 8 over by subtracting it so we're gonna have the x squared we're gonna have the 8x and then what is 23 minus 8 over here we subtract 23 minus 8 we're gonna get a positive 15 and that's gonna be equal to 0 so let me check 8x squared I'm sorry x squared plus looks like I have a sign error somewhere you can see where it is I could erase it and tell you guys oh yeah you just don't don't do this again but I'm sorry I'm human as well we have negative 14 I added 6 to it that can't be a positive 8 it's a negative 8 all right so this actually is just a simple sign error it happens to everybody believe me so you have a negative here and you have a negative here let me double check myself again so I have 8 sorry x squared minus 8x plus 15 equals 0 that's correct so I just when I added the positive 6 I accidentally made a positive warrant should be negative so now we have to solve this right if I just gave you that polynomial something something equals 0 how do you solve it you factor it and you have it set equal to 0 so on the next board I'm all I'm gonna do is rewrite that polynomial it's x squared minus 8x plus 15 and that's equal to 0 how did we solve those we factor it right we have open parenthesis and we set it equal to 0 we have an X square so we know it's got to be an x times an X for 15 you have choices you could do 3 times 5 you could do 1 times 15 let's try 3 times 5 which is nice because I know that 3 plus 5 is also 8 so the way it's gonna work is negative times negative and double check yourself right x times X is x squared that's negative 3x that's negative 5x add them together you get the negative 8x multiply these negatives that gives you the positive 15 so then this can be equal to 0 or this can be equal to 0 equally well drives the whole thing to 0 so X minus 3 can be equal to 0 or X minus 5 can also be equal to 0 notice we're going to get two answers because we have a square and the variable X will be equal to three we can add the 3 and then X will be equal to 5 we add this guy so we basically circle it we say we're done but then you got to caution yourself this is the big gotcha been trying to mention since the beginning right anytime you multiply an equation left in right hand side by variables in order to clear denominators you have to be careful right you have to look at your solutions and verify that they're not going to blow up your equation so let's take a look at it X is equal to 5 is one of our answers X is equal to 3 is one of our answers if we look at the original equation the original equation was this but in factored form the original equation was this plus 2 equals this okay if I put 3 in here I'll get which is one of the answers 3 minus 2 that's a number 3 minus 5 thats a number that's cool you can multiply and that's going to be a number three minus five that's a number but notice that one of the solutions was five you can see right away you're gonna have a problem because if you put five minus five in here you get something over zero which means this whole term goes to infinity and you have the same problem here five minus five zero no matter what is going on here zero is gonna multiply give you a zero the whole fraction is gonna go to infinity so basically what's going on here is if I were to give you this function ahead of time and let's take a look at its domain which are the range of its the the domain of the valid values of X that we can use in this function X is equal to five is not even allowed to be a part of this function because it goes straight to infinity at that point it's a vertical asymptote so we got an answer of x is equal to five only because we multiplied by something that cancelled what was on the bottom so it kind of like it allowed us to make the equation simpler to solve but because we've kind of removed the denominators then we have to go back and check to make sure that the answers we get don't actually cause problems in those denominators because we killed them as part of the solution process it's always wondered and teachers would tell me that will you go to check your solutions make sure they're okay nobody really tells you why it's simply because you multiplied by something that got rid of them so the new equal we got which was this one right we we say it's equivalent to it but what we really have done is we've eliminated all problem areas in the domain by multiplying it because this equation here right there are no problems that X is equal to 5 right there's no denominators at all so the bottom line is when we multiply by something to give a simpler equation to solve it's totally fine to do it but in the process of clearing those denominators if some of the answers that you get exists in those denominators then they can't be valid right because we kind of used a trick to get the answer make it easy we have to check our answers to make sure that they're okay so what we found is that 5 is not even in the domain of this function it cannot be a solution of this function so what we say is that this is extraneous solution it doesn't mean you erase it it doesn't mean you did anything wrong it means you used a mathematical trick to eliminate the fractions which made this problem much much easier to solve but the trade-off is we've got to go I'm gonna draw like a little X through here even though it's I don't want you to race it on your paper it means it's not a valid solution it's an extraneous solution right so leave it on your paper tell the teacher it's an extraneous solution it's not correct it's not valid the only valid solution to this equation is this one right here because we put the number 3 in and three had no issues anywhere okay and if you were to put three in here for X and three in here for this and you would do three and three and just go through it all you will find that the left-hand side equals the right-hand side with no problems I'm not going to go through that because I've done it on my paper but I can it's a good exercise for you put three in here three in here multiply and add the two three in here three in here you're gonna find out that it all works out fine all right so here we have introduced the concept of solving these fractional equations so we talked about in the beginning that they're very similarly related to what we did before fractional coefficients those the variables are in the top but the bottoms of the fractions or numbers here in fractional equations the denominators can have variables too so what you have to do is multiply by left and right often containing the variables that you're trying to clear but the trade-off of that is that you're going to get answers that you always always have to check and make sure they don't cause any problems I said don't cause any problems going to infinity then the solution is totally valid but if they do cause any problems like we've done like we've had in that last problem then you get a market as an extraneous solution it's not correct solution but it was just part of our technique to get our answer that made it easy to solve but we have to throw that solution away it's not correct math even though it mathematically fell out and we circle the lien on extraneous solutions if you solve these problems you're gonna find that probably the extraneous solutions pop up in about half the problems maybe about a third of the problems you'll have an extraneous solution a lot of times you won't have it at all it just depends on the problem and how it was set up so make sure you can solve all of these I want you to do all of these yourself and then follow me onto the next lesson we have a lot more fractional equations to solve to get your skills to where they need to be in algebra

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Keywords: fractional equations, solve fractional equations, solving fractional equations, how to solve fraction equations, learn how to solve fractions, solve equations, solving equations, learn to solve equations, how to solve equations, equations with fractions, algebra 2, algebra for beginners, fractional equations with variables, fractional equations algebra 2, fractional equations examples, learn algebra, algebra course, algebra online course, tutorial, algebra help

Id: LMS1NR4gZN8

Length: 32min 9sec (1929 seconds)

Published: Thu Apr 25 2019