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### Solving Rational Equations

- To solve a rational expression first find the LCD of all the terms, and multiply by it. This will transform the equation so that it no longer contains fractions.
- Always check the solutions you find when working rational equations. A possible solution may not work in the original because it makes us divide by zero.
- Some formulas contain rational expressions. When solving these for a specific variable, work to isolate it on one side.

### Solving Rational Equations

[x/(x + 1)] = [(x âˆ’ 4)/(x âˆ’ 5)]

- x( x âˆ’ 5 ) = ( x + 1 )( x âˆ’ 4 )
- x
^{2}âˆ’ 5x = x^{2}âˆ’ 3x âˆ’ 4 - âˆ’ 2x = âˆ’ 4

[y/(y + 8)] = [(y + 6)/(y âˆ’ 7)]

- y( y âˆ’ 7 ) = ( y + 8 )( y + 6 )
- y
^{2}âˆ’ 7y = y^{2}+ 14y + 48 - âˆ’ 21y = 48

[4/(x âˆ’ 2)] = [(x + 6)/(x âˆ’ 1)]

- 4( x âˆ’ 1 ) = ( x âˆ’ 2 )( x + 6 )
- 4x âˆ’ 4 = x
^{2}+ 4x âˆ’ 12 - 8 = x
^{2}

[y/(2y âˆ’ 1)] + [3y/(y âˆ’ 5)] = 8

- ( 2y âˆ’ 1 )( y âˆ’ 5 )g[y/(2y âˆ’ 1)] + [3y/(y âˆ’ 5)]( 2y âˆ’ 1 )( y âˆ’ 5 ) = 8( 2y âˆ’ 1 )( y âˆ’ 5 )
- y( y âˆ’ 5 ) + 3y( 2y âˆ’ 1 ) = 8( 2y
^{2}âˆ’ 11 + 5 ) - y
^{2}âˆ’ 5y + 6y^{2}âˆ’ 3y = 16y^{2}âˆ’ 88y + 40 - 0 = 9y
^{2}âˆ’ 80y + 40 - x = [( âˆ’ b Â±âˆš{b
^{2}âˆ’ 4ac} )/2a] - y = [( âˆ’ ( âˆ’ 80 ) Â±âˆš{( âˆ’ 80 )
^{2}âˆ’ 4( 9 )( 40 )} )/2( 9 )] - y = [(80 Â±âˆš{4960} )/18]
- y = [(80 Â±4âˆš{310} )/18]
- y = [(40 Â±2âˆš{310} )/9]
- Extraneous Solutions: 2y âˆ’ 1 = 0
- 2y = 1
- y = [1/2]
- y âˆ’ 5 = 0

- ( n + 3 )( n âˆ’ 4 )[n/(n + 3)] + ( n + 3 )( n âˆ’ 4 )[n/(n âˆ’ 4)] = 7( n + 3 )( n âˆ’ 4 )
- n(n âˆ’ 4) + n(n + 3) = 7(n
^{2}âˆ’ n âˆ’ 12) - n
^{2}âˆ’ 4n + n^{2}+ 3n = 7n^{2}âˆ’ 7n âˆ’ 84 - 5n
^{2}âˆ’ 6n âˆ’ 84 = 0 - n = [( âˆ’ ( âˆ’ 6) Â±âˆš{( âˆ’ 6)
^{2}âˆ’ 4(5)( âˆ’ 84)} )/2(5)] - n = [(6 Â±âˆš{1716} )/10]
- n = [(6 Â±2âˆš{429} )/10]
- n = [(3 Â±âˆš{429} )/5]
- Extraneous Solutions: n + 3 = 0
- n = âˆ’ 3
- n âˆ’ 4 = 0

- 3r(2r + 1) + 10r(r âˆ’ 2) = (r âˆ’ 2)(2r + 1)
- 6r
^{2}+ 3r + 10r^{2}âˆ’ 20r = 2r^{2}âˆ’ 3r âˆ’ 2 - 14r
^{2}âˆ’ 14r + 2 = 0 - r = [( âˆ’ ( âˆ’ 14) Â±âˆš{( âˆ’ 14)
^{2}âˆ’ 4(14)(2)} )/2(14)] - r = [(14 Â±âˆš{84} )/28]
- r = [(14 Â±2âˆš{21} )/28]
- r = [(7 Â±âˆš{21} )/14]
- Extraneous Solutions: r âˆ’ 2 = 0
- r = 2
- 2r + 1 = 0

- [1/20] + [1/15] = [1/x]
- [1/20](20)(15)(x) + [1/15](20)(15)(x) = [1/x](20)(15)(x)
- 15x + 20x = 300
- 35x = 300

- [1/3] + [1/5] = [1/x]
- [1/3](5)(3)(x) + [1/5](5)(3)(x) = [1/x](5)(3)(x)
- 5x + 3x = 15
- 8x = 15

^{2}âˆ’ 4)] = 1

- LCM = (a âˆ’ 2)(a + 2)
- [3a/(a + 2)](a âˆ’ 2)(a + 2) + [(a âˆ’ b)/(a
^{2}âˆ’ 4)](a âˆ’ 2)(a + 2) = (a âˆ’ 2)(a + 2) - (3a)(a âˆ’ 2) + (a âˆ’ 6) = a
^{2}âˆ’ 4 - 3a
^{2}âˆ’ 6a + a âˆ’ 6 = a^{2}âˆ’ 4 - 2a
^{2}âˆ’ 5a âˆ’ 2 = 0 - a = [( âˆ’ ( âˆ’ 5 ) Â±âˆš{( âˆ’ 5 )
^{2}âˆ’ 4( 2 )( 2 )} )/2( 2 )]

- LCM = (3p + 2)(p âˆ’ 1)
- [p/(3p + 2)](3p + 2)(p âˆ’ 1) âˆ’ [5p/(p âˆ’ 1)](3p + 2)(p âˆ’ 1) = 5(3p + 2)(p âˆ’ 1)
- p(p âˆ’ 1) âˆ’ 5p(3p + 2) = 5(3p
^{2}âˆ’ p âˆ’ 2) - p
^{2}âˆ’ p âˆ’ 15p^{2}âˆ’ 10p = 15p^{2}âˆ’ 5p âˆ’ 10 - 29p
^{2}+ 6p âˆ’ 10 = 0 - p = [( âˆ’ 6 Â±âˆš{6
^{2}âˆ’ 4( 29 )( âˆ’ 10 )} )/2( 29 )] - p = [( âˆ’ 6 Â±âˆš{1196} )/58]
- p = [( âˆ’ 6 Â±2âˆš{299} )/58]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Solving Rational Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Objectives 0:07
- Solving Rational Equations 0:23
- Isolate the Specified Variable
- Example 1 1:58
- Example 2 5:00
- Example 3 8:23
- Example 4 13:25

### Algebra 1 Online Course

### Transcription: Solving Rational Equations

*Welcome back to www.educator.com.*0000

*In this lesson we are going to work on solving rational equations.*0003

*Two things that we will look at is I can solve a equation that contains a rational expression.*0009

*I will throw in an example or we deal with a formula that also involves a rational expression.*0016

*Recall that when you are working with equations that have a lot of fractions in them we will often use the least common denominator*0025

*to go ahead and clear out all of those fractions.*0032

*These particular types of equations you definitely want to make sure that you check your solutions.*0036

*This will be especially important for these types of equations this is because we are dealing with expressions that involve fractions,*0041

*we will have some restricted values, values that we cannot have.*0051

*Some of these values may try and show up as our solution.*0056

*We simply have to get rid of them.*0059

*They are not valid and they will make our denominator 0.*0061

*When working with these expressions that have a lot of fractions we like to keep the least common denominator throughout the entire process*0068

*or that way we can go back in and make sure that makes sense in the original.*0076

*Let us go ahead.*0081

*When you see my example for formulas here again and know that the same process plays out.*0086

*We like to use that least common denominator to go ahead and try and clear things out*0092

*Of course, our goal is to isolate which ever variable we are trying to solve for.*0096

*When we are all done with that formula we still have an equation of these words*0103

*and for more variables still floating around in there.*0107

*That is okay as long as we get that variable we are looking for completely isolated.*0109

*That is how we will know it is solved.*0114

*Let us take a look at one of these.*0119

*I have 1-2 / x + 1 = 2x / x +1 and we can see that this is a rational equation because looking at all these rational expressions I have,*0121

*think of those fractions.*0131

*To help out Iâ€™m going to work on getting a common denominator.*0134

*Already these 2 parts of it have an x + 1 in the bottom.*0139

*We are just going to focus on trying to give an x + 1 to the one over there on the left.*0144

*I can do that by taking 1 / 1 and then multiplying the top and bottom of that by x + 1.*0151

*(x + 1)(x +1) - 2 / (x + 1) = 2x /(x +1)*0162

*On the left side that allows us to finally combine things now that we have a common denominator.*0175

*In doing so, we just put the top together.*0182

*x + 1 - 2 / x +1*0184

*Doing a little bit of simplifying let us go ahead and subtract 2 from 1, x - 1 now x + 1 = 2x / x +1.*0196

*At this stage notice how we have two fractions that are set equal to one another and the bottoms are exactly the same.*0211

*Since the bottoms are the same then I know that the tops of these must also be the same.*0219

*Let us just focus on the top part for a little bit and see if we can find a solution out of that.*0225

*x - 1 = 2x this one is not so bad to solve.*0230

*Let us just go ahead and subtract an x from both sides and I think we will be able to get a possible solution.*0236

*It looks like a possible solution is that -1 = x.*0245

*Be careful this can sometimes happen with these rational equations.*0249

*It looks like we have done all of our steps correctly and it looks like we have found a solution, but this guy does not work.*0253

*To see why it does not work, take this possible solution all the way back to one of our original denominators*0260

*and you can see that if you try and plug it in and it makes the bottom 0.*0267

*We have to get rid of that possible solution.*0272

*Since we do not have any other possibilities left, we can say that this particular equation has no solution.*0275

*With these types of equations, even when you find something that looks like it is a solution you have to go back*0284

*and check to make sure that it make sense in the original.*0291

*Oftentimes you may end up getting rid of things that will simply make that denominator 0.*0294

*Let us try a different one.*0301

*This one is 2 /(p ^{2} - 2p) = 3 /(p^{2} â€“ p)*0302

*Before working with this, I need to see what is in the denominator.*0311

*To get a better idea we are going to go ahead and factor.*0316

*In this side I can see that they both have a p in common, we will go ahead and take that out.*0321

*I get p Ã— p -2.*0326

*On the right side here, it looks like it has a p in common as well, p Ã— p -1.*0330

*They almost have the same denominator, but I need to essentially give each one its missing factor.*0339

*Let us give the left side p -1 and we will give the right side p -2.*0346

*Just copy this down and see where the extra pieces are coming from.*0353

*On the left I will put in p -1 on the bottom and on the top.*0365

*The one on the right we will give p - 2 in the bottom and p -2 on top.*0374

*Distributing this will give us (2p -2) / and now we have this nice common denominator equals (3p â€“ 6) / (p) (p -1) (p -2)*0385

*This is right back to that situation we had before.*0406

*The bottoms of each of these fractions are exactly the same I know that the tops of the fractions will also end up being the same.*0409

*Let us extract out just the tops and focus on those.*0418

*If we have to solve this what can we do?*0425

*I will go ahead and add 2 to both sides 2p = 3p - 4 and let us go ahead and subtract 3p from both sides, -p = -4.*0428

*One last step, let us go ahead and multiply both sides by -1.*0448

*My possible solution is that p = 4.*0455

*We cannot necessarily just assume that that is the solution, not until we go back to the original*0459

*and make sure that it is not going to give us 0 in the bottom.*0464

*Let us check to see what factors are in the bottom.*0468

*Some of our restricted values that we simply cannot have is 1p = 0, we cannot have that one.*0473

*We also cannot have p = 2 and we cannot have p = 1.*0480

*Those three values are restricted.*0487

*Fortunately when we look down at what we got, it is not any of these restricted values.*0490

*We will go ahead and keep it as our solution.*0497

*The solution to this one is p = 4.*0499

*Let us look at a little bit large one, one that has a few more things in the denominator.*0505

*This one is 1 / x -2 + 1/5 =2/5 Ã— (x ^{2} â€“ 4)*0511

*Quite a bit going on the bottom and it looks like we definitely need to factor one of our pieces*0518

*just so we can see what total factors we have in here.*0523

*1/x â€“ 2 + 1/5 = 2/ I have 5 and then the x ^{2} â€“ 4 that happens to be the difference of squares.*0528

*(x + 2)(x â€“ 2)*0545

*It looks like this will be our LCD and we are going to give it to these other fractions here with missing pieces.*0550

*The 1/x â€“ 2 and see what we can put into that one.*0560

*It needs to have a 5, we will put that down below and up top and it also needs to have an x + 2 and below and up top.*0569

*Onto the next fraction here, this is 1/5 and it looks like it is missing an x - 2 piece so we will throw that in there.*0585

*We are missing x + 2 missing both of those pieces.*0597

*On the other side of our equal sign it already has the LCD, so we just leave it just as it is.*0607

*We spread out quite a bit but now we are going to focus on the tops of everything since all the bottoms are now exactly the same.*0616

*We will also use our distribution property a little bit to help us out by multiplying 5 Ã— 2*0626

*and we will take the two binomials over here and we will foil them so we can put those together.*0633

*5x + 10 would be that first piece on the top of the first fraction.*0641

*We have our first terms x ^{2}, outside terms 2x, inside terms â€“ 2x and last terms -4.*0649

*We want that all equal to 2.*0658

*You can notice this is just the tops of all of our fractions.*0661

*Let us go ahead and combine some things.*0666

*My + 2x and my - 2x will go ahead and cancel each other out and we will have x ^{2}, I will take care of that one.*0669

*5x took care of that one and combining the 10 and -4, +6 = 2.*0680

*We are almost there. Let us just go ahead and subtract 2 from both sides and see that we need to solve a quadratic.*0689

*We are going to solve x ^{2} + 5x + 4 is all equal to 0.*0698

*This quadratic is not too bad I can end up factoring it using reverse foil without too much problem.*0703

*Our first terms x and x and two terms that need to multiply to give us 4 but add to give us 5.*0710

*I am thinking of 1 and 4, those will do it, both positive.*0719

*This will give us two possible solutions.*0725

*When I take x + 1 = 0 and x + 4 = 0, x could equal -1 or x could equal -4.*0727

*There are two possible things that I'm worried about.*0739

*Before we put our seal of approval on them,*0744

*let us go back to some of those original fractions and see what some of our restricted values are.*0746

*Looking at this first one, the bottom would be 0 when x is equal to 2 so I know one of my restricted values x can not equal 2.*0752

*The bottom of this one is a number 5, so that one is not is not going to be 0.*0762

*Over here I can see that when x =-2 I will run into a 0 on the bottom, x cannot equal -2.*0768

*This factor here is exactly the same as the first one.*0776

*We already have that list as one of our restricted values.*0780

*As long as I know my possible solutions are 2 and -2, it looks like we are okay.*0783

*Since I'm looking at -1 and -4, both of these are going to be valid solutions, none of them are restricted.*0789

*The process of solving these rational equations you just have to work on combining them and focus on the tops for a bit.*0796

*Let us look at that formula as I promised.*0807

*To see how you could solve the formula involving some these rational expressions.*0809

*To make things interesting, let us go ahead and solve for y.*0814

*In order to get rid of a lot of these things in the bottom we will identify our lowest common denominator.*0822

*One has an x another one has y and the other one has z.*0830

*It looks like we will need all three of those parts, x, y and z in our new denominator.*0833

*We will go through and give each of them their missing pieces.*0840

*Here we have our originals, let us throw in the missing.*0846

*The first one it already has an x in the bottom, we will give it y and z on the top and the bottom.*0852

*The next one already has a y, so give it the x and z.*0860

*For the last one we have z so xy.*0868

*This will ensure that all of the bottoms are exactly the same.*0878

*We will go ahead and turn your attention to just the tops of all of these and we will continue moving forward.*0882

*2yz = 1 Ã— x Ã— z that will be just xz + xy.*0889

*We are trying to get these y isolated all by himself and looks like we have two of them, one on the left side and one on the right side.*0899

*Let us get them on the same side by subtracting an xy from both sides.*0907

*2yz â€“ xy*0914

*We can factor out that common y that way we will have 1y to deal with.*0922

*Let us go ahead and factor out the common y out front.*0928

*y and then left over we will have 2z â€“ x.*0931

*We are almost free with this one.*0941

*To get the y completely isolated, let us finally divide both sides by that 2z â€“ x.*0946

*y = xz / 2z - x and this one is solved because we have completely isolated the variable we are looking for.*0952

*In a sense we have created a new type of formula.*0970

*Be careful in solving many of these different types of rational equations.*0973

*Make sure you always check your solutions and use that least common denominator to help yourself out.*0976

*Thank you for watching www.educator.com.*0982

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