## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Adding and Subtracting Rational Expressions

- To add or subtract: first, find the LCM of the denominators. Then adjust the numerator and denominator of each fraction so that its denominator is the LCM. Finally, add or subtract the numerators and simplify the result.
- Most of the time, the result will not simplify.

### Adding and Subtracting Rational Expressions

- List the prime factors of 2x and 10x to find the LCD.
- 2x =
- 10x =
- 2x = 2*x
- 10x = 2*5*x
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = 2*5*x = 10x
- Change each rational expression into an equivalent expression with the LCD.
- [1/2x] − [2/10x] = ( [5/5] )[1/2x] − [2/10x] =
- [5/10x] − [2/10x] =
- [(5 − 2)/10x] =

- List the prime factors of 7x and x to find the LCD.
- 7x =
- x =
- 7x = 7*x
- x = x
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = 7*x = 7x
- Change each rational expression into an equivalent expression with the LCD.
- [1/7x] + [2/x] = [1/7x] + [2/x]( [7/7] ) =
- [1/7x] + [14/7x] =
- [(1 + 14)/7x] =

^{2})] + [5/(y

^{2})]

- List the prime factors of xy
^{2}and y^{2}to find the LCD. - xy
^{2}= - y
^{2}= - xy
^{2}= x*y*y - y
^{2}= y*y - Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = x*y*y = xy
^{2} - Change each rational expression into an equivalent expression with the LCD.
- [10/(xy
^{2})] + [5/(y^{2})] = [10/(xy^{2})] + [5/(y^{2})]( [x/x] ) = - [10/(xy
^{2})] + [5x/(xy^{2})] =

^{2})] =

^{3})] − [7/a]

- List the prime factors of a
^{3}and a to find the LCD. - a
^{3}= - a =
- a
^{3}= a*a*a - a = a
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = a*a*a = a
^{3} - Change each rational expression into an equivalent expression with the LCD.
- [9/(a
^{3})] − [7/a] = [9/(a^{3})] − [7/a]( [(a^{2})/(a^{2})] ) = - [9/(a
^{3})] − [(7a^{2})/(a^{3})] =

^{2})/(a

^{3})] =

- List the prime factors of 3x + 6 and x + 2 to find the LCD.
- 3x + 6 =
- x + 2 =
- 3x + 6 = 3*(x + 2)
- x + 2 = (x + 2)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = 3*(x + 2)
- Change each rational expression into an equivalent expression with the LCD.
- [2/(3x + 6)] + [5/(x + 2)] = [2/(3(x + 2))] + [5/(x + 2)]( [3/3] ) =
- [2/(3(x + 2))] + [15/(3(x + 2))] =
- [(2 + 15)/(3(x + 2))] =
- [17/(3(x + 2))]
- or

- List the prime factors of 2x − 8 and x − 4 to find the LCD.
- 2x − 8 =
- x − 4 =
- 2x − 8 = 2*(x − 4)
- x − 4 = (x − 4)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = 2*(x − 4)
- Change each rational expression into an equivalent expression with the LCD.
- [7/(2x − 8)] − [2/(x − 4)] = [7/(2(x − 4))] − [2/((x − 4))]( [2/2] ) =
- [7/(2(x − 4))] − [4/(2(x − 4))] =
- [(7 − 4)/(2(x − 4))] =
- [3/(2(x − 4))]
- or

- List the prime factors of x + 1 and 4x + 4 to find the LCD.
- x + 1 =
- 4x + 4 =
- x + 1 = (x + 1)
- 4x + 4 = 4(x + 1)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = 4*(x + 1)
- Change each rational expression into an equivalent expression with the LCD.
- [2x/(x + 1)] + [x/(4x + 4)] = ( [4/4] )[2x/((x + 1))] + [x/(4(x + 1))] =
- [8x/(4(x + 1))] + [x/(4(x + 1))] =
- [(8x + x)/(4(x + 1))] =
- [9x/(4(x + 1))]
- or

^{2}− 16)] + [2/(x + 4)]

- List the prime factors of x
^{2}− 16 and x + 4 to find the LCD. Factoring may be required. - x
^{2}− 16 = - x + 4 =
- x
^{2}− 16 = (x + 4)(x − 4) - x + 4 = (x + 4)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = (x + 4)(x − 4)
- Change each rational expression into an equivalent expression with the LCD.
- [7x/(x
^{2}− 16)] + [2/(x + 4)] = [7x/((x + 4)(x − 4))] + [2/(( x + 4 ))]( [((x − 4))/((x − 4))] ) = - [7x/((x + 4)(x − 4))] + [(2(x − 4))/(( x + 4 )(x − 4))] =
- [(7x + 2(x − 4))/((x + 4)(x − 4))] =
- [(7x + 2x − 8)/((x + 4)(x − 4))]
- [(9x − 8)/((x + 4)(x − 4))]
- or

^{2}− 16)]

- List the prime factors of x
^{2}− 16 and x + 4 to find the LCD. Factoring may be required. - x
^{2}− 16 = - x + 4 =
- x
^{2}− 16 = (x + 4)(x − 4) - x + 4 = (x + 4)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = (x + 4)(x − 4)
- Change each rational expression into an equivalent expression with the LCD.
- [7x/(x
^{2}− 16)] + [2/(x + 4)] = [7x/((x + 4)(x − 4))] + [2/(( x + 4 ))]( [((x − 4))/((x − 4))] ) = - [7x/((x + 4)(x − 4))] + [(2(x − 4))/(( x + 4 )(x − 4))] =
- [(7x + 2(x − 4))/((x + 4)(x − 4))] =
- [(7x + 2x − 8)/((x + 4)(x − 4))]
- [(9x − 8)/((x + 4)(x − 4))]
- or

^{2}− 16)]

^{2}+ x − 20)] + [2/(x + 5)]

- List the prime factors of x
^{2}+ x − 20 and x + 5 to find the LCD. Factoring may be required. - x
^{2}+ x − 20 = - x + 5 =
- x
^{2}+ x − 20 = (x + 5)(x − 4) - x + 5 = (x + 5)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = (x + 5)(x − 4)
- Change each rational expression into an equivalent expression with the LCD.
- [3x/(x
^{2}+ x − 20)] + [2/(x + 5)] = [3x/((x + 5)(x − 4))] + [2/(x + 5)]( [((x − 4))/((x − 4))] ) = - [3x/((x + 5)(x − 4))] + [(2(x − 4))/((x + 5)(x − 4))] =
- [(3x + 2(x − 4))/((x + 5)(x − 4))] =
- [(3x + 2x − 8)/((x + 5)(x − 4))]
- [(5x − 8)/((x + 5)(x − 4))]
- or

^{2}+ x − 20)]

*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

### Adding and Subtracting Rational Expressions

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
- Least Common Multiple (LCM) 0:27
- Examples: LCM of Numbers
- Example: LCM of Polynomials
- Adding and Subtracting 7:55
- Least Common Denominator (LCD)
- Example: Numbers
- Example: Rational Expressions
- Equivalent Fractions
- Simplifying Complex Fractions 21:19
- Example: Previous Lessons
- Example: More Complex
- Example 1: Find LCM 28:30
- Example 2: Add Rational Expressions 31:44
- Example 3: Subtract Rational Expressions 39:18
- Example 4: Simplify Rational Expression 38:26

### Algebra 2

### Transcription: Adding and Subtracting Rational Expressions

*Welcome to Educator.com.*0000

*Today, we are going to be working on adding and subtracting rational expressions.*0002

*And recall that rational expressions are algebraic fractions; so this is more complicated than multiplying or dividing rational expressions.*0006

*But we are just going to apply the same techniques that we used to add or subtract fractions involving numbers to add or subtract rational expressions.*0017

*Let's start out by reviewing some concepts from working with fractions.*0027

*Recall the idea of the least common multiple, because we are going to use that to get a common denominator.*0032

*As you know, if you have a common denominator with your fractions (such as 2/4 + 1/4), it is very easy to add.*0038

*You just add the numerators (2 + 1 is 3) and use the common denominator.*0046

*It is the same with rational expressions: if you already have a common denominator, add the numerators*0052

*and use the common denominator as the denominator of the rational expression that you end up with from adding or subtracting.*0058

*However, just like with fractions, when you don't have a common denominator,*0069

*you have to find a common denominator and convert to equivalent fractions before you can add or subtract.*0073

*Let's talk about an example using numbers: if I want to add 1/10 and 2/45, you might just look at this,*0081

*know what the common denominator is, quickly convert it, possibly even in your head, and add it.*0091

*But I am going to bring out each step of that consciously, so we can use that to work with adding and subtracting rational expressions.*0096

*Let's look at the denominators: let's find the least common multiple of these two numbers, 10 and 45.*0105

*In order to find a common multiple, you need to factor.*0113

*And whether we are working with a number, or if the denominator is a polynomial (as in a rational expression),*0118

*the idea is still that you have to factor.*0123

*So, let's factor this out: and I get 2 times 5 here; for 45, I might say it's 9 times 5, but I can go farther with my factoring.*0127

*9 factors into 3 times 3; I could also rewrite this as 3 ^{2} times 5.*0137

*So, I have 2 times 5 for 10, and then I have 3 ^{2} times 5 as my prime factorization for 45.*0144

*The least common multiple is going to be the product of the unique factors that I see here,*0152

*to the highest power that they appear for any one number (or, when we are talking about polynomials, for any one polynomial).*0161

*So, the unique factors: 2 is a factor, and I am going to take the product of that and the other factors I see.*0170

*5 is here once only; and it is here once only; so the greatest number of times that 5 appears in any one set of factors is once.*0179

*So, I am not going to write it twice: I am just going to write it once.*0188

*I am done with this one; here I still have another factor that isn't accounted for; I have a 3.*0192

*But it appears twice, so I am going to put 3 ^{2}.*0197

*Now, if I had 5 twice up here, then I would have written 5 ^{2}; I only had it once.*0201

*All right, this equals...2 times 5 is 10, times 3 ^{2}...so 10 times 9 is 90.*0206

*The least common multiple of these two is 90; and I am going to end up using that as my common denominator.*0215

*But as you know, I can't just change this to 1/90 and add that to 2/90, because those fractions, then, won't be the same.*0221

*So, we are going to talk about, once you get the common denominator that you are going to use, what to do with the rest of the fraction.*0228

*And we will talk about that in a minute; but first, let's apply this concept to finding the least common multiple of polynomials.*0234

*If I was working with something like 1 over x ^{2} plus 7x, plus 14, and I wanted to add that*0241

*to 1 over 2x ^{2} - 2x - 12, then I need to find the common multiple of these two denominators.*0250

*So, to find the LCM, I am going to factor out both of these denominators, just the way I did up here with the numbers.*0260

*OK, so this is going to give me x, and then x; I have all plus signs, so I know that these are going to be positive.*0272

*Factors of 14...actually, let's make this a 10...let's go ahead and make that a 10 for much easier factorization.*0282

*This is going to be factors of 10: 1 and 10, and 2 and 5; and I need them to add up to 7, and 2 + 5 = 7.*0298

*OK, so this is factored out as far as I can go.*0313

*Now, down here, I am aware that I have a common factor of 2;*0319

*so I am going to pull that out to get 2(x ^{2} - x...and that is going to give me - 6).*0325

*Now, work on factoring this: 6 has factors that are 1 and 6, and 2 and 3.*0341

*I have an x here; I have an x here; and since this is a negative, I have to have a positive and a negative.*0351

*And I want them to add up to -1, so I need to find factors that are close together.*0359

*And I need to make the larger factor negative, so that this will add up to -1.*0364

*Let's try 2 - 3; that equals -1; so I am going to make the 2 positive and the 3 negative.*0368

*OK, now I factored out both denominators; and this is going to then give me my least common multiple.*0377

*I am going to look for each factor that appears.*0386

*And I see an x + 2; it appears here once, and it only appears here once.*0389

*So, the highest power to which it appears is just 1, so it is x + 2.*0394

*I also have an x + 5; it appears once, and I am going to put it once.*0401

*This is done; I accounted for this; I also need an x - 3.*0412

*Therefore, the least common multiple of these denominators is going to be (x + 2)(x + 5)(x - 3)...*0416

*and let's not forget about the 2; the 2 needs to come along as well, so 2(x + 2)(x + 5)(x - 3).*0427

*If I had a situation where I factored something out, and I got (x + 3)(x + 3)(x - 1), and then I had down here*0438

*(x + 3)(x - 5), then the LCM would be...I have an (x + 3), but it appears twice in this polynomial, so it would be (x + 3) ^{2}.*0448

*And then, I would have my x - 1 and x - 5.*0461

*So, the least common multiple will allow you to find a common denominator when you are adding or subtracting rational expressions.*0468

*To add or subtract, first find a common denominator.*0476

*The LCM of the denominators is the smallest value that we can use as a common denominator.*0479

*And we are going to refer to that as the least common denominator.*0484

*The least common denominator is just the LCM of the two denominators.*0487

*OK, so let's go back to our example with numbers.*0492

*And I had 1/10, and I asked you to add 2/45.*0497

*We factored out 10 to get 2 times 5; we factored out 45 to get 3 ^{2} times 5.*0505

*The LCM...a 2 appears up here; 5 appears, at most, once; and 3 appears twice, so it is 3 ^{2}.*0516

*This equals 90; now, I am going to use this 90 as the least common denominator; so, the least common denominator for these two fractions is 90.*0528

*But I can't simply say, "OK, I am going to make this 1/90 and 2/90"; those won't be the same fractions.*0538

*I need to convert these to equivalent fractions, but with my new denominator.*0544

*Again, this is something you just know how to do; but we need to bring it out and look at each step.*0548

*So, let's talk about converting 1/10: if I rewrite this as 1/(2 times 5), it is factorization.*0552

*That allows me to look at the LCM and determine what I am missing--what factor am I missing?*0559

*I have a 2; I have a 5; but I am missing the 3 ^{2}.*0566

*The thing is that I can't just multiply the denominator alone; what I am allowed to do*0572

*is multiply both the numerator and the denominator by the same number, because this is just 1--*0577

*this cancels out to 1--and I am allowed to multiply by 1.*0584

*So, this is 1 times 9; this is 2 times 5 is 10, times 9 is 90; and of course, that simplifies to 1/10, so I know I have a fraction that still has the same value.*0587

*2/45 can be written as 2 over (3 ^{2} times 5); what is missing from this denominator?*0602

*Well, I have a 5; and I have a 3 ^{2}; I look down here--the 2 is missing.*0611

*So, I need to multiply both the numerator and the denominator by 2.*0621

*2 times 2 is 4, over 9 times 5 is 45, times 2...so that is 4/90.*0627

*Now, I can add these two: I have 9/90 + 4/90; that is 13/90.*0633

*We are doing the same thing when we add rational expressions or subtract rational expressions.*0641

*We are going to take the same steps: we are going to find the least common multiple, and that will be our common denominator.*0646

*Then we are going to convert to fractions that are equivalent to the original fractions, but with the new denominator.*0651

*And then, we can add or subtract those once they have the same denominator.*0658

*OK, looking at an example using rational expressions: here is my first rational expression, and I am going to add that to (x + 4)/(2x - 6).*0663

*My first step is to find the LCD: factor out the denominators, and then find the least common multiple to use that as the denominator.*0678

*So, I have my two denominators: this factors into, since it has a negative, (x + something) (x - something).*0692

*The only factors of 3 that I am going to have to work with are 1 and 3.*0703

*Since I want to end up with a negative 2x, I am going to make the larger number negative: 1 - 3 is -2.*0708

*So, I am going to make this a 2: (x + 2) (x - 3), and (x + 1) (x - 3); so that is*0717

*x ^{2} - 3x + 1x (so that is -2x), and then the last terms multiply out to -3.*0728

*OK, and this should be 2x - 6, so I am going to look at the common facto that I have, which is 2.*0738

*So, this factors out to 2(x - 3); the LCM is the product of the factors of both of these.*0756

*I have (x + 1), and that only appears once; that is the highest power to which it appears in any of these polynomials.*0771

*So, I am just going to put (x + 1).*0780

*I also have an (x - 3), and that is present once here and once here, so I represent it once.*0782

*Down here, I also have a 2; I am going to pull that out in front; the LCM is 2(x + 1) (x - 3).*0789

*And I am going to use it as the least common denominator.*0797

*We found the common denominator; next we need to find the equivalent fractions, using that common denominator (using the LCD as their denominator).*0805

*Once I have done that, all I have to do is add or subtract the numerators of the fractions, and then simplify.*0814

*Let's continue on with the example I started in the previous slide, which was 1/(x ^{2} - 2x - 3) + (x + 4)/(2x - 6).*0822

*This factored out to 1/(x - 3) (x + 1).*0839

*This, you recall, factored into 2(x - 3).*0859

*So, I factored these out; I haven't done anything else with them yet--I have left them the same, but I have factored them out.*0869

*So, I have them factored: the LCD is going to be the product of the unique factors.*0874

*Remember, I had an (x - 3) as part of that; I had an (x + 1), which appears here; I already have my (x - 3) accounted for; and I need a 2.*0882

*So, I am going to rewrite this with a 2 out in front; and I am going to use my least common multiple as my denominator, my LCD.*0893

*2 times (x - 3) times (x + 1)...*0904

*The hard part can be converting to equivalent fractions.*0909

*You just have to be careful and make sure that you multiply the numerator and the denominator by the same thing.*0912

*Otherwise, you will not end up with an equivalent fraction.*0917

*To convert to equivalent fractions is my next step.*0922

*I am going to take this first fraction, and I am going to say, "OK, what is lacking?"*0935

*I want to turn this denominator into this.*0939

*I have my (x - 3); I have my (x + 1); but I am missing a 2.*0944

*Therefore, I am going to multiply both the numerator and the denominator by 2/2, which is just multiplying this fraction by 1.*0949

*The second fraction: to turn this denominator into what I have here, I have to figure out what I am missing.*0960

*I have a 2; I have an (x - 3); I am missing the (x + 1).*0974

*So, I am going to multiply both the numerator and the denominator by (x + 1).*0978

*Again, this is just multiplying by 1.*0984

*This is going to give me...1 times 2 is 2, over (x - 3) (x + 1) (2)...and I am pulling the 2 out in front.*0988

*Here, I am going to add that; and in the numerator, I am going to have (x + 4) (x + 1) for the second fraction, for the numerator.*1006

*In the denominator, I am going to have that same common denominator: 2/(x - 3)(x + 1).*1022

*So, I have just multiplied the numerator and the denominator by whatever was missing from the denominator to form the LCD.*1031

*Now that I have a common denominator, I just add or subtract the numerator of these fractions.*1041

*So, since I am adding, this is going to become 2 + (x + 4)(x + 1), all over my common denominator.*1047

*In the same way, if I am adding 3/4 and 5/4, I just add the numerators and put them over the common denominator.*1065

*The same idea here--I added the numerators and put them over the common denominator.*1076

*Finally, I am going to simplify: I am going to need to multiply this out, and this is going to give me 2 +...let's work up here:*1084

*(x + 4) times (x + 1): using FOIL, this is going to give me x ^{2}, and then the outer term is going to be + x;*1094

*then inner term is going to be 4x, so that is 5x; and then, the last term is going to be 4.*1106

*So, this is going to give me x ^{2} + 5x + 4, all over 2 times (x - 3)(x + 1).*1111

*So, I am working right over here, just to do that multiplication.*1129

*(x - 3) times (x + 1)...and then we are going to have to multiply all of that by 2.*1133

*That is going to give me...actually, we are going to go ahead and leave that in factored form, because that will make it easier to simplify.*1141

*Let's just go ahead and leave that one in factored form.*1151

*So, the top gives me x ^{2}, and...I have 5x; I leave that alone; and this is 2 + 4, to give me 6, all over this.*1157

*Now, at the top, I went ahead and multiplied this out, so I could add it to this.*1174

*For the denominator, I didn't need to do that, because it is already factored out; and you will see why in a second.*1179

*OK, can I factor this? Well, let's go ahead and see.*1185

*I have x and x, and right here, I have positives; so I am going to make that a + and make that a +.*1189

*6...what are my factors? 1 and 6, 2 and 3.*1201

*So, I am looking for factors of 6 that are going to end up giving me 5; and I can see that 2 + 3 = 5, so this is going to be (x + 3) (x + 2).*1207

*That is x ^{2} + 2x + 3x (is 5x) + 6.*1219

*OK, now you can see why I just left the denominator in factored form.*1228

*I look here, and I see if I can simplify--are there any common factors?*1232

*And there are actually not--I don't have any common factors in the numerator or denominator.*1235

*And it is perfectly fine to leave the expression like this.*1239

*You might want to go on and do something else with it, and then it is already factored.*1243

*For something like we had in the numerator, where we ended up with 2 plus all of this, we need to actually combine that.*1246

*So, I needed to multiply this out, add, and then factor.*1254

*This was already factored out; so I left it, and this was my final answer.*1258

*So again, I found the LCD of these two by factoring.*1262

*This was the LCD; I converted the first and the second fractions into equivalent fractions, and then I just added and tried to simplify.*1268

*OK, we have talked before about simplifying complex fractions, but this time*1279

*we are going to talk about complex fractions that also may require you to add or subtract in the numerator or denominator.*1284

*So, first I want to explain the difference between what we did in a previous lesson and what we are going to do now.*1290

*In the previous lesson, we talked about things like this, rational expressions that are complex fractions: (3xy/(x ^{2} - 16))/ ((2x + 1)/xy).*1297

*So, the steps for this were to rewrite this as division, because I know that this fraction bar is telling me to take the numerator and divide it by the denominator.*1314

*Once you got to this point, you recognized that this is simply the first rational expression, times the reciprocal of the second.*1330

*And I am not going to do the whole multiplication right now; I just wanted to show you the setup on that.*1352

*So, if I had a complex fraction, such as this, I just took the numerator, set it as divided by the denominator,*1357

*and then converted it to multiplication of the first rational expression times the inverse of the second.*1364

*OK, what we are talking about here is actually different.*1373

*It is more complex, and requires one more step before you can go from here to here.*1377

*Let's say I have something like this: (1/x + 2/y)/(3/x - 1/y).*1385

*Now, not only do I have a complex fraction (I have a situation where I have a fraction in the numerator and a fraction in the denominator),*1400

*but I have a sum and/or a difference (here I have a sum; here I have a difference).*1408

*I have a fraction up here that I am adding to another fraction; I have a fraction down here that I am subtracting from another fraction.*1413

*Up here, I just had one fraction; I could just go ahead and simplify by multiplication; down here, I just had one fraction.*1419

*So, the difference here is: we are talking about simplifying complex fractions*1426

*in which you need to add or subtract in the numerator or the denominator of the complex fraction.*1430

*What you are going to do is handle these separately.*1437

*You are going to handle the numerator; you are going to handle the denominator; and then you are going to put it all together.*1439

*So, to simplify a complex fraction, add or subtract the fractions in the numerator and the denominator separately, and then simplify.*1443

*I am going to start out with the numerator.*1451

*In the numerator, I have 1/x + 2/y; I need to find a common denominator.*1457

*And since all I have in the denominator is an x, and all I have in the denominator over here is a y, then my LCD (or my LCM) is going to be xy.*1464

*Now, I need to convert these to equivalent fractions.*1477

*I see that what I am lacking from this denominator is a y, so I need to multiply this times y/y.*1485

*Now, I have an xy in the denominator, and I multiplied the numerator by the same thing, so that I end up with equivalent fractions.*1495

*Over here, I want to get the LCD of xy; what I am lacking in the denominator is an x.*1501

*So, I am going to multiply both the numerator and the denominator by x.*1508

*This is going to give me y/xy, plus 2x/xy.*1513

*Now, I can add those, because they have the same denominator.*1526

*And it is just going to be (y + 2x)/xy.*1528

*Now, I go back here, and what I have in the numerator now is (y + 2x)/xy.*1534

*I no longer have the sum, where I have two separate fractions: I have one fraction in the numerator.*1546

*OK, denominator: the same thing--the denominator is 3/x - 1/y.*1556

*Again, I have an LCD down here--my LCD is just going to be xy, because this is the only factor here; this is the only factor here.*1577

*I need to convert these to equivalent fractions, so what I am going to do is say, "All right, what am I lacking from this?"*1588

*I am lacking a y in the denominator, so I am going to multiply this times y/y.*1598

*And I am subtracting that from 1/y, and what I am lacking in the common denominator is an x.*1606

*So, I am going to multiply this times x/x; this is going to give me 3y/xy - x/xy.*1612

*I now have a common denominator: this gives me (3y - x)/xy: this is my denominator, (3y - x)/xy.*1629

*That was the hard part: once you get to here, you are working with this situation.*1643

*I could handle this by rewriting this as a division problem: (y + 2x)/xy--I am going to take that;*1649

*it is going to be like this first rational expression; and I am going to say "divided by" this whole thing.*1657

*Once you get to that point, you use our usual method of taking the first rational expression and multiplying it by the inverse of the second.*1663

*This is the difficult step: and the way to really look at this is to handle the numerator and the denominator separately.*1672

*Your goal is to get the numerator to look like this (a single fraction) and the denominator to look like this.*1678

*This is an addition problem with a rational expression; this is an addition problem with a rational expression.*1684

*And once you take care of the numerator and you take care of the denominator, then you can proceed as usual.*1693

*This is a complex problem: it just has a lot of steps, and you just need to take it one at a time and keep track of what you are working with.*1702

*OK, with the examples, we are going to start out just finding the LCM; but this time, it is going to be of three polynomials.*1712

*So, when we find the LCM, no matter if it is 2, 3, 4, or more, we need to factor.*1718

*I am going to factor the first polynomial, the second polynomial, and the third polynomial.*1728

*OK, so this is x; and I have a negative sign here, but I have a positive sign here.*1743

*That clues me into the fact that I have a negative and a negative.*1750

*I have factors of 16, which are 1 and 16, 2 and 8, and 4 and 4; and I need factors of 16 that will add up to -8.*1755

*And I can see that this one is correct, (x - 4) (x - 4), because that is going to give me a middle term of -8x.*1767

*OK, this second set of factors is going to be x and x, and everything is positive.*1782

*And factors of 4 that would add up to 4 would be 2 and 2: so, x ^{2} + 2x + 2x (that is going to give me 4x) + 4.*1792

*Finally, I have a negative here; so I am going to do a plus here and a negative here.*1806

*I want factors of 8 that add up to -2: well, 1 and 8 are too far apart, so I have 2 and 4.*1814

*And I want it to be a -2, so I am going to make the 4 negative.*1824

*This is going to give me a 2 here and a 4 here.*1828

*OK, I could actually rewrite this, also, as (x - 4) ^{2}; and I am going to rewrite this as (x + 2)^{2}.*1832

*So, when I look for my LCM, I am going to look for each factor that I have.*1844

*And the first one is (x - 4), and the highest power I have it to is 2; I have an (x - 4) here, but this is really only to the first power.*1849

*So, I am going to write this as (x - 4) ^{2}.*1858

*I took care of that factor that is right here, also.*1865

*Now, I have another unique factor of (x + 2).*1869

*The highest power in any one polynomial that I find it in is squared; I have an (x + 2) here,*1874

*but it is to a lower power, so I am not going to worry about that; it is covered under this.*1879

*The least common multiple of these three polynomials is (x - 4) ^{2} (x + 2)^{2}.*1884

*Factor out each polynomial, and then take the product of their factors and use the power*1891

*that is the highest power that any factor is present in, in one of the polynomials.*1897

*Here is addition: adding rational expressions with different denominators.*1905

*The first step is to get a common denominator.*1910

*To achieve that, I am going to factor the denominators.*1915

*There is a negative here, so this is plus; this is minus.*1925

*I need factors of 6 that add up to -1; I am going to look at these two, and if I make the 3 negative, then I am going to get the correct middle term.*1930

*So, I am going to put the 3 here and the 2 here.*1943

*OK, so I factored out that first denominator; let's look at the second one.*1946

*Leave the numerator alone for now, and concentrate on the denominator.*1952

*I can see, pretty quickly, that I have a common factor of 4; so that will become x, and this will become minus 3.*1956

*So, this is what I want to add; and I need to find the LCD, so I am going to look at all of these factors that I have in the denominator and find their product.*1965

*I have (x + 2), and that only appears once, so I just leave it as the first power.*1978

*I have (x - 3) here and here, and the highest number of times it appears is once and once; so it is (x - 3).*1984

*Over here, I also have a 4; so the LCD...I am going to rewrite this with a 4 in front...is going to be 4(x + 2) (x - 3).*1994

*Now, I need to rewrite these as equivalent fractions with this denominator.*2007

*I look at the denominator, and I see what I am lacking.*2014

*In this first denominator, I have (x + 2) (x - 3), but I am lacking a 4.*2018

*So, I am going to multiply both the numerator and the denominator by 4 to form an equivalent fraction.*2025

*This is going to give me 4(2x - 3)/(x + 2)(x - 3).*2036

*Here, I am going to end up with...let's work with this one right down here...(3x ^{2} - 2x)/(4(x - 3)).*2048

*What is lacking from the denominator? (x + 2)*2061

*I have my 4; I have my (x - 3); I am lacking an (x + 2).*2065

*So, I am going to multiply both the numerator and the denominator by that.*2070

*This is going to give me (x + 2)(3x ^{2} - 2x), all over...oops, I need a 4 down there, as well;*2078

*this 4 should be over there...times 4, times (x - 3)(x + 2).*2093

*OK, I have a common denominator--it is a slightly different order that I wrote it in, but I still have a 4, an (x + 2), and an (x - 3).*2100

*Now, I can add these: I am going to go ahead--I have my equivalent fractions, and I am going to add these.*2106

*4(2x - 3) divided by 4(x + 2) (x - 3), plus (x + 2)(3x ^{2} - 2x) over this common denominator,*2113

*4...I am going to write this in the same order as I wrote this one...(x + 2) (x - 3).*2133

*Now, once you have a common denominator, all you need to do is add the numerators and put these over the common denominator.*2139

*And the common denominator here is 4(x + 2) (x - 3).*2160

*Here, this is all factored out; I need to add this and then see what I have an if there are common factors.*2170

*I need to go ahead and multiply this all out.*2176

*In the numerator, this is 4 times 2x (that is 8x) minus 12; here I have x times 3x ^{2}, gives me 3x^{3}.*2179

*x times -2x gives me -2x ^{2}; I took care of this and this using the distributive property.*2197

*2 times 3x ^{2} is 6x^{2}; 2 times -2x is -4x.*2212

*This is all, again, over that common denominator.*2220

*Now, I can do a little more simplifying, because I can add like terms.*2226

*My denominator is taken care of; let's look at this numerator.*2231

*Starting with the largest power: I only have one x ^{3} term, so that is 3x^{3}.*2237

*For x ^{2} terms, I have 6x^{2} - 2x^{2}; that is going to give me 4x^{2}.*2243

*For x terms, I have...let's see...8x - 4x; that is going to give me 4x; for constants, I have -12.*2249

*Now, what I am left with is this rational expression.*2266

*And in order to simplify this, the way that you have to go about it is to use synthetic division.*2269

*And there actually turn out to be no common factors, so I am going to leave it as it is.*2276

*But you could check it by synthetic division; and the way to proceed would be to use synthetic division, and to divide this polynomial by (x + 2).*2282

*And remember from the remainder theorem: if the remainder is 0, then (x + 2) is a common factor, and you would be able to cancel that out.*2292

*I know 4 is not a common factor; I could also use synthetic division to determine that (x - 3) is not a factor of this.*2299

*But if you did have common factors, then the final step would be to cancel those out.*2306

*So, this was a pretty complicated problem; but proceeding as usual, factor the denominators, finding the least common denominator.*2312

*Then convert to equivalent fractions by multiplying the numerator and denominator by whatever was lacking from the denominator to form the LCD.*2321

*And this was 4 before, times this first rational expression.*2337

*Down here, I had to multiply the second one times (x + 2), divided by (x + 2).*2343

*I ended up with these two, with a common denominator; I rewrote them up here, and then I just added the numerators and did simplification.*2349

*OK, in this example, we are going to be subtracting rational expressions with different denominators.*2358

*The first thing to do is factor out the denominators to find the LCD.*2365

*Factoring the first one: (x - 7) divided by...I am going to have an x here and an x here...*2369

*Now, this is a negative sign; so I am going to have + and -.*2376

*Factors of 12 are 1 and 12, 2 and 6, 3 and 4; factors of 12 that would add up to -4 would be these two, 2 - 6.*2381

*So, I am going to put my 2 by the positive sign, and the 6 by the negative sign.*2397

*And I am going to leave some space here for when I convert these to equivalent fractions.*2404

*I am going to put my negative sign right here; and this is going to give me 2x + 3, and I am going to go ahead and factor this.*2409

*This one is a little bit more complicated to factor, because the leading coefficient is not 1.*2417

*So, I am going to get 2x here and an x here.*2422

*Now, I have a negative sign here, but I have a positive sign in front of the constant.*2426

*And that tells me that I am working with a negative and a negative, which will give me a positive here and a negative here.*2431

*This is more complicated, because I have to take into account this 2.*2440

*So, let's think about factors of 18, first of all, which are 1 and 18, 2 and 9, and 3 and 6.*2445

*Now, when I am working with a leading coefficient other than 1, I like to start out with smaller numbers,*2456

*because the one that is being multiplied with the 2 is going to become large.*2461

*So, let's start out with 3 and 6 and look at different combinations.*2466

*If I have 2x (let's put the 3 first) - 3, times (x - 6), that is going to give me 2x ^{2};*2471

*the outer terms give -12x; the inner terms give -3x; and the last terms give 18.*2482

*Since -12x and -3x add up to -15x, this is the correct factorization.*2490

*OK, now thinking about the LCD (least common denominator): I look at the factors I have, and I have an (x + 2).*2499

*And it is only present once, so it is just a power of 1.*2510

*I also have (x - 6), and I have one here, and it is only present once here; so again, I am just going to represent that once.*2516

*And so, this one is taken care of; over here, I also have (2x - 3).*2525

*Therefore, the LCD is going to be (x + 2) (x - 6) (2x - 3).*2530

*Now, I need to convert these to equivalent fractions.*2536

*To do that, I am going to multiply the numerator and the denominator by what is lacking from the denominator.*2539

*Here, the factor that is missing is 2x - 3, so I need to multiply both the numerator and the denominator by that.*2547

*Over here, I have the (2x - 3); I have (x - 6); but I need to multiply both the numerator and the denominator by (x + 2).*2559

*Now, when I multiply these out, I am going to end up with a common denominator.*2570

*I am going to take care of that multiplication: (x - 7) times (2x - 3), over (x + 2) times (x - 6) times (2x - 3);*2574

*minus this entire second fraction, which is going to be (2x + 3) times (x + 2), divided by the common denominator.*2595

*I am going to go ahead and write this in the same order as this one: (x + 2) first; (x - 6); and then (2x - 3).*2611

*OK, I am just checking to make sure that I have everything accounted for.*2629

*Now that I have a common denominator, I can subtract; so this is going to become (x - 7) times (2x - 3).*2632

*And I need to be careful with the signs; it is going to be subtracting, so this whole thing is going to be the opposite signs...over the common denominator.*2644

*OK, what is left is to simplify: this is already factored out, but I need to multiply this out,*2666

*add together the like terms, and then factor and see if I can simplify.*2673

*So, starting out right here, this is x times 2x; that gives me 2x ^{2}; x times -3...that is -3x;*2680

*here I get -14x (that is -7 times 2x); and then -7 times -3 is 21.*2699

*OK, minus what is in here: so, 2x times x is 2x ^{2}; 2x times 2 is 4x; now, the second term:*2711

*3 times x is 3x; 3 times 2 is 6; all over the common denominator, (x + 2) (x - 6) (2x - 3).*2732

*The next thing to do is take care of these signs: this equals 2x ^{2}...and do some simplifying.*2749

*-3x and -14x is -17x, plus 21; here I am going to have a negative; that gives me -2x ^{2}.*2757

*Inside here, I have 4x and 3x, so that is 7x; but I need to take the negative of that--the opposite; this is actually -7x.*2772

*For the constant I have 6, and the opposite of that is -6, all over the common denominator.*2780

*Now, I have some like terms that can be combined.*2790

*So, I am going to go down here; and this gives me 2x ^{2} - 2x^{2}; these cancel, so the x^{2} terms are gone.*2793

*I have -17x and -7x combined, to give -24x; that leaves me with the constants: 21 - 6 is 15...over the common denominator.*2806

*OK, looking at this, I can see that I don't have any common factors, so I can't simplify.*2834

*You could pull a 3 from up here--you could factor out a 3; but that is not going to leave you with any common factors.*2846

*There is no (x + 2), (x - 6), or (2x - 3) that is going to be left behind; so I can just leave this as it is.*2856

*This was a pretty complex problem--pretty lengthy.*2863

*Since it is subtraction, you have to be careful with the signs.*2867

*Again, you are factoring the denominators of both, finding the LCD right here, then converting to equivalent fractions.*2869

*This first fraction was lacking the (2x - 3) in the denominator; so I multiplied both by that.*2878

*The second fraction needed an (x + 2) multiplied by both the numerator and the denominator.*2886

*Once I did that, then it was a matter of subtracting, and I had the same denominator.*2892

*I had to do some multiplying, combining, and simplifying to end up with the final answer that I have here.*2898

*OK, here I have a complex fraction; and not only is it a complex fraction,*2906

*but the rational expressions that we see are being added or subtracted in the numerator and the denominator.*2911

*Again, the way to handle this is to handle the numerator and the denominator separately.*2917

*So, first the numerator: I want to subtract this.*2922

*The common denominator: well, my LCD is going to be x ^{2}y^{3}.*2934

*I look at what I have and what is lacking: well, what is lacking from this denominator, x ^{2}, is a y^{3}.*2943

*So, I am going to multiply both the numerator and the denominator by y ^{3}, minus 1/y^{3}.*2952

*What is lacking here from the denominator is the x ^{2}, so I am multiplying both the numerator and the denominator by x^{2}.*2963

*This is going to give me y ^{3} right here, over x^{2}y^{3}, minus x^{2} over x^{2}y^{3}.*2972

*Since I now have a common denominator, I can then subtract.*2988

*So, it is y ^{3} - x^{2}, over this common denominator.*2994

*This is my numerator; so I am going to go over here and write this new numerator that I have.*3000

*And this is much easier to work with, because now I just have a fraction.*3006

*I have a rational expression; I don't have two rational expressions and subtraction.*3010

*The denominator: that was the numerator--let's now work with the denominator.*3016

*I am adding, and I am being asked to add (y/x ^{3}) + (x/y^{2}).*3027

*The LCD is x ^{3}y^{2}; to convert this, I am going to have to multiply this fraction*3034

*by y ^{2}/y^{2}, because that is what is lacking.*3051

*And I am going to add that to x/y ^{2}; and what is lacking from this denominator is x^{3}.*3065

*I multiply this; this is x ^{3} over x^{3}.*3072

*This gives me y times y ^{2} (is y^{3}), over the common denominator, x^{3}y^{2},*3077

*plus x times x ^{3} (is going to give me x^{4}), over x^{3}y^{2}.*3086

*Since these now have a common denominator, I am going to add y ^{3} + x^{4}, over x^{3}y^{2}.*3102

*OK, I handled this as two different problems: a subtraction problem up here to get the numerator,*3112

*and an addition problem adding rational expressions down here, to find the denominator,*3118

*which is y ^{3} + x^{4}, over x^{3}y^{2}.*3123

*Once I am to this point, I just use my usual rules for dividing rational expressions.*3134

*So remember: we are going to rewrite this as a division problem, because this fraction bar is just telling me to divide.*3140

*This is going to give me...I am going to rewrite this down here as (y ^{3} - x^{2}), divided by x^{3}y^{2}.*3147

*This entire rational expression is being divided by this one: y ^{3} + x^{4} divided by x^{3}y^{2}.*3158

*Dividing one rational expression by another is simply multiplying the first by the inverse of the second.*3171

*So, I am going to rewrite this as x ^{3}y^{2}, divided by y^{3} + x^{4}.*3184

*Now, multiplication: the next step is always to simplify.*3194

*So, let's look for common factors: I have x ^{3} here and x^{2} down here--get rid of the x^{2};*3198

*this becomes x, because I took out that factor of x ^{2}.*3208

*I have a y ^{2} here and a y^{3} here; that cancels out, and this just becomes y, because I took out a y^{2}.*3213

*Now, let's see what I have left and multiply that.*3222

*I have an x, times this whole thing, which is y ^{3} - x^{2}, divided by...*3224

*I have a y left here, and I have y ^{3} + x^{4}.*3239

*And now, I have simplified it as far as I can.*3246

*And this took many steps; you have to be careful and make sure that you keep track of everything.*3252

*But start out by simplifying the numerator, by subtracting to get this numerator.*3257

*Simplify the denominator by adding to get this for the denominator.*3264

*Then, treat this as a regular complex fraction, where we are going to take this numerator and divide by the denominator.*3272

*And we handle that by taking the first rational expression and multiplying by the reciprocal of the second.*3283

*I found common factors; I canceled those out; and this is what I ended up with.*3289

*And it cannot be simplified any more.*3295

*That concludes this lesson on adding and subtracting rational expressions.*3298

*Thanks for visiting Educator.com!*3303

0 answers

Post by Dr Carleen Eaton on May 24, 2012

Julius, if I understand the question, at 15:00, it is easier to factor out the 2. Otherwise your LCD is (x-3)(x 1)(2x-6). No reason it shouldn't work but it is a little more complicated and to get the most simplified form you'll have to do a difficult factoring of a cubic equation at the end.

0 answers

Post by julius mogyorossy on May 21, 2012

Dr. Eaton, in the example you gave us I did not factor out the two first, I just factored it, did I do it the wrong way, it seems so, when I substitute 1 in for x it seems I get a different solution than you, but it really seems the way I did it should work 2, if you must factor out the 2 first, why so please?