Algebra 2 Adding and Subtracting Rational Expressions

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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 numerators0052

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 3² times 5.0137

So, I have 2 times 5 for 10, and then I have 3² 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 3².0197

Now, if I had 5 twice up here, then I would have written 5²; I only had it once.0201

All right, this equals...2 times 5 is 10, times 3²...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 x² plus 7x, plus 14, and I wanted to add that0241

to 1 over 2x² - 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(x² - 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 here0438

(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 3² times 5.0505

The LCM...a 2 appears up here; 5 appears, at most, once; and 3 appears twice, so it is 3².0516

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 3².0566

The thing is that I can't just multiply the denominator alone; what I am allowed to do0572

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 (3² times 5); what is missing from this denominator?0602

Well, I have a 5; and I have a 3²; 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 is0717

x² - 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/(x² - 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 x², 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 x² + 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 x², 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 x² + 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 time1279

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/(x² - 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 fractions1426

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, x² + 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 power1891

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...(3x² - 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)(3x² - 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)(3x² - 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 3x², gives me 3x³.2179

x times -2x gives me -2x²; I took care of this and this using the distributive property.2197

2 times 3x² is 6x²; 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 x³ term, so that is 3x³.2237

For x² terms, I have 6x² - 2x²; that is going to give me 4x².2243

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 2x²;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 2x²; 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 2x²; 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 2x²...and do some simplifying.2749

-3x and -14x is -17x, plus 21; here I am going to have a negative; that gives me -2x².2757

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 2x² - 2x²; these cancel, so the x² 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 x²y³.2934

I look at what I have and what is lacking: well, what is lacking from this denominator, x², is a y³.2943

So, I am going to multiply both the numerator and the denominator by y³, minus 1/y³.2952

What is lacking here from the denominator is the x², so I am multiplying both the numerator and the denominator by x².2963

This is going to give me y³ right here, over x²y³, minus x² over x²y³.2972

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

So, it is y³ - x², 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/x³) + (x/y²).3027

The LCD is x³y²; to convert this, I am going to have to multiply this fraction3034

by y²/y², because that is what is lacking.3051

And I am going to add that to x/y²; and what is lacking from this denominator is x³.3065

I multiply this; this is x³ over x³.3072

This gives me y times y² (is y³), over the common denominator, x³y²,3077

plus x times x³ (is going to give me x⁴), over x³y².3086

Since these now have a common denominator, I am going to add y³ + x⁴, over x³y².3102

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 y³ + x⁴, over x³y².3123

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 (y³ - x²), divided by x³y².3147

This entire rational expression is being divided by this one: y³ + x⁴ divided by x³y².3158

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 x³y², divided by y³ + x⁴.3184

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

So, let's look for common factors: I have x³ here and x² down here--get rid of the x²;3198

this becomes x, because I took out that factor of x².3208

I have a y² here and a y³ here; that cancels out, and this just becomes y, because I took out a y².3213

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

I have an x, times this whole thing, which is y³ - x², divided by...3224

I have a y left here, and I have y³ + x⁴.3239

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

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