Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Rational equations. We're going to solve for the unknown variable in these equations and there are many ways to do this, but the easiest is to find the common denominator and multiply it through by every single term. So I actually would encourage to start with looking at your denominators and figuring what the variable can never equal. So X can never ever ever equal 0 here, because if it was, we get a zero in that denominator and we can't divide by zero. So then the common denominator here is going to be X8 and five or 40X. So if I multiply each and every term by 40X, what'll happen is in one step all of my denominators will go away. So the X and the X cancel here because remember we could think of 40X as over one. So the XS cancel and I get 48 goes into 45 * 5 goes into 48 times. So all of a sudden I got rid of every fraction in one step. I get 40 equaling five X -, 24 X. Well, five X -, 24 X is -19 X. So solving for X we can see X is -40 nineteenth. You could reduce that if you wanted to -2 and two nineteenths. So our answer is X is -2 and two nineteenths didn't contradict the X not equaling zero. Our next example, XY can never ever equal 0, and our common denominator is going to be two Y, so I'm going to have two y * 5 / 2 Y plus two y * 8 / y equaling 2Y times one. We have to multiply absolutely everything, so the two YS cancel leaving us 5 plus the YS cancel leaving us 16 equaling 2Y. So 21 equal to Y, or Y is 21 halves. If we wanted to, we can think of that as 10 1/2. It doesn't contradict the original. This next one, our common denominator. Actually T is not going to equal 2 or -4 this time, because if it was 2, it'd give me a 0 here and -4 it'd give me a 0 here. The common denominator is going to be t - 2 * t + 4. So when I multiply through, I get t + 5 left. So if I have t - 2 T +4, I multiply that by t + 5 / t - 2 and then I come over here and I do t - 2 * t + 4 and I also multiply it by the t - 2 / t + 4. So here the T minus twos are going to cancel, leaving me t + 4 * t + 5. If I go ahead and distribute that out, I can get t ^2 + 9 T plus 20. Looking at the right hand side, the T plus fours are going to cancel. So if I take t - 2 * t - 2 and distribute it t ^2 - 4 T +4 solving, I can see the T squareds on each side would cancel taking all the TS to one side 13 T and all the constants to the other -16 T is -16 thirteenths or -1 and 3 thirteenths. This next one we want to factor everything first. So we're going to pull out A2 from this denominator two X - 3. So we can see that X can never equal 3 halves. That's what would make our denominators go to zero. And our common denominator here is going to be 4 * 2 X -3. The way we got that is we look at our common denominator. So or we look at our denominators and figure out what the common is. So when I have two and I have 4, the two and the four are going to have four and then the 2X minus threes. So if I multiply everything through by 4 * 2 X -3 and I reduce, we're going to simplify this out. I'm trying to squeeze it all in and getting a little sloppy. I'm sorry. OK, so here the 2X minus threes cancel and two goes into 4 twice. So the first piece is going to reduce to six. Here are the fours are going to cancel and we get 1 * 2 X -3 or just two X -, 3. And here the 2X minus threes are going to cancel leaving us 4 * 5. So we went from a equation that has all these fractions and in one step by getting our common denominator, we have something that doesn't have any SO2X. If I take the +3 to the other side by subtracting it, we get 17 X is 17 halves or two goes into 17/8 and 1/2. Now the next one's going to get a little more complex and a little longer. So let's start by getting common denominator by factoring everything X + 2 X -5 X + 2, X -2 and X -, 2 X -5 SX is never going to equal to -2 five, two or that's it actually. So our common denominator is X + 2, X -5 and X - 2. I list everything that anything is in common and then I list everything else. So if I multiply each term by the common denominator, this first one, the X + 2 X minus fives are going to cancel. So I'm going to get left two X + 1 * X - 2. On the next one, the X + 2 X -2 is going to cancel, so I'm going to get X - 1 * X - 5. And finally on the last one, three X -, 1's going to get multiplied by X + 2. Lots and lots of foiling to do here. So we get two X ^2 -4 positive 1 so -3 X -2 X squared minus six X + 5 three X ^2 + 6 X -1 X. So plus five X -, 2, two X squared and X ^2 and three X squared are all going to cancel because we get 3X squared on each side. So negative nine X + 3 equal 5 X -2. Taking all the X's to one side, taking all the whole numbers to the other, and dividing, we get X equal 5 fourteenths. So for my next example, this is going to factor into y + 1 Y -1 and this last term is almost y -, 1. If I just multiplied the top and the bottom by a -1, I'd get this to turn into y - 1. So I'm going to end up with 3 - 2 Y over y + 1 - 10 / y + 1 Y times y - 1 equaling -2 Y -3 / y - 1. So the common denominator is going to be y + 1 Y -1 and Y can never ever ever equal -1 or one. Oh. So if I multiply everything through by y + 1 Y -1, we get 3 - 2 Y times y - 1 - 10 because the Y + 1 Y -1 cancel and the -2 Y -3 * y + 1. Lots of foiling again. So we get -2 Y squared, Let's see +3 Y and +2 Y. So five y -, 3 - 10 equaling -2 Y squared -5 Y -3. So the -2 Y squareds on each side are going to cancel. We get five y -, 13 equal -5 Y -3 10 Y equal 10 Y equal 1, but Y couldn't equal one way way back here. So this is actually a no solution problem. All that work to find out that we don't have anything that would make this a true statement. On the next page. We have several for you to try. Thank you and have a wonderful day.