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[SOUND]. Let's look at solving rational equations.

Â For example, let's solve this equation for x.

Â The first thing we'll do, is multiply both sides of the equation, by this

Â denominator here, which gives us 3 times x minus 4 is equal to negative 6 divided

Â by x minus 4 times x minus 4. And the x minus 4 on the right will

Â cancel as long as x does not equal 4. Which would leave us with 3 times X minus

Â 4 is equal to negative 6. And now we distribute the 3, to both of

Â these 2 terms, which gives us 3x-12=-6.

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Now, when solving rational equations, it's very important to check your answer

Â and make sure that it doesn't make the denominator here 0.

Â And it doesn't, the only excluded value is x=4. So x=2 would work and would be

Â our answer. Alright, let's look at another example.

Â [SOUND] Let's solve this equation for y. Again, we'll begin in a similar way and

Â multiply both sides of the equation, by this denominator here, y - 2.

Â Which gives us, (y-2)(1- 1-3/y-2) = (y-2)(2y-3/y-2).

Â And now we'll distribute the y-2, to both of these terms,

Â which gives us (y-2)(1)+(y-2) times negative y minus 3 divided by y minus 2,

Â is equal to y minus 2 times 2y minus 3 divided by y minus 2.

Â And again we can cancel these common factors of (y-2) on both the left and

Â right hand side assuming of course that y does not equal 2.

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Which leaves us with (y-2)(1) which is (y-2).

Â And then, we have minus this whole quantity, y-3=2y-3.

Â And distributing our negative gives us y-2-y+3=2y-3,

Â and combining like terms on the left, we have y-y is 0, and -2+3 is 1.

Â So we have 1=2y-3. Or, adding 3 to both sides you get 2y=1+3

Â or 4, which means y=2. However remember, that we have to check

Â our answer and make sure, it's not the excluded value up here and sure enough it

Â is. If y=2, then we'd be dividing by 0.

Â So we need to cross off, y=2 as a possibility here,

Â which means that there is no solution here.

Â Alright, let's see one more example.

Â [SOUND]. Let's solve the following equations for

Â w. Now what are we going to be multiplying

Â both sides of this equation by? What we need to do, is determine what the least

Â common multiple of these denominators is, or the least common denominator,

Â which we can do by factoring these denominators first.

Â So, doing this gives us 2w-1, divided by the first denominator factors into

Â (w+3)(w+3)=1 divided by, and now here, we can factor out a w.

Â And we're left with w + 3. And then we still have this + 2/w.

Â So the least common denominantor then, or the least common multiple of these

Â denominators is w(w+3)(w+3). Which is what we're going to need to

Â multiply both sides of this equation by. That is, we have

Â w(w+3)(w+3)[(2w-1)/(w+3)(w+3)]=w(w+3)(w+3)[(1/w(w+3)+2/w]. And now we'll distribute this entire

Â product here, to both of these terms, which gives us

Â w(w+3)(w+3)(2w-1/(w+3)(w+3). is equal to,

Â w(w+3)(w+3)(1/w(w+3))+w(w+3)(w+3)(2/w). And again we can cancel these common

Â factors, both on the left and right, assuming of course that w does not equal

Â negative three. And as long as w is not zero, we can also

Â cancel these common factors here. Here.

Â Which leaves us with, don't forget this w out here, w(2w-1)=(w+3)1 or w+3, then

Â plus, then we have, 2(w+3)(w+3). And then distributing the w gives us

Â 2w^2-w=w+3+2 times, and then foiling this out gives us w^2+6w+9 or 2w^2-w=w+3, and

Â then distributing the 2, gives us +2w^2+12w+18.

Â And now the two w squareds will cancel. And now we can bring all the w terms to

Â the left hand side, which gives us negative w.

Â And then minus w is negative two w. And then minus, twelve w, gives us

Â negative 14w is equal to 3+18, which is 21. So w=-21/14, or -3/2. But remember,

Â we have to check and make sure that this does not make our denominators up there

Â 0. And it doesn't, the only excluded values,

Â would be w = 0, or -3. So this is our answer then, w=-3/2.

Â And this is how we solve rational equations.

Â Thank you and we'll see you next time. [MUSIC]

Â