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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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).

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

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

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

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

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

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

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

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

which means that there is no solution here.

Alright, let's see one more example.

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

w. Now what are we going to be multiplying

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

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

which we can do by factoring these denominators first.

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

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

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

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

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

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

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

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

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

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

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

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

cancel these common factors here. Here.

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

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

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

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

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

the left hand side, which gives us negative w.

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

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

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

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

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

And this is how we solve rational equations.

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