Let's solve some rational equations.
Let's say I had 5 over 3x minus 4 is equal to--
actually, let me write it.
Let me scoot it over a little bit.
So let's say I have 5 over 3x minus 4 is equal to
2 over x plus 1.
So your first reaction might be, geez, I've
never seen this before.
I have x's in the denominators, how do I
actually solve this equation?
And the easiest way to proceed, and there's other
ways to do this, is to try to multiply both sides of this
equation by expressions that will get rid of it these x's
in the denominator.
So if we multiply both sides of the equation by x plus 1,
it's going to get rid of this x plus 1 in the denominator.
Of course, we have to do it to both sides.
We can't just do it to one side.
But we also want to get rid of this expression right here.
So let's multiply both sides of the equation by that, by 3x
minus 4, so you're multiplying by 3x minus 4 on
this side as well.
So I'm just multiplying both sides of the equation times
this and times this.
So whatever I multiply on this side, I also have
to do on this side.
What is it going to simplify to?
Well, this 3x minus 4 is going to cancel out with this 3x
minus 4, so the left-hand side is just going to be x plus 1
times 5, or 5 times x plus 1.
And my right-hand side, this x plus 1 is going to cancel with
that x plus 1.
It's just going to equal 2 times 3x minus 4.
And if you just look at it, if you didn't look at what we
actually did, it looks like we did something called
cross-multiplying.
When you have something over something is equal to
something else over something else, notice, the end product
when we multiplied both sides by both of these expressions
was 2 times 3x minus 4 is equal to 5 times x plus 1.
So it looks like we cross-multiplied.
5 times x plus 1 is equal to 2 times 3x minus 4.
But the reality is we didn't do anything new.
We just multiplied both sides by this expression and that
expression.
But now we just have a straight-up linear equation.
We can just simplify and solve.
So the left-hand side becomes 5x plus 5, just distribute the
5, is equal to 6x minus 8.
Now, let's say we subtract 5x from both sides.
If we subtract 5x from both sides, the left-hand side just
becomes a 5.
The right-hand side, we are left with x minus 8, and now
we could add 8 to both sides.
The left-hand side becomes 13, the right-hand side is just an
x, and we're done.
x is equal to 13 solves this equation.
You could even try it out.
5 over-- 3 times 13 is 39.
39 minus 4 is 35, and this should be equal to 2 over 13
plus 1, or 14.
And they both equal 1/7.
So it checks out.
Let's do a more involved one.
Let's do a more involved problem.
So let's say I had-- this one's pretty
involved right here.
Let's say I have negative x over x minus 2 plus-- I'm
going to give it some space here-- 3x minus 1 over x plus
4 is equal to 1 over x squared plus 2x minus 8.
So once again, if we want to get rid of all of these x
terms in the denominator, we want to multiply essentially
by the least common multiple of this expression, this
expression, and this expression.
So this one looks like it can be factored.
So maybe it already has some of these guys in it.
So let's try to figure that out.
So this right here, this is what?
This is x plus 4 times x minus 2, so it does, indeed, have
both of these in it.
So lets them multiply both sides of this equation by this
thing or by x plus 4 and x minus 2.
So if we multiplied-- so this thing, it could just be
rewritten as that, so I just did that.
Now let's multiply both sides of the equation by that.
So let's see, we're going to multiply by x plus 4
times x minus 2.
If we do it on the right-hand side, we have to do it on the
left-hand side.
Scoot over a little bit.
So this term, I'm going to have to multiply by x plus 4
times x minus 2.
Same thing for this term.
That's why I left some space.
X plus 4 times x minus 2.
I'm just multiplying every term in this equation by this
right there.
Now, on the left-hand side, what do we get?
This x minus 2 cancels with this x minus 2.
We have nothing left in the denominator.
This term right here will become negative x
times x plus 4.
That's that term right there.
Now, this term and this term right here, this x plus 4 will
cancel with that x plus 4.
So we're a left with plus 3x minus 1 times x minus 2.
And then the right-hand side, that cancels with that, that
cancels with that.
And we've essentially multiplied both sides by this
denominator or by the inverse of this whole expression.
So we're just left with this 1.
And now, we have to simplify it, or we have to multiply
things out.
So negative x times negative x is negative x squared.
Negative x times 4 is minus 4x.
And then we have 3x times x is plus 3x squared.
3x times negative 2 is minus 6x.
Negative 1 times x is minus x or negative x.
Negative 1 times negative 2 is plus 2.
All of that's going to be equal to 1.
Now let's add the same degree terms. So we have a second
degree term and a second degree term.
3x squared minus x squared.
That gives us 2x squared, and you have a negative 4x, a
negative 6x and a negative x.
So what is that going to be equal to?
Negative 4 and a negative 6 is negative 10.
Minus another 1 is a negative 11x.
And then finally, we just have that constant term out here,
plus 2 is equal to 1.
Let me make sure I got all the terms. Yeah, I got all the
terms. Now we could subtract 1 from both sides and the
equation becomes 2x squared minus 11x plus
1 is equal to 0.
So it just becomes a straight-up, traditional
quadratic equation.
And if we're just looking for the roots, we set it equal to
zero, we're just looking for the x's that satisfy this, we
can use the quadratic formula.
So the solutions are x is going to be equal to negative
b. b is negative 11.
So negative negative 11 is positive 11 plus or minus the
square root of b squared.
Negative 11 squared is 121 minus 4 times a, which is 2,
times c, which is 1, all of that over 2 times a, all of
that over 4.
So x is equal to 11, plus or minus the square root of-- 4
times 2 times 1 is 8.
So 121 minus 8.
So what is that?
That's 113; is that right?
The square root of 113, all of that over 4.
Can I factor 113 at all?
Did I get that right?
4 times 2 times 1.
Yeah, that looks right.
Let's get our calculator out to actually figure out what
these values are.
Let me see, so we want to turn the calculator on.
Go to my main screen, and we want the square root
of 121 minus 8.
10.63.
So this is equal to 11 plus or minus 10.63 over 4.
Those are two answers.
And if we want to find the particular answers, it would
be 11 plus 10.63 over 4, which would be what?
That would be 21.63/4.
We could try to calculate that in a second.
And then if you subtract it, 11 minus 10.63 is-- so 11
minus 10.63 over 4, this is equal to what?
Like 0.37, 0.37/4, if I'm doing my math correctly.
So what do we get?
So let's see, 21.63 divided by 4 is 5.51, so
this is equal to 5.41.
And then the other solution, 0.37 divided by 4 is going to
be like 0.09 something.
So then we have the other solution is 11 minus 10.63.
Yep, we got our 0.37 divided by 4 is equal to 0.0925 So
this is equal to 0.0925.
So I lost some precision.
This isn't the exact, because this wasn't 10.63.
It was 6301.
It just kept going.
It's an irrational number, but this should get close.
So let's check.
Let's verify that these actually work.
So let's try the 5.41 solution first. So if this is true, if
we take-- so let me put 5.4-- let me just do it.
So if we do-- so we're going to start off with 5.41.
So our original equation was negative x, so negative 5.41
divided by x minus 2.
So 5.41 minus 2 is 3.41.
So that's that term.
And then we have plus 3 times 5.41 minus 1 divided by-- 5.41
plus 4 is 9.41-- 9.41.
So the left-hand side gives us 0.0319.
So that's what the left-hand side equals.
Now let's see what the right-hand side equals when we
substitute x equals 5.41.
And of course, we've lost some precision here.
We've lost some of the zeroes, but we
should get pretty close.
So the right-hand side, if we take 1 divided by 5.41 squared
plus 2 times 5.41 plus-- this is a minus 8, I think.
I lost that.
That's a negative 8 right there, so minus 8.
That is that, and they are very, very, very close, at
least up to three digits, 0.31.
So it does work out.
And these other things, they don't equal past that because
we weren't precise enough.
If we added the trailing zeroes here when we took the
square root of 113, we would have gotten the right answer.
If you actually kept it in this form you would have
gotten the exact right answer.
So I'll leave it up to you to verify that this is also
another solution.
But, hopefully, you found that slightly instructive.