### Video Transcript

In this video, weβll learn how to
solve rational equations. These are equations that involve
fractions. Weβll recap how to find common
denominators of a pair of algebraic fractions and learn how this process can help us
to rearrange and thus solve these kinds of equations. We begin by recalling that
algebraic fractions obey the same rules for arithmetic that numerical fractions
do. When multiplying algebraic
fractions, we multiply the numerator of the first fraction by the numerator of the
second. And then we repeat that process
with the denominators.

And what about dividing algebraic
fractions? Well, to divide by an algebraic
fraction or any fraction, we multiply by the reciprocal of that fraction. So here, we multiply by one over π₯
plus one over four or four over π₯ plus one. And then we multiply as normal. Weβll consider two types of example
for adding and subtracting algebraic fractions.

Consider one over three π₯ plus two
over π₯ squared. To add these fractions, we need to
find a common denominator. In this case, we see that the
highest power of π₯ is two. But we also have this coefficient
of three to consider. So the common denominator weβre
going to use will be three π₯ squared. We multiply the numerator and
denominator of our first fraction by π₯. And we get π₯ over three π₯
squared. And multiplying the numerator and
denominator of our second fraction by three gives us six over three π₯ squared. Once we have a common denominator,
we simply add the numerators. And so this fraction becomes π₯
plus six over three π₯ squared. So sometimes, we can find a common
denominator by using observation.

But what about π₯ over π₯ plus one
minus four over π₯ plus three? In this case, itβs not instantly
obvious what the common denominator is. So we find a denominator that
absolutely must be common to both. This is found by multiplying one
denominator by the other, giving us a common denominator of π₯ plus one times π₯
plus three. We multiply the numerator and
denominator of our first fraction by π₯ plus three. So the numerator becomes π₯ times
π₯ plus three, which is π₯ squared plus three π₯. And then we multiply the numerator
and denominator of our second fraction by π₯ plus one, which gives us four π₯ plus
four.

Weβre now ready to subtract the
numerators, being really careful to take into account that weβre subtracting. And so weβre going to subtract both
four π₯ and four. This gives us π₯ squared minus π₯
minus four over π₯ plus one times π₯ plus three. So letβs look at how to apply this
process to help us solve algebraic equations.

Solve the equation a half equals
two over three π₯ minus one.

When asking us to solve, this
question wants us to rearrange to make π₯ the subject. So how are we going to do that? There are two techniques, and weβll
consider both. We have a bit of an issue here. And that is that the π₯ is on the
denominator of our fraction. So we might decide to begin by
multiplying everything by three π₯ to counter this. If we do that, we need to multiply
every single term in this equation by three π₯. So instead, letβs look to deal with
the negative one to begin with. We begin by adding one to both
sides of our equation. That gives us a half plus one
equals two over three π₯. We can write one as one over
one. And then to add it to one-half, we
create a common denominator of two by multiplying the numerator and denominator of
this fraction by two.

So we get a half plus two over two
equals two over three π₯. A half plus two over two is three
over two. So our equation is three over two
equals two over three π₯. Weβre now going to deal with that
awkward three π₯ on the denominator of our fraction. And weβre going to counter this by
multiplying both sides of our equation by three π₯. On the left-hand side, we get three
over two times three π₯, which we can write as three over two times three π₯ over
one. And then we simply multiply the
numerators and the denominators to get nine π₯ over two. Then on the right-hand side, weβre
simply left with two. Note here that a common mistake
would be to think that weβre left with six π₯. But remember, multiplying by three
π₯ is the inverse to dividing by three π₯. So what happens is these three π₯βs
cancel.

Next, weβll multiply by two to get
rid of this fraction on the left-hand side. Nine π₯ divided by two times two is
nine π₯. And two times two is four. Remember, weβre trying to make π₯
the subject. So our last step is to divide
through by nine. And that gives us π₯ is equal to
four-ninths. Remember, at this stage, we could
check our solution for π₯ by substituting it back into the original equation. Instead, weβre going to consider an
alternative technique to help us solve this problem. It involves writing the number
negative one as negative one over one. And then weβre going to create a
common denominator on that right-hand side.

The common denominator of three π₯
and one is three π₯. So weβre going to multiply the
numerator and denominator of our second fraction by three π₯. When we do, we get two over three
π₯ minus three π₯ over three π₯. And since the denominators of these
fractions are now equivalent, we can simply subtract the numerators. And our equation becomes a half
equals two minus three π₯ over three π₯. Like we did in our previous method,
weβre going to multiply through by three π₯ to get rid of this denominator. And on the left-hand side, we get
three π₯ times a half which is three π₯ over two. Then on the right-hand side, this
simply eliminates our denominator. So we end up with two minus three
π₯.

Next, weβll multiply by two to get
rid of the denominator on the left-hand side. We make sure we multiply everything
on the right-hand side by two. And our equation is now three π₯
equals four minus six π₯. Then, whenever weβre solving an
equation with π₯βs on both sides, we look to deal with these smallest value of
π₯. Here, thatβs negative six π₯. So the inverse operation is to add
six π₯ to both sides. That gives us nine π₯ equals
four. And then our last step is to divide
through by nine. Once again, we find π₯ is equal to
four-ninths.

In our next example, weβll consider
how we can solve an algebraic equation when the fraction is equal to zero.

Find π₯ given π₯ minus 20 over π₯
plus 10 is equal to zero.

Letβs consider whatβs really going
on in our equation. We have some fraction, and itβs
equal to zero. Now, this tells us nothing about
the denominator of our fraction. In fact, we donβt really care about
the denominator. For our fraction to be equal to
zero, the numerator itself must be equal to zero, since zero divided by any number
is zero. So to solve this equation, we
simply need to solve π₯ minus 20 equals zero. And to solve this equation, we just
add 20 to both sides, giving us π₯ is equal to 20. We might even wish to check our
solution by substituting it back into the original expression. When π₯ is equal to 20, π₯ minus 20
over π₯ plus 10 is 20 minus 20 over 20 plus 10. Thatβs zero divided by 30, which
is, of course, zero. So we know our solution π₯ equals
20 is correct.

Weβll now consider how to solve a
rational equation that contains two algebraic fractions.

Given that seven π₯ over π₯ minus
three is equal to 16π₯ over π₯ plus three minus nine, find the value of π₯.

To solve this equation, weβre going
to begin by looking simply at the right-hand side. We have the sum of a rational
expression, thatβs the 16π₯ over π₯ plus three, and an integer. Thatβs negative nine. So weβre going to begin by
subtracting nine from our fraction by creating a common denominator. Now, currently, the denominator of
nine is one. Remember, weβve got nine ones. And weβre going to multiply the
numerator and denominator of this fraction by π₯ plus three. That will create a denominator of
just π₯ plus three. That gives us 16π₯ over π₯ plus
three minus nine π₯ plus three over π₯ plus three.

Once their denominators are equal,
we can simply subtract their numerators. So our numerator becomes 16π₯ minus
nine lots of π₯ plus three. Weβll now distribute these
parentheses, remembering that weβre multiplying everything inside by negative
nine. When we do, we get 16π₯ minus nine
π₯ minus 27 over π₯ plus three, which simplifies to seven π₯ minus 27 over π₯ plus
three. Weβll now bring down the left-hand
side of our equation. And we can see that seven π₯ over
π₯ minus three is equal to seven π₯ minus 27 over π₯ plus three. And we have a couple of options
here. We could subtract one of our
fractions so that we end up with some expression equal to zero. We then create a common denominator
of π₯ plus three times π₯ minus three.

In this case, though, weβre not
going to do that. Weβre just going to look to get rid
of the denominators of our fractions. To get rid of the denominator on
the left-hand side, weβre going to multiply by π₯ minus three. That leaves us simply with seven π₯
on the left-hand side. And on the right, weβre really
multiplying by π₯ minus three over one. So we get seven π₯ minus 27 over π₯
plus three times π₯ minus three over one. We multiply as normal by
multiplying the numerators and then separately multiplying the denominators. And the right-hand side is now
seven π₯ minus 27 times π₯ minus three over π₯ plus three.

Next, weβll multiply by π₯ plus
three to get rid of the denominator on the right-hand side. And we now have seven π₯ times π₯
plus three equals seven π₯ minus 27 times π₯ minus three. Note though instead of multiplying
by π₯ minus three in one step and then π₯ plus three in another, we could have
initially cross multiplied by π₯ minus three and π₯ plus three and achieved the same
result. Weβre now going to distribute both
pairs of parentheses. On the right-hand side, thatβs
achieved by multiplying everything inside by seven π₯ to get seven π₯ squared plus
21π₯. And on the right-hand side, we
multiply everything in the second set of parentheses by seven π₯ and then by
negative 27. And that gives us seven π₯ squared
minus 21π₯ minus 27π₯ plus 81. This on the right-hand side
simplifies to seven π₯ squared minus 48π₯ plus 81.

Weβre aiming to solve for π₯. So letβs begin by subtracting seven
π₯ squared from both sides of our equation. This has the effect of just
eliminating any π₯ squareds. And weβre left with 21π₯ equals
negative 48π₯ plus 81. We then add 48π₯ to get 69π₯ equals
81. And finally, we divide through by
69. In doing so, we find π₯ is equal to
81 over 69 which simplifies by dividing through by three to 27 over 23. And so the value of π₯ that
satisfies our equation is 27 over 23.

Weβll now look at how we can solve
an algebraic equation that involves three fractions.

Solve three over π squared minus
four plus one over π plus two equals two over π minus two.

We have three algebraic
fractions. And there are a number of ways we
can solve this. The key to any of these methods,
though, is spotting that π squared minus four could be factored. We can write it using the
difference of two squares. We write it as π minus two times
π plus two. And then we notice that this is the
product of our other two denominators. So we could create a common
denominator of π minus two times π plus two and gather all our terms on the
left-hand side. Alternatively, we could subtract
one over π plus two from both sides of our equation. Letβs see what that would look
like.

Our equation becomes three over π
squared minus four equals two over π minus two minus one over π plus two. Next, weβll multiply the numerator
and denominator of our first fraction by π plus two and of our second by π minus
two, creating a common denominator of π minus two times π plus two. The numerators on the right-hand
side become, respectively, two times π plus two and one times π minus two. And in fact, we saw we could write
the denominator on the left-hand side as π minus two times π plus two.

Now, the denominators are equal on
the right-hand side. Weβre going to subtract one times
π minus two from two times π plus two. And the expression on the
right-hand side becomes two times π plus two minus one times π minus two over π
minus two times π plus two. Notice now that the denominators of
these two fractions are equal. And weβre told the fractions
themselves are equal. So this means our numerators must
be equal also. So three must be equal to two times
π plus two minus one times π minus two.

Letβs distribute these parentheses,
remembering that, on our second set of parentheses, weβre multiplying everything by
negative one and our equation becomes three equals two π plus four minus π plus
two. We simplify the right-hand side to
get π plus six. And we now see we have quite a
simple equation that we can solve for π. We subtract six from both sides,
giving us π equals negative three. So the solution to our equation is
π equals negative three.

Remember, we could check the
solution by substituting it back into the original equation and making sure that
both sides are then equal.

In our final example, weβll look at
how to solve an equation where we have rational expressions being divided.

Solve π₯ plus one over π₯ minus one
divided by two π₯ plus 10 over π₯ plus one equals one-half.

Thereβs quite a lot going on
here. But notice the left-hand side is
the division of two algebraic fractions. So letβs deal with that first. We recall that to divide by a
fraction, we multiply by the reciprocal of that fraction. The reciprocal of two π₯ plus 10
and π₯ plus one is π₯ plus one over two π₯ plus 10. So weβre going to work out π₯ plus
one over π₯ minus one times π₯ plus one over two π₯ plus 10. Note that an alternative method we
have for dividing algebraic fractions is to create a common denominator and then
simply divide the numerators. In this case, that might be
slightly more long winded.

Once we have it in this form, once
weβre essentially multiplying two fractions, we simply multiply their numerators and
then individually multiply their denominators. So the fraction is now π₯ plus one
times π₯ plus one over π₯ minus one times two π₯ plus 10. Letβs distribute these
parentheses. In doing so, we get π₯ squared plus
two π₯ plus one over two π₯ squared ~~minus~~ [plus] eight π₯ minus 10. Letβs bring down the right-hand
side of our equation, and so that our entire rational expression is equal to
one-half.

We now want to eliminate the
denominator of our fractions. Weβre going to begin by multiplying
both sides by two π₯ squared plus eight π₯ minus 10. Now, we could if we wanted at the
same time multiply both sides by two. But we donβt need to do that
here. And thatβs because the right-hand
side of our equation becomes a half times two π₯ squared plus eight π₯ minus 10. Notice that every term in our
quadratic expression here is a multiple of two. So letβs multiply each term
individually by one-half. And so the right-hand side becomes
π₯ squared plus four π₯ minus five. Weβll subtract π₯ squared from both
sides of our equation.

And since weβre solving, we want to
make π₯ the subject. So letβs deal with the smallest
number of π₯. Letβs subtract two π₯ from both
sides. That gives us one is equal to two
π₯ minus five. Then we add five to both sides to
get six is equal to two π₯. And finally, we divide through by
two, giving us a solution π₯ is equal to three.

In this video, we recapped the fact
that algebraic fractions obey the same rules for arithmetic that numerical fractions
do. We saw that we can use these rules
to help us solve equations involving algebraic fractions by rearranging to make π₯
the subject. Finally, we saw that if weβre
working with an algebraic fraction thatβs equal to zero, we can simply set its
numerator equal to zero and solve for π₯.