Adding and Subtracting Rational Expressions 2

U11_L1_T3_we2 Adding and Subtracting Rational Expressions 2

Find the sum.

Express the answer as a
simplified rational

expression, and state
the domains.

We have these two rational
expressions, or two fractions,

if you will.

And whenever we add fractions,
we need to find a common

denominator, and the common
denominator has to be

something that's divisible by
both of these denominators.

In general, we want to find
the least common, or the

smallest, multiple of these
numbers, or the smallest

number that's divisible
by both.

When you look at it immediately,
it might pop out

at you that 6 is divisible by
3, and x to the fourth is

definitely divisible by x
squared, so 6x to the fourth

is definitely divisible
by 3x squared.

Of course, it's divisible by
itself, so this actually is

the least common multiple-- this
is the smallest number,

or the smallest expression, I
guess, that is divisible by

both 6x to the fourth and 3x
squared, so let's make that

the common denominator.

So this sum is going to be
equal to 5 over 6x fourth

plus-- and now, what we want
to do is write this with 6x

fourth as the denominator, so
let me just write it again.

So plus 7 over 3x squared.

So how do we make a 3x squared
into a 6x to the fourth?

Well we're going to have to
multiply it times 2 to make

the 3 into a 6, and then we're
going to have to multiply it

times another x squared, so
we're going to have to

multiply it by 2x squared.

Now we can't just multiply only
the denominator by 2x

squared-- that'll fundamentally
change the value

of this expression.

We can only multiply it by 1,
so let's multiply it by 2x

squared over 2x squared-- and
we're assuming here that is x

is not equal to 0.

x does not equal to 0.

And that was actually a safe
assumption to make, that 0 is

not a member of our domain right
from the get go, because

that would've made either of
these expressions undefined.

If we assume x is not equal to
0, we can multiply by 2x

squared over 2x squared, and
then that will give us the

expression 5 over 6x to the
fourth, plus-- this become 7

times 2 is 14.

Fourteen x squared over 3 times
2 is 6, x to the squared

times x squared is x to the
fourth, so now we have a

common denominator.

The common denominator is 6x to
the fourth, and we can just

add the numerators, so it's 5
plus 14x squared-- or, I like

to write the higher degree
term first-- or 14x

squared plus 5.

And we are assuming that x does
not equal 0, because this

would make the expression
undefined.

That's about as simple as we can
make it-- we can't divide.

14 is divisible by 2, and so
6, but 5 isn't, so we can't

divide everything by 2, and
then, there's x squared x to

the fourth, but 5 has no x term
on it, so we can't divide

everything by x or
a power of x.

We're done: it's 14x squared
plus 5 over 6x to the fourth,

and x cannot be equal to zero.