Add and Subtract Rational Expressions

How to add and subtract rational expressions: 2 examples and their solutions.

Example 1

Example

Solution

Find the least common multiple, LCM,
of the denominators:
(LCM) = x(x + 2).

Change the denominator of 3/x,
x,
to the LCM: x(x + 2).

(x + 2) factor is missing.
So multiply (x + 2)
to both of the numerator and the denominator.

3/x
= 3(x + 2)/[x(x + 2)]

Change the denominator of +x/(x + 2),
(x + 2),
to the LCM: x(x + 2).

x factor is missing.
So multiply x
to both of the numerator and the denominator.

+x/(x + 2)
= +[x⋅x]/[x(x + 2)]

So 3/x + x/(x + 2)
= 3(x + 2)/[x(x + 2)] + [x⋅x]/[x(x + 2)].

3(x + 2) = 3x + 6

x⋅x = x2

The denominators of the fractions are the same.
So add the fractions.

Arrange the terms of the numerator
in descending order.

So [x2 + 3x + 6]/[x(x + 2)]
is the answer.

Example 2

Example

Solution

Find the LCM of the denominators.

Factor the denominator of the first fraction
x2 - 1.

x2 - 1
= x2 - 12
= (x + 1)(x - 1)

Factor the Difference of Two Squares: a2 - b2

Factor the denominator of the second fraction
x2 + x.

x2 + x
= x(x + 1)

Arrange the same factors vertically.

Then the LCM of the denominators is
x(x + 1)(x - 1).

x2 - 1 = (x + 1)(x - 1)
x2 + x = x(x + 1)

So (given) = 2x/[(x + 1)(x - 1)] - 5/[x(x + 1)].

Change the denominator of 2x/[(x + 1)(x - 1)],
(x + 1)(x - 1),
to the LCM: x(x + 1)(x - 1).

x factor is missing.
So multiply x
to both of the numerator and the denominator.

2x/[(x + 1)(x - 1)]
= 2x⋅x/[x(x + 1)(x - 1)]

Change the denominator of -5/[x(x + 1)],
x(x + 1),
to the LCM: x(x + 1)(x - 1).

(x - 1) factor is missing.
So multiply (x - 1)
to both of the numerator and the denominator.

-5/[x(x + 1)]
= -5(x - 1)/[x(x + 1)(x - 1)]

So 2x/[(x + 1)(x - 1)] - 5/[x(x + 1)]
= 2x⋅x/[x(x + 1)(x - 1)] - 5(x - 1)/[x(x + 1)(x - 1)].

2x⋅x = 2x2

5(x - 1) = 5x - 5

The denominators of the fractions are the same.
So subtract the fractions.

2x2 - [5x - 5]
= 2x2 - 5x + 5

So [2x2 - 5x + 5]/[x(x + 1)(x - 1)]
is the answer.