Monday, Jan 13th Adding and Subtracting Rational Expressions

Post date: Jan 13, 2014 9:30:41 PM

Notes

Let's refresh by looking at an example with regular fractions:

Simplify the following:

2/5 + 3/25 + 1/10

To find the common denominator, I first need to find the least common multiple (LCM) of the three denominators.

LCM: 5×5×2 = 50

To convert each fraction to the common denominator, you multiply each denominator by what it needs in order to turn it into "50". For instance, in the 2/5, the denominator needs to be multiplied by 10, since 10×5 = 50. To keep things fair, you multiply the top by 10 as well. This is because 10/10 = 1, and multiplying things by 1 doesn't actually change them. So you get:

2/5 = (2/5)×(10/10) = 20/50

Converting the other fractions, you get:

2/5 + 3/25 + 1/10 = 20/50 + 6/50 + 5/50 = 31/50

The process works similarly for rationals.

2/x + 3/(x^2) + 1/(2x)
LCM: x × x × 2 = 2x^2

My common denominator will be 2x2. To convert the "2/x" to the common denominator, I will need to multiply by 2x/2x, since the denominator already has one copy of x but needs a 2 and another x:

2/x = (2/x)×(2x/2x) = (4x)/(2x^2)

Similarly, for the 3/x2, I will multiply by 2/2; and for the 1/2x, I will multiply by x/x. This gives me:

2/x + 3/(x^2) + 1/(2x) = (4x)/(2x^2) + 6/(2x^2) + (1x)/(2x^2) = (5x + 6)/(2x^2)

Then the answer is:

(5x + 6)/(2x^2)

Examples:

Let's practice!

Simplify the following:

4x/(2x - 1) - 5/(x - 6)

The two denominators have no common factors, so the common denominator will be

(2x – 1)(x – 6).

4x/(2x - 1) - 5/(x - 6) = (4x^2 - 34x + 5)/[(2x - 1)(x - 6)]

The numerator doesn't factor, so there is no chance of anything cancelling off. It is customary to leave the denominator factored like this, so, unless your instructor says otherwise, don't bother multiplying the denominator out. The answer is:

(4x^2 - 34x + 5)/[(2x - 1)(x - 6)]

Simplify the following:

(3x + 5)/(x + 5) - (x + 1)/(2 - x) - (4x^2 - 3x - 1)/(x^2 + 3x - 10)

First I'll factor the quadratic in the third denominator:

x2 + 3x – 10 = (x + 5)(x – 2)

Note that these factors almost match the other denominators, but the second fraction's denominator is "backwards". How can I fix that? I can fix it by remembering the following:

5 – 3 = 2

3 – 5 = –2

The point of these two subtractions is that, when I reversed the subtraction, I got the same answer except for the sign. So I can reverse the subtraction in the second fraction's denominator, as long as I remember to also reverse the sign. This is what that looks like:

(3x + 5)/(x + 5) - (x + 1)/(2 - x) - (4x^2 - 3x - 1)/(x^2 + 3x - 10) = 4(2x - 1)/[(x + 5)(x - 2)]

I factored the numerator, but nothing cancels out. As you can see, I had to factor a denominator, multiply two of the fractions to get a common denominator, multiply those two fractions' numerators, add, simplify, and then factor again. You should expect to see some problems that are at least this involved. They're not as much "complicated" as they are "long and annoying". Work them out step-by-step as I did above, and you'll get the right answers fairly regularly. In this case, the answer is:

4(2x - 1)/[(x + 5)(x - 2)]