how to add fractions with variables
23
ADDING ALGEBRAIC FRACTIONS
Different denominators -- The LCM
2nd level
THERE IS ONE RULE for adding or subtracting fractions: The denominators must be the same -- just as in arithmetic.
| a c | + | b c | = | a + b c |
Add the numerators, and place their sum
over the common denominator.
| Example 1. | 6x + 3 5 | + | 4x − 1 5 | = | 10x + 2 5 |
The denominators are the same. Add the numerators as like terms.
| Example 2. | 6x + 3 5 | − | 4x − 1 5 |
To subtract, change the signs of the subtrahend, and add.
| 6x + 3 5 | − | 4x − 1 5 | = | 6x + 3 − 4x + 1 5 | = | 2x + 4 5 |
Problem 1.
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| a) | x 3 | + | y 3 | = | x + y 3 | b) | 5 x | − | 2 x | = | 3 x |
| c) | x x − 1 | + | x + 1 x − 1 | = | 2x + 1 x − 1 | d) | 3x − 4 x − 4 | + | x − 5 x − 4 | = | 4x − 9 x − 4 |
| e) | 6x + 1 x − 3 | − | 4x + 5 x − 3 | = | 6x + 1 − 4x − 5 x − 3 | = | 2x − 4 x − 3 |
| f) | 2x − 3 x − 2 | − | x − 4 x − 2 | = | 2x − 3 − x + 4 x − 2 | = | x + 1 x − 2 |
Different denominators -- The LCM
To add fractions with different denominators, we must learn how to construct the Lowest Common Multiple of a series of terms.
The Lowest Common Multiple (LCM) of a series of terms
is the smallest product that contains every factor of each term.
For example, consider this series of three terms:
pqprps
We will now construct their LCM -- factor by factor.
To begin, it will have the factors of the first term:
LCM = pq
Moving on to the second term, the LCM must have the factors pr. But it already has the factor p -- therefore, we need add only the factor r:
LCM = pqr
Finally, moving on to the last term, the LCM must contain the factors ps. But again it has the factor p, so we need add only the factor s:
LCM = pqrs.
That product is the Lowest Common Multiple ofpq, pr,ps. It is the smallest product that contains each of them as factors.
Example 3. Construct the LCM of these three terms:x,x 2,x 3.
Solution. The LCM must have the factor x.
LCM = x
But it also must have the factors of x 2 -- which are x · x. Therefore, we must add one more factor of x :
LCM = x 2
Finally, the LCM must have the factors of x 3, which are x · x · x. Therefore,
LCM = x 3.
x 3 is the smallest product that contains x,x 2, andx 3 as factors.
We see that when the terms are powers of a variable -- x,x 2,x 3 -- then their LCM is the highest power.
Problem 2. Construct the LCM of each series of terms.
| a) | ab,bc,cd. abcd | b) | pqr,qrs,rst. pqrst | |
| c) | a,a 2,a 3,a 4. a 4 | d) | a 2 b,a b 2. a 2 b 2 | |
e)ab,cd. abcd
We will now see what this has to do with adding fractions.
| Example 4. Add: | 3 ab | + | 4 bc | + | 5 cd |
Solution. To add fractions, the denominators must be the same. Therefore, as a common denominator choose the LCM of the original denominators. Choose abcd. Then, convert each fraction to an equivalent fraction with denominator abcd.
It is necessary to write the common denominator only once:
| 3 a b | + | 4 b c | + | 5 c d | = | 3 c d + 4 a d + 5 a b a b c d |
To change 
into an equivalent fraction with denominator a b c d , simply multiply a b by the factors it is missing, namely c d . Therefore, we must also multiply 3 by c d . That accounts for the first term in the numerator.
To change 
into an equivalent fraction with denominator a b c d , multiply b c by the factors it is missing, namely a d . Therefore, we must also multiply 4 by a d . That accounts for the second term in the numerator.
To change 
into an equivalent fraction with denominator a b c d , multiply c d by the factors it is missing, namely a b . Therefore, we must also multiply 5 by a b . That accounts for the last term in the numerator.
That is how to add fractions with different denominators.
Each factor of the original denominators must be a factor
of the common denominator.
Problem 3. Add.
| a) | 5 ab | + | 6 ac | = | 5c + 6b abc |
| b) | 2 pq | + | 3 qr | + | 4 rs | = | 2rs + 3ps + 4pq pqrs |
| c) | 7 ab | + | 8 bc | + | 9 abc | = | 7c + 8a + 9 abc |
| d) | 1 a | + | 2 a 2 | + | 3 a 3 | = | a 2 + 2a + 3 a 3 |
| e) | 3 a 2 b | + | 4 a b 2 | = | 3b + 4a a 2 b 2 |
| f) | 5 ab | + | 6 cd | = | 5cd + 6ab abcd |
| g) | _2_ x(x + 2) | + | __3__ (x + 2)(x − 3) | = | 2(x − 3) + 3x x(x + 2)(x − 3) |
| = | _ 2x − 6 + 3x_ x(x + 2)(x − 3) | ||||
| = | _5x − 6_ x(x + 2)(x − 3) | ||||
At the 2nd Level we will see a similar problem, but the denominators will not be factored.
| Problem 4. Add: 1 − | 1 a | + | c + 1 ab | . But write the answer as |
1 − a fraction.
| 1 − | 1 a | + | c + 1 ab | = | 1 − ( | 1 a | − | c + 1 ab | ) |
Example 5. Denominators with no common factors.
When the denominators have no common factors, their LCM is simply their product, mn.
| a m | + | b n | = | an + bm m n |
The numerator then appears as the result of "cross-multiplying" :
an + bm
However, that technique will work only when adding two fractions, and the denominators have no common factors.
| Example 6. | 2 x − 1 | − | 1 x |
Solution. These denominators have no common factors -- x is not a factor of x − 1. It is a term. Therefore, the LCM of denominators is their product.
| 2 x − 1 | − | 1 x | = | 2x − (x − 1) (x − 1)x | = | 2x − x + 1 (x − 1)x | = | _x + 1_ (x − 1)x |
Note: The entirex − 1 is being subtracted. Therefore, we write it in parentheses -- and its signs change.
Problem 5.
| a) | x a | + | y b | = | xb + ya ab | b) | x 5 | + | 3x 2 | = | 2x + 15x 10 | = | 17x 10 |
| c) | 6 x − 1 | + | 3 x + 1 | = | 6(x + 1) + 3(x − 1) (x + 1)(x − 1) |
| = | 6x + 6 + 3x − 3 (x + 1)(x − 1) | ||||
| = | _9x + 3_ (x + 1)(x − 1) | ||||
| d) | 6 x − 1 | − | 3 x + 1 | = | 6(x + 1) − 3(x − 1) (x + 1)(x − 1) |
| = | 6x + 6 − 3x + 3 (x + 1)(x − 1) | ||||
| = | _3x + 9_ (x + 1)(x − 1) | ||||
| e) | 3 x − 3 | − | 2 x | = | 3x − 2(x − 3) (x − 3)x |
| = | 3x − 2x + 6 (x − 3)x | ||||
| = | x + 6 (x − 3)x | ||||
| f) | 3 x − 3 | − | 1 x | = | 3x − (x − 3) (x − 3)x |
| = | 3x − x + 3 (x − 3)x | ||||
| = | 2x + 3 (x − 3)x | ||||
| g) | 1 x | + | 2 y | + | 3 z | = | yz + 2xz + 3xy xyz |
| Example 7. Add:a + | b c | . |
Solution. We have to express a with denominator c.
Therefore,
| a + | b c | = | ac + b c | . |
Problem 6.
| a) | p q | +r | = | p + qr q | b) | 1 x | − 1 | = | 1 − x x |
| c)x − | 1 x | = | x 2 − 1 x | d) 1 − | 1 x 2 | = | x 2 − 1 x 2 |
| e) 1 − | 1 x + 1 | = | x + 1 − 1 x + 1 | = | x x + 1 |
| f) 3 + | 2 x + 1 | = | 3x + 3 + 2 x + 1 | = | 3x + 5 x + 1 |
| Problem 7. Write the reciprocal of | 1 2 | + | 1 3 | . |
| [Hint: Only a single fraction | a b | has a reciprocal; it is | b a | .] |
| 1 2 | + | 1 3 | = | 3 + 2 6 | = | 5 6 |
| Therefore, the reciprocal is | 6 5 | . |
2nd Level
Next Lesson: Equations with fractions
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how to add fractions with variables
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