R7. Addition of Rational Expressions with Unlike Denominators
Before adding fractions with unlike denominators, a warm-up, based upon the
previous lesson is important. First, try the four warm-up exercises below, then
check your results with the results that follow.
Warm-up Exercises.
1. Find the LCM of 2, 3, and 6.
2. Find the LCM of 3x and x2 .
3. Find the LCM of x2 + x − 6 and 2x2 − 3x − 2.
4. Find the LCM of x − 2, x + 2, and x2 − 4.
Solutions to the Warm-up Exercises.
1. Find the LCM of 2, 3, and 6.
2
3
6
LCM
= 2
=
3
= 2 3
= 2 3
2. Find the LCM of 3x and x2 .
3x
x2
LCM
= 3
=
= 3
x
x2
x2
3. Find the LCM of x2 + x − 6 and 2x2 − 3x − 2.
x2 + x − 6 =
2x2 − 3x − 2 =
LCM
=
(x + 3)1
(x + 3)
(x − 2)1
(x − 2)1
(x − 2)
4. Find the LCM of x − 2, x + 2, and x2 − 4.
x−2 = x−2
x+2 =
x+2
x2 − 4 = (x − 2) (x + 2)
LCM = (x − 2) (x + 2)
1
(2x + 1)1
(2x + 1)
In this section the skills of changing the denominator and finding the lowest
common multiple come together. In the four problems above, the problem was
to find the Lowest Common Multiple. When adding unlike fractions, the first
step is to find the common denominator. Finding the common denominator
is the exact same process as finding the LCM of the denominators. When the
LCM is needed to find a common denominator, the result is referred to as the
Lowest Common Denominator (LCD).
Example. Simplify
1
2
+
1
3
+
1
6
The fractions in this problem do not have like denominators. Before fractions
can be added, the fractions must change form to all have the same denominator.
Using the results in warm-up excercise (1) the LCD of 2, 3, and 6 is 6; thus,
the LCD of the fractions is also 6. Using the process learned in Lesson 3:
Changing the Denominator, write each fraction with the LCD of 6.
1
=
2
1
=
3
Then . . .
1 1 1
+ + =
2 3 6
1
2
1
3
3
3
1
= Writing with the LCD.
3
6
2
2
2
1
· = Writing with the LCD.
2
6
3
·
3 2 1
+ +
6 6 6
3+2+1
=
6
6
= Simplify the fraction.
6
= 1.
Example. Simplify
2y
3y
− .
3x x2
The fractions are not like fractions, so the LCD must be determined. In warmup exercise (2), the LCM of 3x and x2 was found to be 3x2 . Thus, the LCD for
this problem is the same as the LCM in the warm-up exercise.
2
2y
=
3x
3y
=
x2
Then . . .
2y
3y
=
−
3x x2
2y
2y x
2xy
. Writing
· =
with the LCD.
3x x
3x2
3x
3y 3
3y
9y
· = 2 . Writing 2 with the LCD.
x2 3
3x
x
2xy
9y
− 2
3x2
3x
2xy − 9y
=
.
3x2
y(2x − 9)
=
3x2
Example. Simplify
Or,
x−5
x+1
+
.
x2 + x − 6 2x2 − 3x − 2
The fractions are unlike, so a common denominator must first be found. Remember that finding the LCD is the same problem as finding the LCM of x2 + x − 6
and 2x2 − 3x − 2. Fortunately, this was accomplished in warm-up exercise (3).
The LCM=LCD=(x + 3)(x − 2)(2x + 1). Each fraction will first be written with
the LCD, then the fractions can be added.
x−5
x−5
=
x2 + x − 6
(x + 3)(x − 2)
x−5
2x + 1
=
·
(x + 3)(x − 2) 2x + 1
2x2 + 1x − 10x − 5
=
Distribute the numerators.
(x + 3)(x − 2)(2x + 1)
2x2 − 9x − 5
Combine like terms.
=
(x + 3)(x − 2)(2x + 1)
x+1
x+1
=
2
2x − 3x − 2
(x − 2)(2x + 1)
x+1
x+3
=
·
(x − 2)(2x + 1) x + 3
x2 + 3x + 1x + 3
=
Distribute the numerators.
(x − 2)(2x + 1)(x + 3)
x2 + 4x + 3
=
Combine like terms.
(x − 2)(2x + 1)(x + 3)
3
Add the fractions using their new forms.
x−5
x+1
2x2 − 9x − 5
x2 + 4x + 3
+
=
+
x2 + x − 6 2x2 − 3x − 2
(x + 3)(x − 2)(2x + 1) (x − 2)(2x + 1)(x + 3)
2x2 − 9x − 5 + x2 + 4x + 3
=
LCD
3x2 − 5x − 2
Combine like terms.
=
LCD
(3x + 1)(x − 2)
=
Factor numerator and denominator.
(x + 3)(x − 2)(2x + 1)
(3x + 1)
(x
−
2)
=
Notice the common factor of x − 2.
(x + 3)
(x
− 2)(2x + 1)
(3x + 1)
Or,
=
(x + 3)(2x + 1)
3x + 1
=
Or,
(x + 3)(2x + 1)
3x + 1
= 2
2x + 7x + 3
Sometimes the numerator will be distributed and written as a sum; most often,
it is left in the factored form. Sometimes the denominator will be distributed
and written as a sum; most often, it is left in the factored form.
Example. Simplify
5
2
x+3
−
+
.
x − 2 x + 2 x2 − 4
Try this problem yourself first. Hint: see warm-up exercise (4).
Solution. The LCM found in warm-up exercise (4) is the LCD. Notice that
the fraction xx+3
2 −4 is in the correct form, i.e., already has the LCD.
5
x+2
5
=
·
x−2
x−2 x+2
5x + 10
= 2
x −4
2
2
x−2
=
·
x+2
x+2 x−2
2x − 4
= 2
x −4
5
2
x+3
(5x + 10) − (2x − 4) + (x + 3)
−
+
=
x − 2 x + 2 x2 − 4
x2 − 4
5x + 10 − 2x + 4 + x + 3
=
Distribute the negative.
x2 − 4
4x + 17
= 2
x −4
4
0.1
Practice Problems
1.
2
3
+
x y
2.
2y
3y
− 2
3x x
3.
2
1
+ 2
2
2
3x y
6y
4.
5
3
−
2
12x
8xy 2
5.
2x − 3 5 − y
+
16x
4y
6.
x+1
−1
x−1
x−1
x+1
8.
3
4
+
x+1 x−3
9.
1
4
−
x−5 x+2
10.
2x
3
−
x−4 x+2
11.
x
7
−
x + 1 2x
12.
x
x
−
x+2 x−2
13.
2x
5
+
x − 3 x2 − 6x + 9
14.
2
3x
−
x + 2 x2 + 2x
15.
4
2x
−
x + 1 x2 + 2x + 1
16.
4
7x
+
x − 5 2x2 − 10x
17.
2x + 1 x − 2
+
x−7
x+3
18.
x − 4 2x − 1
−
x+5
x−3
19.
x−2 x+2
+
x+3 x−4
20.
−4xy
x+y
+
x2 − y 2
x−y
21.
22.
2
y
+
y 2 − y − 20 y + 4
23.
3a + 6
3a − 4
+
4a2 + 9a + 2 4a + 1
24.
1
a−1
+ 2
a + 6 a + 8a + 12
25.
x+5
3
+
2
x −1 x+1
7. 2 −
26.
x2
x+1
x+3
− 2
+ 5x − 14 x + 8x + 7
27.
x2
y3
8y 2
4y
−
− 16y y 2 − 4y
x−5
x+1
+ 2
+ x − 6 2x − 3x − 2
28.
x−2
x−3
−
x2 + 2x − 15 x2 + 3x − 10
29.
x + 12
x − 10
+
x2 − x − 6 x2 − 2x − 8
30.
5
2
y+3
−
+
y − 2 y + 2 y2 − 4
31.
x
3
−
+1
x2 − 10x + 24 x − 6
32.
2x
1
1
−
+
x2 − y 2
x+y y−x
33.
5
2y
8y
−
+
y − 2 y2 − 4 y + 2
34.
y
21
3
−
+
2y + 2 y 2 − 5y − 6 y − 6
5
0.2
Solutions
1.
2y + 3x
xy
2.
2xy − 9y
3x2
3.
4 + x2
6x2 y 2
4.
10y 2 − 9x
24x2 y 2
5.
−2xy − 3y + 20x
16xy
6.
2
x−1
7.
x+3
x+1
8.
7x − 5
(x + 1)(x − 3)
9.
−3x + 22
(x − 5)(x + 2)
12.
−4x
(x − 2)(x + 2)
15.
2(x + 2)
(x + 1)2
2x2 − 7x − 7
2x(x + 1)
10.
2x2 + x + 12
(x − 4)(x + 2)
11.
13.
7x − 15
(x − 3)2
14. −
16.
15
2(x − 5)
17.
3x2 − 2x + 17
(x − 7)(x + 3)
18. −
19.
2x2 − x + 14
(x − 4)(x + 3)
20.
x−y
x+y
21.
22.
3y − 10
(y − 5)(y + 4)
23.
3a − 1
4a + 1
24.
2a + 1
(a + 2)(a + 6)
25.
2(2x + 1)
(x − 1)(x + 1)
26.
1
(x − 2)(x + 1)
27.
3x + 1
(x + 3)(2x + 1)
29.
2x − 9
(x − 3)(x − 4)
31.
x−6
x−4
33.
2y + 5
y+2
28. 0
30.
4y + 17
(y − 2)(y + 2)
32. 0
1
x+2
34.
6
(x − 1)(x + 17)
(x − 3)(x + 5)
4
y+4
y+6
2(y + 1)