1.4
Solving Linear Equations - Fractions
Objective: Solve linear equations with rational coefficients by multiplying by the least common denominator to clear the fractions.
Often when solving linear equations we will need to work with an equation with
fraction coefficients. We can solve these problems as we have in the past. This is
demonstrated in our next example.
Example 1.
7 5
3
x− =
2 6
4
+
7
7
+
2
2
Focus on subtraction
Add
7
to both sides
2
5
6
7
+ 2 . Notice we have a
7 3
21
common denominator of 6. So we build up the denominator, 2 3 = 6 , and we
can now add the fractions:
Notice we will need to get a common denominator to add
21 5
3
x−
=
6
6
4
+
21 21
+
6
6
Same problem, with common denominator 6
Add
21
to both sides
6
3
26
x=
4
6
Reduce
3
13
x=
4
3
Focus on multiplication by
26 13
to
6
3
3
3
4
3
We can get rid of 4 by dividing both sides by 4 . Dividing by a fraction is the
4
same as multiplying by the reciprocal, so we will multiply both sides by 3 .
13 4
4 3
Multiply by reciprocal
x=
3 3
3 4
52
x=
Our solution!
9
While this process does help us arrive at the correct solution, the fractions can
make the process quite difficult. This is why we have an alternate method for
dealing with fractions - clearing fractions. Clearing fractions is nice as it gets rid
of the fractions for the majority of the problem. We can easily clear the fractions
by finding the LCD and multiplying each term by the LCD. This is shown in the
next example, the same problem as our first example, but this time we will solve
by clearing fractions.
1
Example 2.
3
7 5
x− =
4
2 6
(12)7 (12)5
(12)3
x−
=
2
6
4
(3)3x − (6)7 = (2)5
9x − 42 = 10
+ 42 + 42
9x = 52
9
9
52
x=
9
LCD = 12, multiply each term by 12
Reduce each 12 with denominators
Multiply out each term
Focus on subtraction by 42
Add 42 to both sides
Focus on multiplication by 9
Divide both sides by 9
Our Solution
The next example illustrates this as well. Notice the 2 isn’t a fraction in the origional equation, but to solve it we put the 2 over 1 to make it a fraction.
Example 3.
3
1
2
x−2= x+
2
6
3
(6)2
(6)2 (6)3
(6)1
x−
=
x+
3
1
2
6
(2)2x − (6)2 = (3)3x + (1)1
4x − 12 = 9x + 1
− 4x
− 4x
− 12 = 5x + 1
−1
−1
− 13 = 5x
5
5
13
=x
−
5
LCD = 6, multiply each term by 6
Reduce 6 with each denominator
Multiply out each term
Notice variable on both sides
Subtract 4x from both sides
Focus on addition of 1
Subtract 1 from both sides
Focus on multiplication of 5
Divide both sides by 5
Our Solution
We can use this same process if there are parenthesis in the problem. We will first
distribute the coefficient in front of the parenthesis, then clear the fractions. This
is seen in the following example.
Example 4.
3 5
4
x+
=3
2 9
27
5
2
x+ =3
6
9
Distribute
3
through parenthesis, reducing if possible
2
LCD = 18, multiply each term by 18
2
(18)2 (18)3
(18)5
x+
=
9
9
6
(3)5x + (2)2 = (18)3
15x + 4 = 54
−4 −4
15x = 50
. 15 15
10
x=
3
Reduce 18 with each denominator
Multiply out each term
Focus on addition of 4
Subtract 4 from both sides
Focus on multiplication by 15
Divide both sides by 15. Reduce on right side.
Our Solution
While the problem can take many different forms, the pattern to clear the fraction is the same, after distributing through any parentheses we multiply each term
by the LCD and reduce. This will give us a problem with no fractions that is
much easier to solve. The following example again illustrates this process.
Example 5.
3
1 1 3
7
x − = ( x + 6) −
4
2 3 4
2
1
Distribute , reduce if possible
3
1 1
7
3
x− = x+2−
2 4
2
4
LCD = 4, multiply each term by 4.
(4)3
(4)1 (4)1
(4)2 (4)7
x−
=
x+
−
4
2
4
1
2
Reduce 4 with each denominator
(1)3x − (2)1 = (1)1x + (4)2 − (2)7
3x − 2 = x + 8 − 14
3x − 2 = x − 6
−x
−x
2x − 2 = − 6
+2 +2
2x = − 4
2
2
x=−2
Multiply out each term
Combine like terms 8 − 14
Notice variable on both sides
Subtract x from both sides
Focus on subtraction by 2
Add 2 to both sides
Focus on multiplication by 2
Divide both sides by 2
Our Solution
World View Note: The Egyptians were among the first to study fractions and
linear equations. The most famous mathematical document from Ancient Egypt
is the Rhind Papyrus where the unknown variable was called “heap”
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)
3
1.4 Practice - Fractions
Solve each equation.
21
3
1
1) 5 (1 + p) = 20
5
6
3
4
7)
635
72
5
− 4m =
5
9
9) 2b + 5 = −
11)
6)
113
24
= − 2( −
11
4
29
3
8
11
4
+ 4r =
163
32
16
9
4 5
8) −
+ x)
11
5
3 7
3
( n + 1) = 2
2 3
3
= − 3 ( 3 + n)
7
10)
3
2
12)
41
9
9
− 4v = − 8
5
2
1
7
1 2
15)
55
6
3
2
17)
16
9
4
19) − =
11
3
5
(r
4
4
3
5
5
+ 2 b = 2 (b − 3 )
3
5
45
16
3
7
27) 2 (v + 2 ) = − 4 v −
29)
47
9
3
13
8
83
7
22)
7
6
4
10
3
53
= − 18
5
7
= 3 x + 3 (x − 4 )
4
3
3
− 3 n = − 2 n + 2(n + 2 )
149
16
−
11
7
r= − 4 r
3
5
4
− 4 ( − 3 r + 1)
7 5
1
26) − 2 ( 3 a + 3 ) =
11
25
a+ 8
4
19
7
3
=−
3
+ 2 n = 4 n − 16
3
1
12
24) −
23) − ( − 2 x − 2 ) = − 2 + x
25)
20)
3
)
2
−
9
18) 3 (m + 4 ) −
= − 3( − 3n − 3)
5
8
21) −
4
10
k
3
16) − 2 ( 3 x − 4 ) − 2 x = − 24
5 3
5
( p − 3)
2 2
=−
1
= 2 (x + 3 ) − 3 x
14) 3 ( − 4 k + 1) −
5
8
19
( − 3 a + 1) = − 4
4
13) − a −
3
4) 2 n − 3 = − 12
3) 0 = − 4 (x − 5 )
5)
3
2) − 2 = 2 k + 2
8
1
4
2
28) − 3 − 2 x = − 3 x − 3 ( −
19
6
1
30) 3 n +
29
6
4
13
x + 1)
4
2
= 2( 3 n + 3 )
5 5
+ 2 x = 3 ( 2 x + 1)
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)
4
1.4
Answers to Solving with Fractions
1)
3
4
4
2) − 3
11) 0
22) − 1
4
3
23) − 2
12)
3
3)
6
5
13) − 2
4)
1
6
14)
5) −
6)
19
6
9
24) − 4
1
2
15) −
25) 16
4
3
1
26) − 2
16) 1
25
8
5
27) − 3
17) 0
7
7) − 9
1
8) − 3
9) − 2
10)
3
2
5
18) − 3
3
28) − 2
19) 1
20) 1
21)
1
2
29)
4
3
30)
3
2
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)
5