Algebra
3.6
Clearing Fractions and Decimals
Clearing the fractions

It is easier to deal with whole numbers in an
equation than with fractions.

To clear the fractions out of the equation,
follow these STEPS:
1.
2.
3.
4.
Identify all the fractions in the equation
that are not inside grouping symbols.
Find the LCD of these fractions.
Multiply both sides of the equation by the
LCD.
Solve as usual.
Clearing the fractions
3
x  6  2 x 8
5
Clearing the fractions
3
x  6  2 x 8
5
To clear the fraction, multiply
both sides of the equation by
the denominator of 5.
Clearing the fractions
3
x  6  2 x 8
5
3

5  x  6    2 x  8 5
5

To clear the fraction, multiply
both sides of the equation by
the denominator of 5.
Clearing the fractions
3
x  6  2 x 8
5
3

5  x  6    2 x  8 5
5

3 x  30  10 x  40
To clear the fraction, multiply
both sides of the equation by
the denominator of 5.
The fraction is cleared.
Solve as usual
Clearing the fractions
3
x  6  2 x 8
5
3

5  x  6    2 x  8 5
5

3 x  30  10 x  40
70  7x
To clear the fraction, multiply
both sides of the equation by
the denominator of 5.
The fraction is cleared.
Solve as usual
Clearing the fractions
3
x  6  2 x 8
5
3

5  x  6    2 x  8 5
5

3 x  30  10 x  40
70  7x
x  10
To clear the fraction, multiply
both sides of the equation by
the denominator of 5.
The fraction is cleared.
Solve as usual
Clearing the fractions
3
1
x4 x 8
5
2
Find the LCD of the fractions
in the equation. What is the
LCD of 5 and 2?
Clearing the fractions
3
1
x4 x 8
5
2
3
 1

10  x  4    x  8 10
5
 2

Find the LCD of the fractions
in the equation. What is the
LCD of 5 and 2?
Multiply both sides of the
equation by the LCD of 10.
You must distribute to
each term on both
sides!
Clearing the fractions
3
1
x4 x 8
5
2
3
 1

10  x  4    x  8 10
5
 2

6 x  40  5 x  80
Find the LCD of the fractions
in the equation. What is the
LCD of 5 and 2?
Multiply both sides of the
equation by the LCD of 10.
You must distribute to
each term on both
sides!
The fractions are cleared.
Now, solve as usual.
Clearing the fractions
3
1
x4 x 8
5
2
3
 1

10  x  4    x  8 10
5
 2

6 x  40  5 x  80
x  40
Find the LCD of the fractions
in the equation. What is the
LCD of 5 and 2?
Multiply both sides of the
equation by the LCD of 10.
You must distribute to
each term on both
sides!
The fractions are cleared.
Now, solve as usual.
Clearing the fractions
2
3
(x  4 )  2  x
3
4
Clearing the fractions
2
3
(x  4 )  2  x
3
4
The LCD of 3 and 4 is 12
Clearing the fractions
2
3
(x  4 )  2  x
3
4
3 
2
 
12  ( x  4 )    2  x  12
4 
3
 
The LCD of 3 and 4 is 12
Multiply both sides of the
equation by the LCD of 12.
Clearing the fractions
2
3
(x  4 )  2  x
3
4
3 
2
 
12  ( x  4 )    2  x  12
4 
3
 
8( x  4)  24  9 x
The LCD of 3 and 4 is 12
Multiply both sides of the
equation by the LCD of 12.
The fractions are cleared.
Now, solve as usual.
Clearing the fractions
2
3
(x  4 )  2  x
3
4
3 
2
 
12  ( x  4 )    2  x  12
4 
3
 
8( x  4)  24  9 x
8 x  32  24  9 x
The LCD of 3 and 4 is 12
Multiply both sides of the
equation by the LCD of 12.
The fractions are cleared.
Now, solve as usual.
Clearing the fractions
2
3
(x  4 )  2  x
3
4
3 
2
 
12  ( x  4 )    2  x  12
4 
3
 
8( x  4)  24  9 x
8 x  32  24  9 x
x8
The LCD of 3 and 4 is 12
Multiply both sides of the
equation by the LCD of 12.
The fractions are cleared.
Now, solve as usual.
Clearing the decimals

It is easier to deal with whole numbers in an
equation than with decimals.

To clear the decimals out of the equation,
follow these STEPS:
1.
2.
3.
4.
Identify all the decimals in the equation
that are not inside grouping symbols.
Find the term with the most digits to the
right of the decimal point.
Multiply both sides of the equation by the
power of 10 that will make that term a
whole number.
Solve as usual.
Clearing the decimals
4x –.24 = .56 – .8x
There are 2 places to
the right of the
decimal point in 2 of the
terms. So, multiply both
sides of the equation by
100.
100 [4x - .24] = [.56 – .8x] 100
400x – 24 = 56 – 80x
480x
= 80
x = 1/6
Clearing the decimals
.005x + .02 = .01x – .025
There are 3 places to
the right of the
decimal point so
multiply both sides of
the equation by
1000.
1000 [.005x + .02] = [.01x – .025] 1000
5x + 20 = 10x – 25
45 = 5x
x=9
Try these
1.
1
1
x 5  x
3
2
Try these
1.
1
1
x 5  x
3
2
Solution: x = -6
Try these
1.
1
1
x 5  x
3
2
Solution: x = -6
2.
. 75x + 5 = 2.5x – 2
Try these
1.
1
1
x 5  x
3
2
Solution: x = -6
2.
. 75x + 5 = 2.5x – 2
Solution: x = 4
Homework
Next year do some more on the hw with
binomials In the numberator