Sec. 2.3, page 8 #4(d):
Solve
0.6(10x – 3) = 1.5(x + 2) – 0.3
There are two ways that we can proceed. We can either covert the decimals to fractions and eventually
clear the fractions, or we can keep them as decimals and eventually clear the decimals.
Either way, we need to first clear the parentheses by distributing.
As decimals:
0.6(10x – 3) = 1.5(x + 2) – 0.3
0.6 · 10x – 0.6 · 3 = 1.5 · x + 1.5 · 2 – 0.3
Distribute to clear the parentheses.
Multiply to simplify the terms.
6x – 1.8 = 1.5x + 3 – 0.3
Before we clear the decimals we must make sure that each term has the same number of decimal places.
Because in all of the terms, the most number of decimal places if one, we can make each of them have
one decimal place:
6.0x – 1.8 = 1.5x + 3.0 – 0.3
Because they are in tenths (one decimal place), we can multiply each side by 10 to clear the decimals:
10(6.0x – 1.8) = 10(1.5x + 3.0 – 0.3)
This means that, when we distribute, we’ll multiply each term by 10 and thereby clear the decimal as we
do so. We get:
60x – 18 = 15x + 30 – 03
We are now ready to continue solving the equation. First, combine any like terms:
60x – 18 = 15x + 27
And I believe you can solve it from here. We’ll eventually get to 45x = 45, so x = 1.
The fraction method is presented on the next page.
As Fractions:
0.6(10x – 3) = 1.5(x + 2) – 0.3
Write each decimal and each whole
number as a fraction.
6 10x 3
15 x
2
3
–
=
+
–




10  1
1
10 1
1
10
Distribute to clear the parentheses.
6 10x
6 3
15 x
15 2
3
10 · 1 – 10 · 1 = 10 · 1 + 10 · 1 – 10
60x
18
–
10
10
15x
30
= 10 + 10
Multiply to get each product as a single fraction.
3
– 10
Now that each term has a denominator (and they are all 10) we can multiply each side by 10 to clear the
denominator:
10 60x
18 
10 15x
30
3 
·
–
=
·
+
–



1
10 
1
10
10 
 10
 10
10 60x
10 18
1 · 10 – 1 · 10
10’s).
10 15x
10
30
= 1 · 10 + 1 · 10
Distribute to clear the parentheses.
–
10
3
1 · 10
Simplify (divide out the
60x – 18 = 15x + 30 – 3
Which is the same equation we got after clearing the decimals and simplifying.
This is a challenging equation to solve. Which method do you prefer?