Math Help for Section 3.2, Page 112

Equations
Involving Fractions
An equation such as 

    $\displaystyle{{x} \over a} + {b \over
c} = d$

that contains one of more fractions can be cleared of
fractions by multiplying each side by the least common multiple (LCM)
of a and c.

For example, the equation

    $\displaystyle{{3x} \over 2} – {1 \over
3} = 2$

in Example 3, part (a) can be cleared of fractions by
multiplying each side by the
LCM of the denominators 2 and 3. To find the LCM of 2 and 3, list the
multiples of each number and find the smallest multiple that they have
in common.

    Multiples of 2: 2, 4, 6,
8, …
    Multiples of 3: 3, 6,
9, 12, …

The smallest multiple that they have in common is 6.

Example
2: Check

To check a fractional solution such as ${{14}\over 9},$ it is often
helpful to rewrite
the variable term as a product. This makes the substitution
of ${{14}
\over 9}$ for x easier to calculate.

$\eqalign{\textbf{a.}\quad\quad
{{3x} \over 2} – {1 \over 3} =&
2&{\small\color{red}\quad\quad\text{Write
original equation.}}  \cr {3 \over 2} \bullet x – {1 \over
3} =&
2&{\small\color{red}\quad\quad\text{Rewrite }{{3x} \over 2}\text{ as
a  product.}} \cr {3 \over 2} \bullet \color{red}{{14} \over
9}\color{black} – {1 \over 3}\overset{?}{=}&
2&{\small\color{red}\quad\quad\text{Substitute }}\small\color{red}{{14} \over 9}{\text{ for }x.} 
\cr
{{42} \over {18}} – {1 \over 3}\overset{?}{=}&
2&{\small\color{red}\quad\quad\text{Multiply.}}  \cr {7
\over 3} – {1 \over 3}\overset{?}{=}&
2&{\small\color{red}\quad\quad\text{Reduce
fraction.}}  \cr {6 \over 3}\overset{?}{=}&
2&{\small\color{red}\quad\quad\text{Subtract.}}  \cr
2=& 2&{\small\color{red}\quad\quad\text{Solution
checks. }\checkmark}
\cr}$

$\eqalign{\textbf{b.}\quad\quad\quad\quad {x
\over 5} + {{3x} \over 4}
=& 19{\small\color{red}\quad\quad\text{Write original
equation.}}  \cr {{1 \bullet (\color{red}20\color{black})}
\over 5} + {{3(\color{red}20\color{black})} \over 4}
\overset{?}{=}&
19{\small\color{red}\quad\quad\text{Substitute 20 for }x.}\cr {{20} \over 5} + {{60}
\over 4}\overset{?}{=}&
19{\small\color{red}\quad\quad\text{Multiply.}} \cr 4 +
15\overset{?}{=}& 19{\small\color{red}\quad\quad\text{Reduce fractions.}}
\cr 19 =&
19{\small\color{red}\quad\quad\text{Solution checks. }\checkmark}
\cr} $ 

$\eqalign{\textbf{c.}\quad {2 \over 3}\left( {x +
{1 \over 4}}
\right)=& {1 \over
2}&{\small\color{red}\quad\quad\text{Write original equation.}}  \cr  {2 \over
3}\left({\color{red}{1 \over 2}\color{black} + {1 \over 4}}
\right)\overset{?}{=}& {1
\over 2}&{\small\color{red}\quad\quad\text{Substitute }{1\over 2}\text{ for }x.}  \cr  {2
\over 3}\left( {{2 \over 4} + {1 \over 4}}
\right)\overset{?}{=}& {1
\over 2}&{\small\color{red}\quad\quad\text{Rewrite fraction with common denominator.}}  \cr
 {2 \over 3}\left( {{3 \over 4}}
\right)\overset{?}{=}& {1 \over
2}&{\small\color{red}\quad\quad\text{Add.}}  \cr {1 \over 2} =& {1 \over
2}&{\small\color{red}\quad\quad\text{Solution checks. }\checkmark} \cr} $

Study
Tip:

For an equation that contains a single numerical fraction, such as $2x – {3 \over 4} = 1,$ you
can simply add ${3 \over 4}$ to each side and then solve for x. You do not need to
clear the fraction.

    $\eqalign{ 2x – {3 \over 4}
+ \color{red}{3 \over
4} &= 1 + \color{red}{3 \over
4}&{\small\color{red}\quad\quad\text{Add }{3 \over 4} \text{ to each side.}} \cr 2x &= {7 \over
4}&{\small\color{red}\quad\quad\text{Combine
terms.}}  \cr x &=
{7 \over 8}&{\small\color{red}\quad\quad\text{Multiply by 
}{1
\over 2}.} \cr} $