Math Help for Section 3.5, Page 140

Solving
Work-Rate Problems
In work-rate problems, the work rate is the reciprocal of the time
needed to do the entire job. For instance, if it takes 7 hours to
complete a job, the per-hour work rate is ${1 \over 7}$ job per
hour. Similarly, if it takes $4{1 \over 2}$ minutes to complete a job,
the per-minute rate is

    $\displaystyle{1 \over {4{\textstyle{1
\over
2}}}} = {1 \over {{\textstyle{9 \over 2}}}} = {2 \over 9}$ job
per minute.

Example
5: Tip

Machine 1 takes 3 hours to do one job, so its work rate is the
reciprocal of 3 or ${1 \over 3}$ job per hour. Machine 2 takes $2{1
\over 2}$ hours to do one job, so its work rate is the reciprocal of
$2{1 \over 2}$ or 

    $\displaystyle{1 \over {2{\textstyle{1
\over
2}}}} = {1 \over {{\textstyle{5 \over 2}}}} = {2 \over 5}$ job per
hour. 

The time is t
hours, so the portion done by machine 1
is $\left( {{1 \over 3}} \right)\left( t \right)$ and the portion done
by machine 2 is $\left( {{2 \over 5}} \right)\left( t \right).$

Example 5: Check

    $\eqalign{\left( {{1 \over 3}}
\right)\left( t
\right) + \left( {{2 \over 5}} \right)\left( t \right) &=
1&{\small\color{red}\quad\quad\text{Write original equation.}}
\cr \left( {{1 \over 3}} \right)\left( {\color{red}{{15}\over{11}}} \right) +
\left( {{2\over5}} \right)\left( {\color{red}{{15} \over {11}}} \right) 
&= 1&{\small\color{red}\quad\quad\text{Substitute
}}\small\color{red}{{15} \over {11}}{\text{ for }t.}
\cr {{15}\over{33}}+{{30} \over {55}} &=
1&{\small\color{red}\quad\quad\text{Multiply.}} \cr{{75} \over
{165}}
+ {{90} \over {165}} &=
1&{\small\color{red}\quad\quad\text{Rewrite fractions with
common denominator.}} \cr{{165} \over
{165}} &= 1 &{\small\color{red}\quad\quad\text{Add.}} \cr  1&=1 &{\small\color{red}\quad\quad\text{Solution
checks. }\checkmark} } $