Math Help for Section 3.2, Page 115

Example
6: Tip
You could solve this problem using the method described on page
114—clearing the equation of decimals.

    $\eqalign{\color{red}14\color{black}
=& 0.35t + 9.1
&
{\small\color{red}\quad\quad\text{Substitute 14 for }y\text{
in
original equation.}}
\cr \color{red}100\color{black}\left({14} \right)
=&\color{red}100\color{black}\left(
{0.35t + 9.1} \right)
&{\small\color{red}\quad\quad\text{Multiply each side by 100.}}
\cr
1400 =& 35t + 910 &
{\small\color{red}\quad\quad\text{Distributive Property}} \cr
490 = & 35t &
{\small\color{red}\quad\quad\text{Subtract 910
from each side.}} \cr
14 =& t & {\small\color{red}\quad\quad\text{Divide
each side
by 35.}} \cr} $

Example
6: Check

    $\eqalign{y =& 0.35t +
9.1&{\small\color{red}\quad\quad\text{Write original
equation.}} \cr \color{red}14\color{black}
\overset{?}{=}& 0.35\left( \color{red}{14} \right) +
9.1&{\small\color{red}\quad\quad\text{Substitute 14 for }t\text{
and 14 for }y.} \cr 14\overset{?}{=}& 4.9 +
9.1&{\small\color{red}\quad\quad\text{Multiply.}}  \cr
14 =&
14&{\small\color{red}\quad\quad\text{Solution checks. }\checkmark}
\cr}$

The fact that $t = 14$ years and $y = 14$ million
students is a coincidence. Normally you wouldn’t substitute the same
number for both variables when checking a solution like this.