Math Help for Section 3.1, Page 103

Example
2: Check

$\eqalign{{\textbf{a.}\quad\quad}2x + 18 =&
0&{\small\color{red}\quad\quad\text{Write original equation.}}
\cr 2(\color{red}-9\color{black}) + 18
\overset{?}{=}&
0&{\small\color{red}\quad\quad\text{Substitute – 9
for }x.} \cr – 18 +
18 \overset{?}{=}&
0&{\small\color{red}\quad\quad\text{Multiply.}} \cr 0 =&
0&{\small\color{red}\quad\quad\text{Solution
checks. }\checkmark}
\cr} $

$\eqalign{{\textbf{b.}\quad\quad\quad}5x -12 =&
0&{\small\color{red}\quad\quad\text{Write original
equation.}} \cr 5\left(\color{red}{12 \over 5}\color{black}\right)
-12 \overset{?}{=}&
0&{\small\color{red}\quad\quad\text{Substitute }{12
\over 5}\text{
for }x.} \cr 12 – 12 \overset{?}{=}&
0&{\small\color{red}\quad\quad\text{Multiply.}} \cr 0 =&
0&{\small\color{red}\quad\quad\text{Solution
checks. }\checkmark}
\cr} $

$\eqalign{{\textbf{c.}\quad\quad}{x\over3} + 3 =&
0&{\small\color{red}\quad\quad\text{Write original
equation.}} \cr {\color{red}-9\color{black}\over 3} + 3
\overset{?}{=}&
0&{\small\color{red}\quad\quad\text{Substitute }-9\text{
for }}\small\color{red}x. \cr -3 + 3 \overset{?}{=}&
0&{\small\color{red}\quad\quad\text{Divide.}} \cr 0 =&
0&{\small\color{red}\quad\quad\text{Solution
checks. }\checkmark}
\cr} $

Solving
Linear Equations in Standard Form

You know that $x=-9$ is a solution of the
equation in
part (a) of Example 2, but at this point you might be asking, “How can
I be sure that the equation does not have other solutions?” The answer
is that a linear equation in one variable always has exactly one
solution. You can show this with the following steps.

    $\eqalign{ax + b =&
0&{\small\color{red}\quad\quad\text{Original equation,
with }a \ne 0.}  \cr ax + b –
\color{red}b\color{black} =& 0
– \color{red}b\color{black}&{\small\color{red}\quad\quad\text{Subtract
}b\text{ from each side.}}  \cr  ax
=&  -b&{\small\color{red}\quad\quad\text{Combine like
terms.}}  \cr{{ax} \over \color{red}a\color{black}}
=& {{-b}
\over \color{red}a\color{black}}&{\small\color{red}\quad\quad\text{Divide
each side by }a.}  \cr x =& {{-b}
\over
a}&{\small\color{red}\quad\quad\text{Simplify.}} \cr} $

It is clear that the last equation has only one
solution, $x=-b/a.$
Because the last equation is equivalent to the original equation, you
can conclude that every linear equation in one variable written in
standard form has exactly one solution.

Study
Tip

To eliminate a fractional coefficient,
it may be easier to multiply each
side by the reciprocal
of the
fraction than to divide by the
fraction itself. Here is an example.

    $\eqalign{ – {2 \over 3}x
=& 4 \cr \left( \color{red}{ – {3 \over
2}\color{black}} \right)\left( { – {2 \over 3}} \right)x =&
\left( { \color{red}-{3 \over 2}}\color{black}
\right)4 \cr x =& – {{12} \over
2} \cr x =& – 6 \cr} $