3.1 Solving linear equations: the multiplication property

Solving $ax=b$ for $x$

$$\begin{array}{rrrr}\hfill ax& \hfill =& \hfill b& \hfill \begin{array}{l}\\ a\text{is associated with}x\text{by multiplication}\text{.}\\ \text{Undo the association by dividing both sides by}a\text{.}\end{array}\\ \hfill \frac{ax}{a}& \hfill =& \hfill \frac{b}{a}& \hfill \\ \hfill \frac{\cancel{a}x}{a}& \hfill =& \hfill \frac{b}{a}& \hfill \\ \hfill 1\cdot x& \hfill =& \hfill \frac{b}{a}& \hfill \frac{a}{a}=1\text{and}1\text{is the multiplicative identity}\text{. 1}\cdot x\text{=}x\end{array}$$

Solving $\frac{x}{a}=b$ for $x$

$\begin{array}{rrrr}\hfill x& \hfill =& \frac{b}{a}\hfill & \hfill \text{This equation is equivalent to the first and is solved by}x\text{.}\\ \hfill \frac{x}{a}& \hfill =& b\hfill & \hfill \begin{array}{l}a\text{is associated with}x\text{by division}\text{. Undo the association}\\ \text{by multiplying both sides by}a\text{.}\end{array}\\ \hfill a\cdot \frac{x}{a}& \hfill =& a\cdot b\hfill & \hfill \\ \hfill \cancel{a}\cdot \frac{x}{\cancel{a}}& \hfill =& ab\hfill & \hfill \\ \hfill 1\cdot x& \hfill =& ab\hfill & \frac{a}{a}=1\text{and}1\text{is the multiplicative identity}\text{. 1}\cdot x\text{=}x\hfill \\ \hfill x& \hfill =& ab\hfill & \hfill \text{This equation is equivalent to the first and is solved for}x\text{.}\end{array}$

Method for Solving $ax=b\text{and}\frac{x}{a}=b$

To solve $ax=b$ for $x$ , divide both sides of the equation by $a$ .
To solve $\frac{x}{a}=b$ for $x$ , multiply both sides of the equation by $a$ .

Solve $5x=35$ for $x$ .

$\begin{array}{rrrr}\hfill 5x& \hfill =& \hfill 35& \hfill \begin{array}{l}\\ 5\text{is associated with x by multiplication}\text{. Undo the}\\ \text{association by dividing both sides by 5}\text{.}\end{array}\\ \hfill \frac{5x}{5}& \hfill =& \hfill \frac{35}{5}& \hfill \\ \hfill \frac{\cancel{5}x}{\cancel{5}}& \hfill =& \hfill 7& \hfill \\ \hfill 1\cdot x& \hfill =& \hfill 7& \hfill \frac{\text{5}}{\text{5}}=1\text{and 1 is multiplicative identity}\text{. 1}\cdot x=x.\\ \hfill x& \hfill =& \hfill 7& \hfill \end{array}$

$\begin{array}{lllll}Check:\hfill & 5(7)\hfill & =\hfill & 35\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill & 35\hfill & =\hfill & 35\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}$

Solve $\frac{x}{4}=5$ for $x$ .

$\begin{array}{llll}\hfill \frac{x}{4}& =\hfill & 5\hfill & 4\text{is}\text{asssociated}\text{with}x\text{by}\text{division}\text{.}\text{Undo}\text{the}\text{association}\text{by}\hfill \\ \hfill & \hfill & \hfill & \text{multiplying}both\text{sides}\text{by}4.\hfill \\ 4\cdot \frac{x}{4}\hfill & =\hfill & 4\cdot 5\hfill & \hfill \\ \cancel{4}\cdot \frac{x}{\cancel{4}}\hfill & =\hfill & 4\cdot 5\hfill & \hfill \\ 1\cdot x\hfill & =\hfill & 20\hfill & \frac{4}{4}=1\text{and}1\text{is}\text{the}\text{multiplicative}\text{identity}.1\cdot x=x.\hfill \\ \hfill x& =\hfill & 20\hfill & \hfill \end{array}$

$\begin{array}{lllll}Check:\hfill & \frac{20}{4}\hfill & =\hfill & 5\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill & 5\hfill & =\hfill & 5\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}$

Solve $\frac{2y}{9}=3$ for $y$ .

Method (1) (Use of cancelling):

$\begin{array}{llll}\hfill \frac{2y}{9}& =\hfill & 3\hfill & 9\text{is}\text{associated}\text{with}y\text{by}\text{division}\text{.}\text{Undo}\text{the}\text{association}\text{by}\hfill \\ \hfill & \hfill & \hfill & \text{multiplying}both\text{sides}\text{by}9.\hfill \\ \hfill (\cancel{9})\left(\frac{2y}{\cancel{9}}\right)& =\hfill & (9)(3)\hfill & \hfill \\ \hfill 2y& =\hfill & 27\hfill & 2\text{is}\text{associated}\text{with}y\text{by}\text{multiplication}\text{.}\text{Undo}\text{the}\hfill \\ \hfill & \hfill & \hfill & \text{association}\text{by}\text{dividing}both\text{sides}\text{by}2.\hfill \\ \hfill \frac{\cancel{2}y}{\cancel{2}}& =\hfill & \frac{27}{2}\hfill & \hfill \\ \hfill y& =\hfill & \frac{27}{2}\hfill & \hfill \end{array}$

$\begin{array}{lllll}Check:\hfill & \hfill \frac{\cancel{2}\left(\frac{27}{\cancel{2}}\right)}{9}& =\hfill & 3\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill & \hfill \frac{27}{9}& =\hfill & 3\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill & \hfill 3& =\hfill & 3\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}$

Method (2) (Use of reciprocals):

$\begin{array}{llll}\hfill \frac{2y}{9}& =\hfill & 3\hfill & \text{Since}\frac{2y}{9}=\frac{2}{9}y,\frac{2}{9}\text{is}\text{associated}\text{with}y\text{by}\text{multiplication}\text{.}\hfill \\ \hfill & \hfill & \hfill & \text{Then,}\text{Since}\frac{9}{2}\cdot \frac{2}{9}=1,\text{the}\text{multiplicative}\text{identity,}\text{we}\text{can}\hfill \\ \hfill \left(\frac{9}{2}\right)\left(\frac{2y}{9}\right)& =\hfill & \left(\frac{9}{2}\right)(3)\hfill & \text{undo}\text{the}\text{associative}\text{by}\text{multiplying}both\text{sides}\text{by}\frac{9}{2}.\hfill \\ \hfill \left(\frac{9}{2}\cdot \frac{2}{9}\right)y& =\hfill & \frac{27}{2}\hfill & \hfill \\ \hfill 1\cdot y& =\hfill & \frac{27}{2}\hfill & \hfill \\ \hfill y& =\hfill & \frac{27}{2}\hfill & \hfill \end{array}$

Solve the literal equation $\frac{4ax}{m}=3b$ for $x$ .

$\begin{array}{llll}\hfill \frac{4ax}{m}& =\hfill & 3b\hfill & m\text{is}\text{associated}\text{with}x\text{by}\text{division}\text{.}\text{Undo}\text{the}\text{association}\text{by}\hfill \\ \hfill & \hfill & \hfill & \text{multiplying}both\text{sides}\text{by}m\text{.}\hfill \\ \hfill \cancel{m}\left(\frac{4ax}{\cancel{m}}\right)& =\hfill & m\cdot 3b\hfill & \hfill \\ \hfill 4ax& =\hfill & 3bm\hfill & 4a\text{is}\text{associated}\text{with}x\text{by}\text{multiplication}\text{.}\text{Undo}\text{the}\hfill \\ \hfill & \hfill & \hfill & \text{association}\text{by}\text{multiplying}both\text{sides}\text{by}4a.\hfill \\ \hfill \frac{\cancel{4a}x}{\cancel{4a}}& =\hfill & \frac{3bm}{4a}\hfill & \hfill \\ \hfill x& =\hfill & \frac{3bm}{4a}\hfill & \hfill \end{array}$

$\begin{array}{lllll}Check:\hfill & \hfill \frac{4a\left(\frac{3bm}{4a}\right)}{m}& =\hfill & 3b\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill & \hfill \frac{\cancel{4a}\left(\frac{3bm}{\cancel{4a}}\right)}{m}& =\hfill & 3b\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill & \hfill \frac{3b\cancel{m}}{\cancel{m}}& =\hfill & 3b\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill & \hfill 3b& =\hfill & 3b\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}$