Solving Literal Equations Solutions

• When you solve an equation in terms of a variable, you are    isolating    that variable by    algebraically    moving terms to the other side of the equation.

• Solving the equation    in terms of x    means isolating x in the equation.

•    Inverse    operations are used to isolate the variable indicated. (i.e., The square of a term and the square root of a term are inverse operations.)

Example 1

Solve the equation for b.

$\frac{m–a+b}{5}=3c$

Plan

$$\begin{gathered} –\left(a\right) \\ +\left(m\right) \\ ÷\left(5\right) \end{gathered}$$

Implement

$\begin{array}{rcl}\left(5\right)\frac{m–a+b}{5}& =& 3c\left(5\right)\\ m–a+b& =& 15c\\ & & –m+a –m+a\\ b& =& a+15c–m \end{array}$

 

Example 2

Solve for the indicated variable.

$\frac{1}{3}\left(2R–r\right)=x ; R$

Plan

$$\begin{gathered} ·\left(2\right) \\ –\left(r\right) \\ ·\left(\frac{1}{3}\right) \end{gathered}$$

Implement

$\begin{array}{rcl}\left(3\right)\frac{1}{3}\left(2R–r\right)& =& x\left(3\right) ; R\\ 2R–r& =& 3x\\ & & +r +r\\ \frac{2R}{2}& =& \frac{r+3x}{2}\\ R& =& \frac{r+3x}{2} \mathrm{or} R=\frac{r}{2}+\frac{3x}{2} \end{array}$