Practice Solutions

For problems 1–2, solve for t.  

1. $R=\frac{pt}{A}$
   
   $\begin{array}{rcl}\left(A\right)R& =& \frac{pt}{A}\left(A\right)\\ \frac{AR}{p}& =& \frac{pt}{p}\\ t& =& \frac{AR}{p}\end{array}$

2. $v=v_0+at$

     $\begin{array}{rcl} v& =& v_0+at\\ & & –v_0 –v_0\\ \frac{v–v_0}{a}& =& \frac{at}{a}\\ t& =& \frac{v–v_0}{a} \mathrm{or} t=\frac{v}{a}–\frac{v_0}{a}\end{array}$

For problems 3–4, use the equation
 $k=\frac{1}{2}m{v}^2$.
Recall, the inverse of squaring a term is taking the square root.

3. Solve for m.
   
   $\begin{array}{rcl}\left(2\right)k& =& \left(\frac{1}{2}m{v}^2\right)\left(2\right) \\ \frac{2k}{v^2}& =& \frac{m{v}^2}{v^2}\\ m& =& \frac{2k}{v^2}\\ & & \end{array}$

4. Solve for v.
   
   $\begin{array}{rcl}\left(2\right)k& =& \left(\frac{1}{2}m{v}^2\right)\left(2\right)\\ \frac{2k}{m}& =& \frac{m{v}^2}{m}\\ \sqrt{\frac{2k}{m}}& =& \sqrt{v^2}\\ v& =& \pm \sqrt{\frac{2k}{m}}\\ & & \end{array}$

Note

You will learn why the +/– symbol is included in the answer in Algebra 2. If this is not included in your answer at this point, that is okay.

Solve for the indicated variable.

5. $Ax+By=C; y$
   
   $$\begin{gathered} Ax+By=C \\ –Ax –Ax \\ \frac{By}{B}=\frac{–Ax+C}{B} \\ y=\frac{–Ax+C}{B} \mathrm{or} y=–\frac{A}{B}x+\frac{C}{B} \end{gathered}$$

6. $y=m\left(x–x_1\right)+y_1 ; x$

     $$\begin{gathered} y=m\left(x–x_1\right)+y_1 \\ –y_{1 } –y_1 \\ \frac{y–y_1}{m}=\frac{m\left(x–x_1\right)}{m} \\ \frac{y–y_1}{m}=x–x_1 \\ +x_1 +x_1 \\ x=\frac{y–y_1}{m}+x_1 \end{gathered}$$

For problems 7–8, solve for h.

7. $S=2\mathrm{\pi }rh+2\mathrm{\pi }{r}^2$
   
   $$\begin{gathered} S =2\mathrm{\pi }rh+2\mathrm{\pi }{r}^2 \\ –2\mathrm{\pi }{r}^2 –2\mathrm{\pi }{r}^2 \\ \frac{S–2\mathrm{\pi }{r}^2}{2\mathrm{\pi }r}=\frac{\mathit{2}\mathit{\pi }\mathit{r}\mathit{h}}{\mathit{2}\mathit{\pi }\mathit{r}} \\ h=\frac{S–2\mathrm{\pi }{r}^2}{2\mathrm{\pi }r} \mathrm{or} h=\frac{S}{2\mathrm{\pi }r}–r \end{gathered}$$

8. $V=\frac{\mathrm{\pi }}{3}h\left(r^2+rR+R^2\right)$

     $\begin{array}{rcl}\left(\frac{3}{\mathrm{\pi }}\right)V& =& \frac{\mathrm{\pi }}{3}h\left(r^2+rR+R^2\right)\left(\frac{3}{\mathrm{\pi }}\right)\\ \frac{3V}{\mathrm{\pi }\left(r^2+rR+R^2\right)}& =& \frac{h\left(r^2+rR+R^2\right)}{\left(r^2+rR+R^2\right)}\\ h& =& \frac{3V}{\mathrm{\pi }\left(r^2+rR+R^2\right)}\end{array}$

Solve for the indicated variable.

9. $F=G\left(\frac{m_1{m}_2}{r^2}\right); r$

     $\begin{array}{rcl}F& =& \frac{G{m}_1{m}_2}{r^2}\\ \frac{F{r}^2}{F}& =& \frac{G{m}_1{m}_2}{F}\\ \sqrt{r^2}& =& \sqrt{\frac{G{m}_1{m}_2}{F}}\\ r& =& \pm \sqrt{\frac{G{m}_1{m}_2}{F}} \end{array}$

Note

Remember a subscript is not an exponent.

10. $T=2\mathrm{\pi }\sqrt{\frac{m}{k}} ; m$

     $\begin{array}{rcl}T& =& 2\mathrm{\pi }\sqrt{\frac{m}{k}} \\ {\left(\frac{T}{2\mathrm{\pi }}\right)}^2& =& {\left(\sqrt{\frac{m}{k}} \right)}^2\\ \left(k\right)\frac{T^2}{4{\mathrm{\pi }}^2}& =& \frac{m}{k}\left(k\right)\\ m& =& \frac{k{T}^2}{4{\mathrm{\pi }}^2}\end{array}$        $\begin{array}{rcl}& \mathrm{OR}& \\ & & \\ \left(k\right){\left(\frac{T}{2\mathrm{\pi }}\right)}^2& =& \frac{m}{k}\left(k\right)\\ m& =& \left(k\right){\left(\frac{T}{2\mathrm{\pi }}\right)}^2\end{array}$

Note

Remember the inverse operation for the square root is squaring a term.

11. $S=2lw+2lh+2wh; w$
    
    $\begin{array}{rcl}S & =& 2lw+2lh+2wh\\ & & –2lh –2lh\\ S–2lh& =& 2lw+2wh\\ \frac{S–2lh}{\left(2l+2h\right)}& =& \frac{w\left(2l+2h\right)}{\left(2l+2h\right)}\\ w& =& \frac{S–2lh}{2l+2h}\end{array}$

12. $y=a{\left(x–h\right)}^2+k; x$
    
    $\begin{array}{rcl}y& =& a{\left(x–h\right)}^2+k\\ \frac{y–k}{a}& =& \frac{a{\left(x–h\right)}^2}{a}\\ \sqrt{\frac{y–k}{a}}& =& \sqrt{{\left(x–h\right)}^2}\\ \pm \sqrt{\frac{y–k}{a}}& =& x–h\\ x& =& h\pm \sqrt{\frac{y–k}{a}}\end{array}$