Solve for Variable "In Terms of" Another Variable Practice AK

Use the equation for the area of a circle: $a=\pi r^2$

1. Solve for $r$ in terms of $a$.

$$\require{cancel}\eqalign{a=&\pi r^2\\\dfrac{a}{\pi}=&\dfrac{\cancel{\pi} r^2}{\cancel{\pi}}\\\sqrt{\dfrac{a}{\pi}}=&\sqrt{r^2}\\\sqrt{\dfrac{a}{\pi}}=&r}$$

Use the equation for the volume of a cylinder: $v=(\pi r^2)h$

2. Solve for $r$ in terms of $v$ and $h$.

$$\eqalign{v=&(\pi r^2)h\\\dfrac{v}{h}=&\dfrac{(\pi r^2)\cancel{h}}{\cancel{h}}\\\dfrac{\dfrac{v}{h}}{\pi}=&\dfrac{\pi r^2}{\pi}\\\dfrac{v}{h \pi}=&r^2\\\sqrt{\dfrac{v}{h \pi}}=&\sqrt{r^2}\\\sqrt{\dfrac{v}{h \pi}}=&r}$$

3. Solve for $h$ in terms of $v$ and $r$.

$$\eqalign{v=&(\pi r^2)h\\\dfrac{v}{(\pi r^2)}=&\dfrac{\cancel{(\pi r^2)}h}{\cancel{(\pi r^2)}}\\\dfrac{v}{(\pi r^2)}=&h}$$

Use the equation for Potential Energy: $PE = mgh$ (m=mass; g=acceleration due to gravity; h=height)

4. Solve for mass in terms of potential energy, gravity, and height.

$$\eqalign{PE=&mgh\\\dfrac{PE}{gh}=&\dfrac{m\cancel{gh}}{\cancel{gh}}\\\dfrac{PE}{gh}=&m}$$

5. Solve for height in terms of potential energy, gravity, and mass.

$$\eqalign{PE=&mgh\\\dfrac{PE}{mg}=&\dfrac{\cancel{mg}h}{\cancel{mg}}\\\dfrac{PE}{mg}=&h}$$

Use the equation for Kinetic Energy: $KE = \dfrac{1}{2}mv^2$ (m=mass; v=velocity)

6. Solve for mass in terms of kinetic energy and velocity.

$$\eqalign{KE=&\dfrac{1}{2}mv^2\\\dfrac{KE}{v^2}=&\dfrac{\dfrac{1}{2}m\cancel{v^2}}{\cancel{v^2}}\\\times 2(\dfrac{KE}{v^2})=&(\dfrac{1}{2}m)\times 2\\\dfrac{2KE}{v^2}=&m}$$

7. Solve for velocity squared in terms of kinetic energy and mass.

$$\eqalign{KE=&\dfrac{1}{2}mv^2\\\dfrac{KE}{m}=&\dfrac{\dfrac{1}{2}mv^2}{m}\\\times 2(\dfrac{KE}{m})=&(\dfrac{1}{2}v^2)\times 2\\\dfrac{2KE}{m}=&v^2}$$

Use the Pythagorean Theory: $c^2 = a^2 + b^2$ 

8. Solve for $a^2$ in terms of $b$ and $c$.

$$\eqalign{c^2=&a^2 + b^2\\c^2-b^2=&a^2}$$

9. Solve for $b^2$ in terms of $a$ and $c$.

$$\eqalign{c^2=&a^2 + b^2\\c^2-a^2=&b^2}$$

Use the equation for Compound Interest: $A = P(1+\dfrac{r}{n})^{nt}$ (A=future value of investment; P=principal investment; r=annual interest rate; n=number of times interest in compounded per year; t=number of years the money is invested):

10. Solve for principal in terms of $A$, $r$, and $n$.

$$\eqalign{A =& P(1+\dfrac{r}{n})^{nt}\\\dfrac{A}{(1+\dfrac{r}{n})^{nt}}=&\dfrac{P\cancel{(1+\dfrac{r}{n})^{nt}}}{\cancel{(1+\dfrac{r}{n})^{nt}}}\\\dfrac{A}{(1+\dfrac{r}{n})^{nt}}=&P}$$