Solve "In Terms of" a Given Variable

Many equations contain several variables, or several unknowns. Oftentimes, we are given or can solve for the value of one or more variables and use those values to solve for the values of other variables. However, other times, you don't have a value -- or don't want one!  You just want to rearrange an equation to solve for one variable "in terms of" another variable. 

The phrase "in terms of" just means that when you solve for a variable, that "in terms of" variable is going to appear in your answer.

Let's take the rate formula as an example. 

Everyone knows that "rate times time equals distance." We write that as:

$rt=d$

As written, the equation solves for $d$ in terms of $r$ and $t$.  What if we want to solve for $r$ in terms of $d$ and $t$? We just need to use algebra rules to rearrange the equation.

$$\require{cancel}\eqalign{rt=&d\\\dfrac{r\cancel{t}}{\cancel{t}}=&\dfrac{d}{t} \qquad \text{Divide both sides by t}\\r=&\dfrac{d}{t}}$$

The equation now solves for $r$ in terms of $d$ and $t$ ($d$ and $t$ are in the answer and $r$ is alone on one side of the equation).

What if we want to solve for $t$ in terms of $d$ and $r$?

$$\eqalign{rt=&d\\\dfrac{r\cancel{r}}{\cancel{r}}=&\dfrac{d}{r} \qquad \text{Divide both sides by r}\\t=&\dfrac{d}{r}}$$

The beauty of solving an equation in terms of other values (in other words, rearranging it) is you only have to remember one formula ($rt=d$).  From that formula, you know three formulas!

Standardized tests will often ask you to solve an equation in terms of other variables.  All that means is rearrange the equation so that the variable you are solving for is alone on one side of the equal sign (the "in terms of" variables will be on the other side).

Sometimes it takes a few steps to rearrange.  Just follow the algebra rule: whatever you do to one side of an equation, you do to the other and work carefully and you'll get to the correct equation. 

Practice Problems:

  • Solve for variable "in terms of" another variable

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

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

     

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

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

     

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

     

    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.

     

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

     

     

     

    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.

     

    7. Solve for velocity squared ($v^2$) in terms of kinetic energy and mass.

     

     

     

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

     

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

     

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

     

    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 $n$.

     

     

    Answer Key:

Test Prep Practice

  • Algebra: Solve for variable "in terms of" another variable

    The formula for centripetal force, the force exerted when objects move in a circle around a center point,  is F=$\dfrac{mv^2}{r}$, where $m=$mass, $v=$velocity, and $r=$ the radius of the circle that the object is traveling.

     

    1. Which of the following expresses the square of the velocity of an object moving in a circle, relative to its centripetal force, mass, and the radius of the circle it's traveling?

    1. $v^2=\dfrac{rF}{m}$
    2. $v^2={Frm}$
    3. $v^2=\dfrac{Fm}{r}$
    4. $v^2=\dfrac{rF}{mr}$

    2. Which of the following expresses the mass of the object, relative to the object's centripetal force, velocity, and the radius of the circle it's traveling?

    1. $m=\dfrac{rF}{v}$
    2. $m=\dfrac{rF}{vr}$
    3. $m=\dfrac{v^2F}{r}$
    4. $m=\dfrac{rF}{v^2}$

    3. Which of the following expresses the area of the circle that the object is traveling, relative to the object's centripetal force, velocity, and mass?

    1. Area=$\dfrac{mF}{v^2}$
    2. Area=$(\dfrac{mF}{v^2})^2\pi$
    3. Area=$2(\dfrac{mv^2}{F})^2\pi$
    4. Area=$(\dfrac{mv^2}{F})^2\pi$

     

     

     

    An astronomy lab uses the following formula to calculate the reduced mass and diameter of Newtonian bodies. 

    $\mu=\dfrac{m_1 m_2}{m_1 + m_2}d$

     

    4. In the formula above $\mu$ represents reduced mass and $d$ represents diameter.  What would the formula for diameter be?

    1. $d=\mu\dfrac{m_1 m_2}{m_1 + m_2}$
    2. $d=2(\dfrac{m_1 m_2}{m_1 + m_2})$
    3. $d=\mu^2\dfrac{m_1 + m_2}{m_1 m_2}$
    4. $d=\mu\dfrac{m_1 + m_2}{m_1 m_2}$

     

     

     

    A physics lab uses the following formula to measure displacement of an object ($s$). $u$ represents initial velocity.  $v$ represents final velocity. $a$ represents acceleration. $t$ represents time.

    $s=ut + \dfrac{1}{2}at^2$

     

    5. Which of the forumlae below represents acceleration in terms of final velocity, initial velocity, time, and displacement?

    1. $a=ut+\dfrac{1}{2}t^2-s$
    2. $a=\dfrac{ut+\dfrac{1}{2}t^2}{s}$
    3. $a=\dfrac{1}{2}\dfrac{s-ut^2}{t}$
    4. $a=\dfrac{2(s-ut)}{t^2}$

     

    6. Which of the forumlae below represents initial velocity in terms of final velocity, acceleration, time, and displacement?

    1. $u=\dfrac{s}{t}-\dfrac{at}{2}$
    2. $u=\dfrac{t+\dfrac{1}{2}at^2}{s}$
    3. $u=\dfrac{s-at^2}{2t}$
    4. $u=\dfrac{at^3}{2s}$

    Answer Key:

Skill:

Common Core Grade Level/Subject