Solve a Formula for a Specific Variable

Use the Distance, Rate, and Time Formula

One formula you will use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant rate. Rate is an equivalent word for “speed.” Do you know what distance you travel if you drive at a steady rate of $60$ miles per hour for $2$ hours? (This might happen if you use your car’s cruise control while driving on the highway.) If you said $120$ miles, you already know how to use this formula!

We will use the Strategy for Solving Applications introduced earlier in this chapter. When the problem requires a formula, we change Step 4: in place of writing a sentence, we write the appropriate formula.

You may want to create a mini-chart to summarize the information in the problem, as in the first example below.

Example. Jamal rides his bike at a uniform rate of $12$ miles per hour for $3\tfrac{1}{2}$ hours. What distance has he traveled?

Example. Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of $520$ miles. If he can drive at a steady rate of $65$ miles per hour, how many hours will the trip take?

We identify that we are looking for time, so let $t =$ time. We know $d = 520$ miles and $r = 65$ mph.

Dividing both sides by $65$ gives $t = 8$. We check by substituting back into the formula: $520 \overset{?}{=} 65 \cdot 8$, and indeed $520 = 520$. ✓ Rey’s trip will take $8$ hours.

Solve a Formula for a Specific Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.

In the two examples above, we used the formula $d = rt$. This formula gives the value of $d$, distance, when you substitute in the values of $r$ and $t$. But in the second example, we had to find the value of $t$ — we substituted values of $d$ and $r$ and then used algebra to solve for $t$. If you had to do this often, you might wonder why there is not a formula that gives the value of $t$ directly when you substitute in the values of $d$ and $r$. We can make a formula like this by solving the formula $d = rt$ for $t$.

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of $1$. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate, and time formula.

Example. Solve the formula $d = rt$ for $t$: (a) when $d = 520$ and $r = 65$ (b) in general.

We write the solutions side-by-side to demonstrate that solving a formula in general uses the same steps as when we have numbers to substitute.

We say the formula $t = \tfrac{d}{r}$ is solved for $t$.

Example. Solve the formula $A = \tfrac{1}{2}bh$ for $h$: (a) when $A = 90$ and $b = 15$ (b) in general.

We can now find the height of a triangle, if we know the area and the base, by using the formula $h = \tfrac{2A}{b}$.

The formula $I = Prt$ is used to calculate simple interest, $I$, for a principal, $P$, invested at rate, $r$, for $t$ years.

Example. Solve the formula $I = Prt$ to find the principal, $P$: (a) when $I = \text{\textdollar}5{,}600$, $r = 4\%$, $t = 7$ years (b) in general.

The principal is $\text{\textdollar}20{,}000$.

Later in this class, and in future algebra classes, you’ll encounter equations that relate two variables, usually $x$ and $y$. You might be given an equation that is solved for $y$ and need to solve it for $x$, or vice versa. In the following example, we’re given an equation with both $x$ and $y$ on the same side and we’ll solve it for $y$.

Example. Solve the formula $3x + 2y = 18$ for $y$: (a) when $x = 4$ (b) in general.

In the examples above, we used the numbers in part (a) as a guide to solving in general in part (b). Now we will solve a formula in general without using numbers as a guide.

Example. Solve the formula $P = a + b + c$ for $a$.

We isolate $a$ on one side of the equation. Both $b$ and $c$ are added to $a$, so we subtract them from both sides of the equation.

Example. Solve the formula $6x + 5y = 13$ for $y$.

Subtract $6x$ from both sides to isolate the term with $y$:

Divide by $5$ to make the coefficient of $y$ equal to $1$:

The fraction is already simplified — we cannot divide $13 - 6x$ by $5$, since $5$ does not divide evenly into both terms of the numerator.

Formulas from geometry can be solved for a specific variable the same way. Some familiar ones are the perimeter of a rectangle, $P = 2L + 2W$, the circumference of a circle, $C = \pi d$, and the volume of a rectangular solid, $V = LWH$.

Key terms

rate — an equivalent word for speed; how fast an object moves per unit of time. solve a formula for a specific variable — to isolate that variable on one side of the equals sign, with a coefficient of $1$, while all other variables and constants are on the other side.

This section is adapted from Elementary Algebra 2e, Section 2.6: Solve a Formula for a Specific Variable by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the worked-example step tables as markdown tables; omitted the Be Prepared quiz, Self Check checklist, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.