Substitute Values into an Equation and Solve for a Variable

Introduction

In this lesson, you will learn how to substitute values for variables within equations and solve for the remaining one variable.

Sometimes you are given an equation, or formula, with multiple variables (letters). When most of these variables are known, you can replace them with the numbers you know they are equal to. Once you do that, you can solve for whichever variable is left.

This video illustrates the lesson material below. Watching the video is optional.

Substitute Values into an Equations and Solve for a Variable (07:40 mins) | Transcript

Solving for Distance, Rate, or Time

Consider the equation distance=(rate)(time) or \(d=rt\). This is the formula that explains how to find the distance an object moves based on the rate it is moving and the time it takes to cross the distance.

Example 1: Distance
A car is moving at a rate of 30 kilometers per hour: km means kilometers; h represents hours. Suppose this car drives at this constant speed for 2 hours. How far has the car traveled at the end of 2 hours?

To solve, substitute the numbers you are given into the equation to find the missing variable. First, you are given the rate of 30 km/h. Time is given as 2 hours. You will solve the equation for distance (\(d\)).

\begin{align*}d &= rt &\color{navy}\small\text{Use the Distance Formula}\\\\d &= (30)(2) &\color{navy}\small\text{Substitute given values}\\\\d &= 60km &\color{navy}\small\text{Multiply} \end{align*}

When you multiply the rate by time, you find that the car moves 60 km in 2 hours.

Example 2: Rate
A car drove 100 miles in 4 hours. What is the rate?

To solve, substitute the numbers you are given into the equation and find the missing variable. In this case, you are given the distance as 100 miles and the time as 4 hours. You will solve the equation for rate by isolating the variable \(r\).

\begin{align*}d &= rt &\color{navy}\small\text{Use the Distance Formula}\\\\100 &= (r)(4) &\color{navy}\small\text{Substitute given values}\\\\100 \color{green}\mathbf{(\frac{1}{4})} &= (r)(4)\color{green}\mathbf{(\frac{1}{4})} &\color{navy}\small\text{Multiplicative inverse of \(4\) is \(\frac{1}{4}\)}\\\\\frac{100}{4}&= r &\color{navy}\small\text{Multiply}\\\\r &= 25km/hr &\color{navy}\small\text{Simplify}\\\\\end{align*}

Example 3: Time
In this scenario, a car needs to get to a destination that is 250 kilometers away. The car travels at a rate of 25 kilometers per hour. How long will it take the car to get from its starting point to its destination?

So solve, substitute the values for rate and distance as given. Solve for time by isolating the variable \(t\).

\begin{align*}d &= rt &\color{navy}\small\text{Use the Distance Formula}\\\\250 &= (25)(t) &\color{navy}\small\text{Substitute given values}\\\\250 \color{green}\mathbf{(\frac{1}{25})} &= (25)(t)\color{green}\mathbf{(\frac{1}{25})} &\color{navy}\small\text{Multiplicative inverse of \(25\) is \(\frac{1}{25}\)}\\\\\frac{250}{25}&= t &\color{navy}\small\text{Multiply}\\\\t &= 10hrs &\color{navy}\small\text{Simplify }\\\\ \end{align*}

It will take the car 10 hours going at a constant rate of 25 kilometers per hour to go 250 kilometers.

Things to Remember

Identify the formula, or equation, needed.
Identify the known information and substitute given values.
Once these variables have been replaced with numbers, you can solve for the remaining variable.

Practice Problems

A train traveled for \(t=5\) hours at a constant speed of \(r=60\) miles per hour. Use the formula \(d=r \cdot t\) to find the total distance (\(d\)) the train traveled (in miles). (
Solution

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Solution: \(d = r \cdot t = 60 \cdot 5 = \mathbf{300 \text{ miles}}\)
Details:
Start by plugging the numbers you’ve been given into the formula for distance.

Plug 60 miles per hour in for the rate \(r\).

Plug 5 hours in for the time \(t\).

\(d = (r)(t)\)

\begin{align*} \text{d}=\left ( \frac{60\text{ miles}}{1\text{ hour}} \right )\left ( \frac{5 \text{ hours}}{1} \right ) \end{align*}

As in unit conversions, the units on the top of the fraction and units on the bottom of the fraction that are the same can cancel out.

\begin{align*} \text{d}=\left ( \frac{60\text{ miles}}{1\cancel{{\color{Red} \text{ hour}}}} \right )\left ( \frac{5\cancel{{\color{Red} \text{ hours}}}}{1} \right ) \end{align*}

Now multiply across. Since the denominators are both 1, and anything divided by 1 is still itself, you just need to multiply 60 times 5.

\begin{align*} \text{d}&=\left ( \frac{60\text{ miles}}{1} \right )\left ( \frac{5}{1} \right ) \\\\\ \text{d} &= {(60 \text{ miles})(5)} \\\\\ \text{d} &= {300 \text{ miles}} \end{align*}

The final solution is: 300 miles.

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A man walked a distance of \(d=15km\) (kilometers) in \(t=3\) hours at a constant rate. Use the formula \(d=r \cdot t\) to find the speed (\(r\)) of the man in km per hour. (
Solution

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Solution: \(r = d \div t = 15 \div 3 = \mathbf{5\;km\text{ per hour}}\)

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A boat traveled a distance of \(d=140\) miles at a constant speed of \(r=70\) miles per hour. Use the formula \(d=r \cdot t\) to find the number of hours (\(t\)) that the trip took. (
Solution

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Solution: \(t = d \div r = 140 \div 70 = \mathbf{2 \text{ hours}}\)
Details:
In this example, you are given the distance and the rate of speed, but you need to find the time.

First, substitute in the numbers that you know.

\(d = r\cdot t\)

\(140 {\text{ miles}} = 70{\text{ miles per hour}}\cdot (t)\)

\begin{align*} \text{d}&=\text{r}\cdot\text{t}=\left(\text{r}\right)\left(\text{t}\right)\\\\\ \color{green}140\text{ miles}\color{green}&=\left( \frac{\color{green}70 \text{ miles}}{\color{green}1 \text{ hour}} \right)\left( \text{t} \right)\end{align*}

Next, solve for the variable t.

In order to get the variable t all by itself on the right-hand side of the equation, multiply by the multiplicative inverse of 70 miles per hour which is 1 hour per 70 miles.

\begin{align*}\left( \frac{\color{green} 1 \text{ hour}}{\color{green} 70\text{ miles}} \right)140\text{ miles}\color{green}=\left( \frac{\color{green} 1 \text{ hour}}{\color{green} 70\text{ miles}} \right) \left( \frac{70 \text{ miles}}{1 \text{ hour}} \right)\left( \text{t} \right)\end{align*}

Next, simplify both sides of the equation.

On the right-hand side, the multiplicative inverses become 1 leaving you with just the variable \(t\).

\begin{align*}\left( \frac{\color{green} 1 \text{ hour}}{\color{green} 70\text{ miles}} \right)140\text{ miles}\color{green}&=\color{green}\underbrace{\color{green}\left( \frac{\color{green} 1 \text{ hour}}{\color{green} 70\text{ miles}} \color{green}\right) \left( \frac{70 \text{ miles}}{1 \text{ hour}} \right)} \color{green} \left( \text{t} \right)\\\\\ \frac{(1\text{ hour})(140\text{ miles})}{70\text{ miles}}&=\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{green}1\color{green}(\text{t})\\\\\ 2 \text{ hours}&= \text{t}\end{align*}

The final solution is: \(t = 2\text{ hours}\).

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A bus traveled for \(t=3.1\) hours at a constant speed of \(r=62\) miles per hour. Use the formula \(d=r \cdot t\) to find the total distance (\(d\)) the bus traveled (in miles). Round your answer to the nearest tenth. (
Video Solution

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Solution: \({\text{d}} = {\text{r}} \cdot {\text{t}} = 62 \cdot 3.1 =\) 192.2 miles
Details:

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A truck traveled a distance of \(d=615km\) (kilometers) over \(t=5.9\) hours at a constant rate. Use the formula \(d=r \cdot t\) to find the speed (\(r\)) of the truck in km per hour. Round your answer to the nearest tenth. (
Solution

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Solution: \(r = d \div t = 615 \div 5.9 = \mathbf{104.2\;km \text{ per hour}}\)

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A bicyclist traveled a distance of \(d=18.6\) miles at a constant speed of \(r=16\) miles per hour. Use the formula \(d=r \cdot t\) to find the number of hours (\(t\)) that the trip took. Round your answer to the nearest tenth. (
Video Solution

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Solution: \({\text{t}} = {\text{d}} \div {\text{r}} = 18.6 \div 16 =\) 1.2 hours
Details:

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