Algebra

Continuously compounding interest formula with examples

Continuously compounding interest is similar to regular compound interest however, interest is not compounded monthly or quarterly but instead, continuously. The continuously compounding interest formula can be used to find the future value of an investment at a given rate or the amount of time it takes to reach a future value given a desired amount. This is commonly taught in college algebra courses or sometimes calculus courses. Below we will look at the formula and some examples of using it.

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Examples of finding the future value with the continuously compounding interest formula

We will start with using the formula to calculate the future value of an investment. Then we will look at how to find the time it takes to reach a given future value.

Remember that $e$ is a constant. This is located on your calculator and you should use that instead of an approximation in order to get the most accurate answer.

Example

An investment of $12,000 is invested at a rate of 3.5% compounded continuously. What is its value after 6 years?

Solution

Determine what values are given and what values you need to find. Then, use the formula.

Investment of $12,000 – this is the principle: $P=12000$
Rate 3.5% – remember to write this as a decimal (divide by 100): $r = 0.035$
Value after 6 years: $t = 6$

You are finding the value 6 years in the future, so you are finding $A$, the future value.

$\begin{align}A &= Pe^{rt}\\ &= 12000e^{0.035\times6}\\ &= \bbox[border: 1px solid black; padding: 2px]{14,804.14}\end{align}$

Answer: The value will be $14,804.14 after 6 years.

On your calculator, if you are using a TI83 or TI 84, you should have the following.

The $e$ is located just above $LN$.

Example of finding the time to reach a certain value using the continuously compounding interest formula

In some problems, you may be given a goal value such as $10,000 and asked how long it will take to reach that value given an initial investment. Here, you will use the fact that $\ln(x)$ is the inverse of $e^x$.

Example

An investment of $5,500 is made at a continuously compounding rate of 2%. How many years will it take to reach a value of $15,000?

Solution

Again, review the values you know and what you are trying to find.

Investment of $5,500 so $P = 5500$
Rate 2% so $r = 0.02$
Reach a value of $15,000 – this is the future value so $A = 15000$

We are asked the time, $t$. Now use the formula. Recall the statement above that $\ln(x)$ is the inverse of $e^x$.

$\begin{align}A &= Pe^{rt}\\ 15000 &= 5500e^{0.02\times t}\\ \frac{15000}{5500} &= e^{0.02\times t}\\ \frac{30}{11} &= e^{0.02\times t}\\ \ln\left(\frac{30}{11}\right) &= \ln\left(e^{0.02\times t}\right)\\ \ln\left(\frac{30}{11}\right) &= 0.02t\\ \frac{\ln\left(\frac{30}{11}\right)}{0.02} &= t \\ \bbox[border: 1px solid black; padding: 2px]{50.17} &= t \end{align}$

Answer: About 50 years

Important! Wait as long as possible to round. Rounding early will make your answer off by even a couple of years. Do your work in the calculator. This applies to finding the future value as well.

Continue your study of interest

Now that you have studied the continuously compounding interest formula, you can study other types of interest. Such as:

What is an integer?

The integers are a set of numbers consisting of all negative whole numbers, all positive whole numbers, and zero. These are a subset of the real numbers.

the set of integers which contain all negative and positive whole numbers along with zero: {...,-3,-2,-1,0,1,2,3,...}

Quick facts about the integers

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Many problems in mathematics involve working with integers. The following are some quick facts about this set of numbers.

Zero is an integer.
All even and positive numbers are integers (since they are whole numbers).
The product of two integers is an integer.
The sum or difference of two integers is an integer.
The symbol $\mathbb{Z}$ is often used to represent the set of integers.

Function notation and evaluating functions

In mathematics, a function is a rule that take an input value (often $x$) and assigns it an output value (often $y$). What makes a function special is that for any given input, there is only one output. Function notation, which is used in all of mathematics, is a way of writing out the rule that relates the input and output values of a function. In this lesson, we will look at how function notation works, how to evaluate a function given the function notation, and how to evaluate a function from its graph.

Table of Contents

Reading function notation
Evaluating functions using function notation
Evaluating functions using a graph

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Reading function notation

Function notation is written using the name of the function and the value you want to find the output for. For example, $f(x)$ is read “$f$ of $x$” and means “the output of the function $f$ when the input is $x$”. Another example is something like $g(2)$. This is read “$g$ of 2” and represents the output of the function $g$ for the input value of 2.

read function notation as "function name of the input". Here f(x) is read "f of x" and represents the output corresponding to the input x.

To help you understand this notation, let’s look at a couple of examples.

Example

When $x = 2$, the value of a function $y = f(x)$ is 10. Use function notation to represent this.

Solution

The function is written as $f(x)$. This means that the input of the function is $x$. Since $x = 2$, we can start by writing:

$f(2)$

The example says that when $x = 2$, the output is 10. Since the notation $f(2)$ represents the output for the input of 2, we can write this as:

$\bbox[border: 1px solid black; padding: 2px]{f(2) = 10}$

Evaluating functions using function notation

When it comes to evaluating functions, you are most often given a rule for the output. To evaluate the function means to use this rule to find the output for a given input.

When evaluating functions, use the rule given by the function notation. For example, is f(x) = x + 4, then the output is always 4 more than the input.

You can do this algebraically by substituting in the value of the input (usually $x$). This is shown in the next couple of examples.

Example

For $f(x) = 2x + 1$, find $f(3)$.

Solution

Start with your function.

$f(x) = 2x + 1$

Let $x = 3$ and then calculate the value of the function. To do this, replace each $x$ in the rule with 3.

$f(3) = 2(3) + 1$

Simplify to find the final answer.

$\begin{align}&= 6 + 1\\ &= \bbox[border: 1px solid black; padding: 2px;]{7}\end{align}$

When evaluating a function, make sure that you replace every $x$ in the rule with the input value. Pay close attention, because there may be more than one $x$ to replace.

Example

For $f(x) = 3x^2 – 5x + 1$, find $f(–2)$.

Start with your function.

$f(x) = 3x^2 – 5x + 1$

Let $x = –2$ by replacing each $x$ in the rule with –2 and then simplify to find your final answer.

$\begin{align}f(-2) &= 3(-2)^2 – 5(-2) + 1 \\ &= 3(4) + 10 + 1\\ &= 12 + 10 + 1 \\ &= \bbox[border: 1px solid black; padding: 2px;]{23}\end{align}$

It is also possible to evaluate a function using an input that is an algebraic expression. This is more complicated since you usually will need simplify your answer, but it still follows the same idea of substitution.

Example

Let $f(x) = –3x + 7$. Find and simplify $f(m + 1)$, where $m$ is a real number.

Solution

Start with your function.

$f(x) = -3x + 7$

Just as you did with numbers, let $x = m + 1$ and evaluate the function. This means replace every $x$ in the rule with “$m + 1$”.

$f(m+1) = -3(m+1) + 7$

Apply the distributive property by multiplying –3 and each term within the parentheses. The combine like terms to simplify.

$\begin{align}&= -3m-3 + 7\\ &= \bbox[border: 1px solid black; padding: 2px;]{-3m+4}\end{align}$

From here, you can get much more complicated with things like the difference quotient, which is used later in calculus. Even with these more complicated examples, you will still apply the same concept we did to each of the examples above.

Evaluating functions using a graph

The graph of a function is a way of viewing the outputs of the function for all possible inputs. Here, we let $y = f(x)$ so that each y-value represents the output and each x-value represents the input. This means any point on the graph is:

$(x, y) = (\text{input}, \text{corresponding output})$

Remember that in the xy-plane, the x-axis is the horizontal axis, and the y-axis is the vertical axis. Using this, we can easily read points off the graph.

The graph pictured passes through the point (4,1). This means that when x = 4, y = 1. Using function notation, this means f(4) = 1.

You can do this for the graph of just about any function. For example, if we wanted to know the value of $f$ when $x = -1$ for the function below, we would just find $x = -1$ on the x-axis and use the graph to find the corresponding y-value.

This function passes through the point (-1, -2), so f(-1) = -2.

There are some types of functions, where you have to be a little more careful. Piecewise defined functions and functions with asymptotes often have more going on in the graph. However, those are studied later on in algebra. For now, you should make sure you can read the graphs of functions like those shown in the examples above.

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Summary

Functions are used in all parts of mathematics, and understanding function notation is necessary for a wide variety of problems. Remember that it is simply a way of relating the input of a function to the value of its output. The notation $f(x)$ is always read “$f$ of $x$” though the function may have a different name like $g$ or $k$ (then we could write $g(x)$ for “$g$ of $x$” or $k(x)$ for “$k$ of $x$”).

Standard form of a line (with examples)

The standard form of a line is simply a special way of writing the equation of a line. You are probably already familiar with the slope-intercept form of a line, y = mx + b. The standard form is just another way to write this equation, and is defined as Ax + By = C, where A, B, and C are real numbers, and A and B are both not zero (see note below about other requirements). As you will see in the lesson below, every line can be expressed in this form.

Note that the requirements on A, B, and C vary a bit from textbook to textbook. While they technically can be any real numbers, it is most common to write them as integers and to keep A positive. It is also common to require that A, B, and C share no common factors. We will follow these rules in this lesson.

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Finding the standard form of a line

We will now look at some examples of writing the standard form of a line, when the line is already given in y = mx + b form.

Definition of the standard form of a line; the standard form is written as Ax + By = C where A,B, and C are real numbers and if possible, integers with A being positive. A and B cannot both be zero.

First, let’s start with one where there aren’t many steps needed.

Example

Write the equation of the line in standard form:
$y = -2x + 4$

Solution

This is the easiest case, because the coefficients on x and y are already integers (negative or positive whole numbers). This means that we just need to bring the x-term to the other side.

Adding 2x to both sides:

$y = -2x + 4$

$2x + y = 4$

The standard form of the line is:
$\bbox[border: 1px solid black; padding: 2px;]{2x + y = 4}$

A more complicated case is when either the slope, the constant, or maybe even both are fractions. Since the coefficients on x and y are preferably written as integers, this means that you have to “clear fractions”. This is shown in the next example.

Example

Write the equation of the line in standard form.
$y = -\dfrac{2}{3}x + \dfrac{1}{4}$

To clear fractions, you can multiply both sides of the equation by a whole number. The whole number should be a multiple of both denominators (ideally it is the least common multiple). Our denominators here are 3 and 4, so we will use 12.

Multiply both sides by 12 to clear fractions.

$\begin{align}y &= -\dfrac{2}{3}x + \dfrac{1}{4}\\ 12(y) &= 12\left(-\dfrac{2}{3}x + \dfrac{1}{4}\right)\\ 12y &= -\dfrac{24}{3}x + \dfrac{12}{4}\\ 12y &= -8x + 3\end{align}$

Now we can add 8x to both sides to get the standard form of the line.

$\bbox[border: 1px solid black; padding: 2px]{8x + 12y = 3}$

Finding the slope from the standard form of a line

You can always find the slope by solving for y and rewriting the equation in slope-intercept form y = mx + b. However, it can be shown that when the equation of a line is written in standard form, the slope is always $–\dfrac{A}{B}$. This quick calculation can allow you to find the slope of any line written in this form without needing to apply extra algebra steps.

Example

What is the slope of the line defined by the equation below?
$5x + 8y = -2$

Solution

The slope is $–\dfrac{A}{B}$, and here A = 5 and B = 8. Therefore:

$\begin{align}\text{slope} = m &= -\dfrac{A}{B}\\ &= \bbox[border: 1px solid black; padding: 2px]{-\dfrac{5}{8}}\end{align}$

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Continue your study of linear equations and graphing

Understanding the equation of a line is one part of the overall study of linear equations and their graphs. You can continue your study with the following lessons.

Using the slope formula to find the slope of a line

The “steepness” of a line in the xy-coordinate plane is called the slope. Slope can be positive (the line rises from left to right), negative (the line falls from left to right), zero (flat line), or even undefined (vertical line). In this lesson, we will look at how to calculate the slope using the slope formula and two points along a line.

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The slope formula

The slope of a line is often thought of as “rise over run”. This comes from the formula, which represents the change in y divided by the change in x. In other words, given two points, you will find the difference between the y-values and divide by the difference between the x-values.

the slope formula is y2 minus y1 over x2 minus x1 where (x1, y1) and (x2, y2) are points on the line

Examples

To understand this formula, let’s look at a couple of examples. First, we will look at the case where you are simply given two points on a line and asked to find the slope. Then, we will see how to find the slope if you are just given the graph instead.

Example (given two points)

Find the slope of a line that passes through the points $(–1, 5)$ and $(1, 9)$.

Solution

Apply the slope formula by letting $(x_1, y_1) = (–1, 5)$ and $(x_2, y_2) = (1, 9)$.

$\begin{align}\text{slope} &= \dfrac{y_2 – y_1}{x_2 – x_1} \\ &= \dfrac{9 – 5}{1 – (-1)} \\ &= \dfrac{4}{2} \\ &= \bbox[border: 1px solid black; padding: 2px]{2}\end{align}$

Therefore, the slope is 2. Often, the slope is labelled m, so you could also write your answer as “$m = 2$”.

Example (using the graph)

Find the slope of the line graphed below.

Solution

First, find two points on the line. You can pick any two points at all, so try to pick points that will be easy to work with or are easy to read from the graph.

two points on the line are (-2, 6) and (0, 1)

Now choose one point to be $(x_1, y_1)$ and the other to be $(x_2, y_2)$. It doesn’t matter which are which, as long as you are consistent with your choice in the formula. Here, we will let $(x_1, y_1) = (–2, 6)$ and $(x_2, y_2) = (0, 1)$.

$\begin{align}\text{slope} = m &= \dfrac{y_2 – y_1}{x_2 – x_1}\\ &= \dfrac{1 – 6}{0 – (-2)}\\ &= \dfrac{-5}{2}\\ &= \bbox[border: 1px solid black; padding: 2px]{-\dfrac{5}{2}}\end{align}$

Other possibilities

It is possible for the slope of a line to be either undefined or to be zero. These are special cases, and explored more in the following lessons:

Zero slope

When you understand that the slope of a line is a way of describing the “steepness” of a line, it is easy to see why a line may have positive or negative slope. But why might a line have zero slope, and what kinds of lines will always have a slope of zero? We will explore these ideas in the following lesson.

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What kinds of lines have zero slope?

Any line of the form $y = c$, where $c$ is some real number, will have a slope of zero. The lines are horizontal (flat) and so have no “steepness” to measure.

graph of the line y = 2, which has a slope of zero

To understand the discussion below, you should be familiar with finding the slope using the slope formula.

Why do horizontal lines have slopes of zero?

Let’s use the line graphed above, $y = 2$, to understand why a line like this will always have zero slope. Pick two points on the line from the graph, say $(1, 2)$ and $(2,2)$. We can let $(x_1, y_1) = (1,2)$ and $(x_2, y_2) = (2,2)$ and apply the slope formula:

$\begin{align}\text{slope} &= \dfrac{y_2 – y_1}{x_2 – x_1}\\ &= \dfrac{2 – 2}{2 – 1} \\&= \dfrac{0}{1} \\&= 0\end{align}$

Every point on the line $y = 2$ has a y-coordinate of 2. So no matter which two points we pick, we will end up with zero in the numerator when applying the slope formula.

For other vertical lines, $y = c$, the y-coordinates will always be $c$. So, the same idea applies and the slope will always end up being zero.

IMPORTANT: the fraction $\dfrac{0}{c}$ is 0 for any nonzero number $c$. But, the fraction $\dfrac{c}{0}$ is always undefined.

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Summary

We have seen how the slope of a line may be zero. The other possibilities when calculating the slope are:

Negative slope – the line falls from left to right.
Positive slope – the line rises from left to right.
Undefined – the line is vertical (of the form $x = c$). (See lesson: undefined slope)

Undefined slope

In the xy-coordinate plane, the slope of a line is a way of measuring its “steepness”. A large positive slope, for example, means that the line rises from left to right very quickly. But considering this, how can we have an undefined slope? The answer is based on how we calculate slope with the slope formula. In this lesson, we will look at what kinds of lines have an undefined slope and at why this is the case.

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What kinds of lines have an undefined slope?

Any vertical line, like the one shown below, will have an undefined slope. These lines are always of the form $x = c$, where $c$ is some number.

example of a line with undefined slope: x = 4

To understand the discussion below, you should be familiar with finding the slope using the slope formula.

Why is the slope undefined for vertical lines?

Let’s use the example of the line $x = 4$, which is graphed above. Two points on this line would be $(4, 1)$ and $(4, 2)$. Since you can use any two points to calculate the slope of the line, we can then apply the slope formula using $(x_1,y_1) = (4,1)$ and $(x_2, y_2) = (4,2)$.

$\begin{align}\text{slope} = m &= \dfrac{y_2 – y_1}{x_2 – x_1} \\&= \dfrac{2 – 1}{4 – 4}\\ &= \dfrac{1}{0}\end{align}$

Now, we can see the issue. Since the two x-values were the same, the denominator of the slope ends up being 0. Division by zero is always undefined. Every point on this line has an x-coordinate of 4, so this will happen regardless of the points picked.

Generalizing this idea a bit, for a vertical line $x = c$, the x-coordinates will always be the number $c$. Therefore, the slope formula will always result in division by zero and therefore the slope will be undefined.

Summary

We have seen how the slope of a line may be undefined. The other possibilities when calculating the slope are:

Negative slope – the line falls from left to right.
Positive slope – the line rises from left to right.
Zero – the line is horizontal (of the form $y = c$). (See lesson: zero slope)

Plotting points (ordered pairs) in the coordinate plane

To plot a point in the xy-coordinate plane (also called the Cartesian coordinate plane), you just have to know how to read the values or numbers in the ordered pair that defines the point. In this lesson, we will look at exactly how to do this!

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Plotting points is all about reading the numbers

Each number in an ordered pair is called a “coordinate”. The first number is called the “x-coordinate” and the second number is called the “y-coordinate”. These two numbers tell you in which part of the coordinate plane your point lies.

the coordinates in an ordered pair are used when plotting points. The first number is the x-coordinate and tells you how many units to move left or right. The second number is the y-coordinate and tells you how many units to move up or down.

Examples

To plot a point, start at the origin, which is the center of the graph (the point (0,0)). Then, the x-coordinate will tell you how many units to move left or right, and the y-coordinate will tell you how many units to move up or down.

to plot the point (4,2), start at the origin and then move 4 units right and 2 units up.

If the x-coordinate is negative, then you will move left instead of right.

plotting points with a negative x-coordinate means that you will move left instead of right. To plot (-3, 5), move 3 units left and 5 units up from the origin (0,0).

It is the same idea with a negative y-coordinate. You will move down from the origin instead of up.

When plotting points with a negative y-coordinate, you will move down from the origin. To plot (3,-2), move 3 units right and then 2 units down.

What if the coordinates are fractions or decimals?

The coordinates in an ordered pair can be any real number. This means that not only can they be positive or negative, but they can also be fractions or decimals. Even in this case, the same general idea applies. But, depending on the decimal or fraction, you might have to estimate the location of the point a little bit.

$When plotting points with fractions or decimals, just move a fraction of the units. For example, to plot (-2.5, -4.5), move 2.5 units left and 4.5 units down.$

Summary

A point is made up of two coordinates: the x-coordinate and the y-coordinate. Each one gives you a piece of information you can use to plot the point in the coordinate plane.

Additional study

Once you have learned how to plot points, you will be able to graph linear equations, or work with ideas like the slope of a line or the distance between two points.

Graphing linear equations

There are three ways you can graph linear equations: (1) you can find two points, (2) you can use the y-intercept and the slope, or (3) you can use the x- and y-intercepts. In the following guide, we will look at all three. Graphing linear equations doesn’t have to be difficult, and can even be fun once you have these methods down!

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Method 1: using two points to graph a linear equation

The graph of any linear equation like $y = 3x + 2$ or $y = -x + 9$ is a line, and only two points are needed to define a line. The idea with this method is to find two points on the line by picking values of $x$.

Example

Graph the linear equation:
$y = \dfrac{1}{3}x – 2$

Solution

To find two points on the line, pick any two values of $x$ that would be easy to work with and then find the corresponding value of $y$. Two easy values here would be 0 and 3 (Since the 3 will cancel with the 3 in the fraction.)

Let $x = 0$:

$\begin{align}y &= \dfrac{1}{3}x – 2 \\ &= \dfrac{1}{3}(0) – 2\\ &= -2\end{align}$

So, one point on the line is $(x,y) = (0, –2)$.

Let $x = 3$:

$\begin{align}y &= \dfrac{1}{3}x – 2\\ &= \dfrac{1}{3}(3) – 2\\ &= 1 – 2\\ &= -1\end{align}$

So, another point on the line is $(x,y) = (3,–1)$.

Once you have your two points, you can plot them and then sketch the line. It is best to use a ruler or something similar to make sure that you draw the best representation of the graph as possible.

graph of y = (1/3)x - 2

TIP: You might find it useful to find three or four points on the graph so that you can sketch it more accurately. To do this, just pick more x-values to find points!

Method 2: Use the slope and y-intercept

A linear equation written in the form $y = mx + b$ is said to be in slope-intercept form. This form shows the slope $m$ and the y-intercept $b$ of the graph. Knowing these two values will let you quickly draw the graph of the linear equation, as you can see in the example below.

Example

Graph the linear equation:
$y = \dfrac{2}{3}x + 4$

Solution

Since this equation is in the form $y = mx + b$, you know that:

The slope is:
$m=\dfrac{2}{3}$
The y-intercept is:
$b = 4$

Let’s look at the y-intercept first.

The y-intercept is the point where the graph crosses the y-axis (the vertical axis). So, you can plot this point as:

point (0,4) which is the y-intercept of the linear function, is plotted

Now consider the slope. Slope can be viewed as a rate of change: it represents the change in $y$ over the change in $x$. Sometimes, this is called “rise over run”.

$\text{slope} = \dfrac{\text{rise}}{\text{run}} = \dfrac{\text{change in }y}{\text{change in }x}$

For this example:

$\text{slope} = \dfrac{2}{3} = \dfrac{\text{change in }y}{\text{change in }x}$

This can be translated to:

$\dfrac{2}{3} = \dfrac{\text{up 2 units}}{\text{for every 3 units right}}$

Therefore, to find another point on the line, start with the y-intercept and go 3 unit right and 2 units up. Do this again, and you will find another point. In fact, you can keep doing this to find as many points as you think you will need to sketch a good graph.

Connect the points, and you have the graph of your linear function!

This method for graphing linear equations can be applied even when the slope is negative or when the slope is not a fraction, even if it doesn’t seem that way. The next example will show you how that works!

Example

Graph the linear equation:
$y = -2x + 1$

Solution

The y-intercept here is 1, so plot this point first.

the y-intercept is located at the point (0,1) for example 3

The slope is –2. While this isn’t a fraction, it can be viewed as one if you let the denominator equal 1.

$\text{slope} = -2 = \dfrac{-2}{1}$

In other words:

$\dfrac{-2}{1} = \dfrac{\text{down two units}}{\text{for every 1 unit right}}$

Now this can be applied to find points on the graph.

points on the graph of y = -2x + 1 found using the slope "down 2 , right 1"

Finally, connect the points to sketch the graph of the linear equation.

graph of linear equation y = -2x + 1

TIP: If the slope is in decimal form, see if you can convert it to a fraction to apply this method. Otherwise, method 1 might be best.

Method 3: Using the x- and y-intercepts

When graphing linear equations that are given in the form $y = mx + b$, it is easiest to just apply method 2. But sometimes, linear equations are given in standard form: $Ax + By = C$, where $A$, $B$, and $C$ are positive or negative whole numbers. In this case, using the x- and y-intercept may be the quickest approach.

Example

Graph the linear equation:
$-3x + 2y = 6$

Solution

To find the x-intercept, which is the point where the graph crosses the x-axis, let $y = 0$ and solve for $x$:

$\begin{align}-3x + 2y &= 6\\ -3x + 2(0) &= 6\\-3x &= 6\\x &= -2\end{align}$

To find the y-intercept, which is the point where the graph crosses the y-axis, let $x = 0$ and solve for $y$:

$\begin{align}-3(0) + 2y &= 6\\2y &= 6\\y &= 3\end{align}$

This gives you two points on the line, which you can plot and then connect to graph the linear equation.

graph of the linear equation -x + 2y = 6

To read more about x-intercepts and y-intercepts, check out the articles Understanding x-intercepts and Understanding y-intercepts.

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Summary

When graphing linear equations, there are many possible approaches. Three common approaches are:

Finding points:
pick simple values of $x$ and find the corresponding values of $y$. Plot these points and use these to graph your line.
Using the slope and y-intercept:
use the concept of “rise over run” and the y-intercept to find points on the graph. This method is especially useful is the line is in slope-intercept form.
Using the x- and y-intercepts:
let $x = 0$ and $y = 0$ to find the intercepts of the graph. Then, use these points to plot the line. This method is useful when the linear equation is in standard form.

Which method you use depends on the form of the linear equation you have and which method you are most comfortable with. No matter what, you can always find points if you get stuck.

Absolute value equations

Solving absolute value equations is based on the idea that absolute value represents the distance between a point on the number line and zero. In this lesson, we will look at a few examples to understand how to solve these equations and also take a bit of a look at this idea of distance as it relates to solving absolute value equations.

Table of Contents

Steps for solving absolute value equations
Step-by-step examples
Absolute value equations with no solutions or one solution
Why does our method work?
Another perspective

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Steps for solving absolute value equations

When given an absolute value equation, you will first need to isolate the absolute value part of the equation. Then, you will write two equations based on the definition of absolute value (though sometimes, there will end up only being one equation). This sounds complicated, but it is only a step or two more than solving the typical linear equation.

the two steps for solving absolute value equations

Let’s try these steps out with some examples!

Examples

In this first example, the absolute value part of the equation is already isolated, so only step two will apply. Whether or not this first step applies or not, you will always have zero, one, or two solutions to any absolute value equation.

Example

Solve the equation:
$|5x – 2| = 13$

Solution

As mentioned, the absolute value part is already isolated. Therefore, we will solve two equations without the absolute value: one where the 13 is positive and one where 13 is negative.

Equation 1:

$\begin{align}5x – 2 &= 13\\ 5x &= 15\\ x &= 3\end{align}$

Equation 2:

$\begin{align}5x – 2 &= -13\\ 5x &= -11 \\ x &= -\dfrac{11}{5}\end{align}$

So, there are two solutions to this equation:
$x = \bbox[border: 1px solid black; padding: 2px]{3, -\dfrac{11}{5}}$

In this next example, there will be a little more work since the absolute value part of the equation is not isolated. In this situation, you will always need to isolate this term before you write your two equations, or you will end up with incorrect answers.

Example

Solve the equation:
$4 + 3|x – 5| = 16$

Solution

Your first step here is to use algebra to isolate the absolute value part of the equation.

$4 + 3|x – 5| = 16$

Subtract 4 from both sides.

$3|x – 5| = 12$

Divide both sides by 3.

$|x – 5| = 4$

Now you can write and solve two equations, one where the 4 is negative and one where the 4 is positive. Remember to drop the absolute value symbol at this step.

Equation 1:

$\begin{align}x – 5 &= 4\\ x &= 9\end{align}$

Equation 2:

$\begin{align}x – 5 &= -4\\ x &= 1\end{align}$

Once again, there are two solutions to the equation:
$x = \bbox[border: 1px solid black; padding: 2px]{1, 9}$

Absolute value equations with one solution or no solutions

In both of our examples above, there were two solutions so you may think that this is always the case. While this is often right, there are cases where there is only one solution and even when there are none. The next two examples will show when this happens.

Example – one solution

Solve the equation:
$6|x – 2| – 1 = -1$

Solution

As usual, we will first isolate the absolute value equation.

$6|x – 2| – 1 = -1$

Add 1 to both sides.

$6|x – 2| = 0$

Divide both sides by 6.

$|x – 2| = 0$

Normally at this stage, we would write two equations without the absolute value bars, but writing 0 with a positive or negative is the same thing. So we only have one equation:

$x – 2 = 0$

Adding 2 to both sides then gives the only solution.

$x = \bbox[border: 1px solid black; padding: 2px]{2}$

We will look more closely at why this happens, but first let’s look at how you might end up with no solutions.

Example – no solutions

Solve the equation:
$|10x – 1| + 3 = -8$

Solution

To isolate the absolute value, subtract 3 from both sides.

$|10x – 1| = -11$

At this step, it can be determined that there are no solutions to the equation. Why? The absolute value of any number is positive. Here, we have the absolute value of something is negative. This is not possible so there are no possible x-values that make this equation true. Therefore, you can write:

Answer: No solutions

The absolute value of any number is always positive. Use this to determine when there are no solutions to an absolute value equation.

Notice that in both examples, the steps were the same as before. You will always follow those two steps when solving any absolute value equation.

Why does this work?

You can think of the absolute value of any number as representing how far it is from zero on the number line. Consider $|3|$ and $|–3|$ below.

number line showing absolute value

This is why the absolute value is always positive – it is representing a distance. Now think of an equation where the absolute value part is isolated, such as $|5x + 1| = 2$. Is the absolute value is 2, then all you know is that $5x + 1$ is 2 units from zero on the number line. This gives two possibilities:

number line showing how absolute value equations work

So this is why we end up with two different equations. In the case of only one solution, you end up with an absolute value expression equal to zero. Since this means that the distance from zero on the number line is zero, you end up with only one equation.

Another perspective

When you study the graphs of absolute value equations, you can see the three cases of one solution, no solution, and two solutions graphically. This is due to the shape of the graph of the absolute value function. This is a bit beyond the scope of this lesson, but in the graph below, you can see the graph of $y = |x – 1|$ and $y = 2$. Notice that the two graphs intersect at two points. These represent the two solutions to the equation $|x – 1| = 2$.

You can probably see how a horizontal line might cross the graph at exactly one point (one solution) or at no points (no solutions). This would be just changing the number on the right hand side of the equation.

Summary

Absolute value equations are always solved with the same steps: isolate the absolute value term and then write equations based on the definition of the absolute value. There may end up being two solutions, one solution, or no solutions. To catch when there is no solution, always remember that absolute values must be positive, but remember to apply this idea only after the absolute value term has been isolated.