What is an Equation?

In an equation with one variable, the variable has a solution, or value, that makes the equation true.

Learning Objectives

Explain what an equation in one variable represents and the reasons for using one

Key Takeaways

Key Points

• An equation is a mathematical statement that asserts the equivalence of two expressions.
• When an equation contains a variable, such as $x$, the variable is considered an unknown value.
• The values of the variables that make the equation true are the solutions of the equation and can be found by solving the equation.
• A solution of an equation can be verified, or checked, by substituting in its value for the variable in the equation.

Key Terms

• solution: A value that can be substituted for a variable to make an equation true.
• unknown: A variable in an equation that needs to be solved for.
• equation: A mathematical statement that asserts the equivalence of two expressions.

An equation is a mathematical statement that asserts the equivalence of two expressions. For example, the assertion that “two plus five equals seven” is represented by the equation $2 + 5 = 7$.

In many cases, an equation contains one or more variables. These are still written by placing each expression on either side of an equals sign ($=$). For example, the equation $x + 3 = 5$, read “$x$ plus three equals five”, asserts that the expression $x+3$ is equal to the value 5.

It is possible for equations to have more than one variable. For example, $x + y + 7 = 13$ is an equation in two variables. However, this lesson focuses solely on equations in one variable.

Solving Equations

When an equation contains a variable such as $x$, this variable is considered an unknown value. In many cases, we can find the possible values for $x$ that would make the equation true. 

For example, consider the equation we were talking about above: $x + 3 =5$. You have probably already guessed that the only possible value of $x$ is 2, because you know that $2 + 3 = 5$ is a true equation. We use an equals sign to show that we know the value of a given variable. In this case, we write $x=2$ (read as “$x$ equals two”).

The values of the variables that make an equation true are called the solutions of the equation. In turn, solving an equation means determining what values for the variables make the equation a true statement.

The equation above was fairly straightforward; it was easy for us to identify the solution as $x = 2$. However, it becomes useful to have a process for finding solutions for unknowns as problems become more complex.

Verifying Solutions

If a number is found as a solution to an equation, then substituting that number back into the place of the variable should make the equation true. Thus, we can easily check whether a number is a genuine solution to a given equation.

For example, let’s examine whether $x=3$ is a solution to the equation  $2x + 31 = 37$.

Substituting 3 for $x$, we have:

$2(3) + 31 = 37 \\ 6 + 31 = 37$

This equality is a true statement. Therefore, we can conclude that $x = 3$ is, in fact, a solution to the equation $2x+31=37$.

Solving Equations: Addition and Multiplication Properties of Equality

The addition and multiplication properties of equalities are useful tools for solving equations.

Learning Objectives

Solve equations using the addition and multiplication properties of equality

Key Takeaways

Key Points

• Equations often express relationships between given quantities (called “knowns”) and quantities that have yet to be determined (called ” unknowns “).
• The addition property of equality states that any real number can be added to both sides of an equation.
• The subtraction property of equality states that any real number can be subtracted from both sides of an equation.The multiplication property of equality states that any real number can be multiplied by both sides of an equation.
• The division property of equality states that any non-zero real number can divide both sides of an equation.

Key Terms

• equality: The state of two or more entities having the same value.

An equation is a mathematical statement that asserts the equivalence of two expressions. In modern notation, this is indicated by placing the expressions on either side of an equal sign (=). For example, $x+3=5$ asserts that $x+3$ is equal to 5.

Equations often express relationships between given quantities (“knowns”) and quantities yet to be determined (“unknowns”). By mathematical convention, unknowns are denoted by letters toward the end of the alphabet $(x,y,z…)$, while knowns are denoted by letters at the beginning of the alphabet $(a,b,c…)$.

The process of expressing an equation’s unknowns in terms of its knowns is called solving the equation. In an equation with a single unknown, a value of that unknown for which the equation is true is called a solution or root of the equation.

If an equation in algebra is known to be true, the following properties may be used to produce another true equation. For each property, both the formal definition and the plain-English definition are provided.

If $a=b$, then $a+c=b+c$.

In other words, any real number can be added to both sides of an equation.

The Subtraction Property of Equality

If $a=b$, then $a-c=b-c$.

In other words, any real number can be subtracted from both sides of an equation.

The Multiplication Property of Equality

If $a=b$, then $ca=cb$.

In other words, any real number can be multiplied to both sides of an equation.

The Division Property of Equality

If $a=b$, and $c\neq 0$, then $\dfrac{a}{c}=\dfrac{b}{c}$.

In other words, any non-zero real number can divide both sides of an equation.

Example 1

The bill for the repair of a car was $458, and the cost of parts was$339.  The cost of labor was $34 per hour. Write and solve an equation to find the number of hours of labor that went into the repair. Let $x$ equal the unknown value: the number of hours of labor. The equation is therefore: $34x+339=458$. In English, the cost of the labor ($34) multiplied by the number of hours of labor $(x)$, plus the cost of the parts ($339), is equal to the total bill for the repair ($458).

To solve for the unknown, first undo the addition operation (using the subtraction property) by subtracting \$339 from both sides of the equation:

\begin{align} 34x+339-339&=458-339 \\ 34x&=458-339 \\ 34x&=119 \end{align}

Then undo the multiplication operation (using the division property) by dividing both sides of the equation by 34:

\begin{align} \dfrac{34x}{34}& =\dfrac{119}{34} \\ x& =\dfrac{119}{34} \\ x&=3.5 \end{align}

This means the car repair labor took 3.5 hours.

Example 2

Solve the following equation using properties of equality:

$\dfrac{1}{8}x-5=3$

First, use the addition property to add 5 to both sides of the equation:

\begin{align} \dfrac{1}{8}x-5+5&=3+5 \\ \dfrac{1}{8}x&=3+5 \\ \dfrac{1}{8}x&=8 \end{align}

Second, use the multiplication property to multiply both sides of the equation by 8:

\begin{align} 8 \cdot \dfrac{1}{8}x&= 8 \cdot 8 \\ x&=64 \end{align}

This is the solution to the equation.

Rational Equations

A rational equation sets two rational expressions equal to each other and involves unknown values that make the equation true.

Learning Objectives

Solve rational equations by finding a common denominator

Key Takeaways

Key Points

• When solving a rational equation, find a common denominator or use the cross-multiplication method.
• If the denominators are the same in a rational equation, the numerators must also be the same. Therefore, use the following strategy: find a common denominator, set the numerators equal to each other, and then solve for the variable, if necessary.

Key Terms

• cross-multiply: To multiply the numerator of each side of an equation by the denominator of the other side.
• rational expression: A set of mathematical terms that can be expressed as the quotient of two polynomials.
• denominator: The number or expression written below the line in a fraction (e.g., the 2 in $1/2$).
• numerator: The number or expression written above the line in a fraction (e.g., the 1 in $1/2$).

Solving a Rational Equation (Same Denominators)

For an equation that involves two fractions or rational expressions, cross-multiplying is a helpful strategy for simplifying the equation or determining the value of a variable.

$\dfrac {x}{8} = \dfrac {3}{8}$

Cross-multiplying yields:

\begin{align} 8x&=8\cdot 3 \\ 8x&=24 \end{align}

Now solve for $x$, by dividing both sides of the equation by $8$:

\begin{align} \dfrac{8x}{8}&=\dfrac{24}{8} \\ x&=3 \end{align}

$\dfrac {x}{8} = \dfrac {3}{8}$

Isolate the variable on the left by multiplying both sides by $8$:

$\left(\dfrac{8}{1} \right)\cdot \dfrac{x}{8}= \left(\dfrac{8}{1} \right) \cdot \dfrac{3}{8}$

The denominators on both sides cancel out, yielding:

$x=3$

You could also come to this conclusion through deductive reasoning. Notice that the rational expressions on both sides of the equal sign have the same denominator. If you have a rational equation where the denominators on either side of the equation are the same, then their respective numerators must also be the same value, even though they might be expressed in different terms. This suggests a strategy: Find a common denominator, set the numerators equal to each other, and solve for any unknowns.

Solving a Rational Equation (Different Denominators)

Several real-life situations can be modeled using equations that set two fractions, or ratios, to be equal to each other—for example, finding unknown dimensions of certain shapes.

Two triangles are said to be “similar” if they have equal corresponding angles. This is the same as the triangles having equal side-length ratios. Similar geometric shapes: Two geometrical objects are similar if they both have the same shape or if one has the same shape as the mirror image of the other. In this image, figures shown in the same color are similar.

The two triangles below are similar. If length $\overline{AC}$ is 10 inches, $\overline{EF}$ is 14 inches, and $\overline{AB}$ is 17 inches, what is the length of $\overline{EG}$? Similar triangles: Corresponding angles in similar shapes are marked with the same symbol. For example, $\angle A$ (i.e., angle $A$) corresponds to $\angle E$, and they are equal.

Let’s start by writing a rational equation:

$\dfrac{\overline{AC}}{\overline{EF}}=\dfrac{\overline{AB}}{\overline{EG}}$

Now let’s plug in the actual numbers:

$\dfrac{10}{14}=\dfrac{17}{x}$

Now cross-multiply:

\begin{align} \dfrac{10}{14}\cdot (x) \cdot (14)&=\dfrac{17}{x} \cdot (x) \cdot (14) \\ 10\cdot x&=14\cdot 17 \\ 10x&=238 \end{align}

Finally, divide both sides of the equation by 10:

$x=23.8$ inches.

Therefore, $\overline{EG}$ is 23.8 inches long.

Equations involving radicals are often solved by squaring both sides.

Learning Objectives

Solve a radical equation by squaring both sides of the equation and checking for false solutions

Key Takeaways

Key Points

• When solving equations that involve radicals, begin by asking: is there an $x$ under the square root ? The answer to this question will determine the way you approach the problem.
• If there is not an $x$ under the square root—if only numbers are under the radicals—the equation can be solved in much the same way as if it contained no radicals.
• However, if there is an $x$ under a square root, then move everything except that radical to one side, and then square both sides of the equation.
• Squaring both sides can potentially introduce extraneous solutions (i.e., false answers), so it is important to check your answers after solving!

Key Terms

• radical: A root (of a number or quantity).
• square: The second power of a number, value, term, or expression.
• extraneous solution: An answer to an equation that emerges from the process of solving the problem but that is not a valid answer to the original problem.
• root: A number that, when plugged into an equation, will produce a zero.

When solving equations that involve radicals, begin by asking: is there an $x$ under the square root? The answer to this question will determine the way the problem is approached. If there is not an $x$ under the square root—if only numbers are under the radicals—the problem can be solved much the same way as if it had no radicals.

Example

Solve this equation:

$\sqrt 2 x+5=7-\sqrt 3 x$

First, isolate the variable on one side of the equation:

\begin{align} \sqrt 2 x+\sqrt 3 x+5-5&=7-5-\sqrt 3 x + \sqrt 3 x \\ \sqrt 2x+\sqrt 3x&=7-5 \end{align}

Next, since both terms on the left hand side of the equation contain $x$, factor out the $x$. (Remember that $\sqrt 2$ and $\sqrt 3$ are unlike terms and cannot be combined.)

$x(\sqrt 2+\sqrt 3)=2$

Now divide both sides of the equation by $(\sqrt 2+\sqrt 3)$ to solve for $x$:

$x=\dfrac 2 {\sqrt 2 + \sqrt 3}$

The key thing to note about problems like this is that both sides of the equation do not have to be squared. $2\sqrt{2}$ may look complicated, but it is just a number—it functions in the equation just the way that the number $10$, or $\frac {1}{3}$, or $\pi$ would.

Steps to Solve a Radical Equation with a Variable Under the Radical

If there is an $x$, or variable, under the square root, the problem must be approached differently. In this case, both sides must be squared to get rid of the radical. However, squaring both sides can introduce extraneous solutions (i.e., false answers), so it is important to check the answers after solving. If no answer checks out, then the solution is “no solution.”

Example

Solve this equation:

$\sqrt{6x-2}-3=7$

\begin{align} \sqrt{6x-2}-3+3&=7+3 \\ \sqrt{6x-2}&=10 \end{align}

Now, to undo the radical symbol (square root), square both sides of the equation (recall that squaring a square root removes the radical):

\begin{align} \left(\sqrt{6x-2}\right)^2&=(10)^2 \\ 6x-2&=100 \end{align}

Finally, solve the remaining equation:

\begin{align} 6x-2+2&=100+2 \\ 6x&=102 \\ \dfrac{6x}{6}&=\dfrac{102}{6} \\ x&=17 \end{align}

Now, let’s go back and check our answer:

\begin{align} \sqrt {6(17)-2}-3&=7 \\ \sqrt {6(17)-2}&=10 \\ \left(\sqrt {6(17)-2}\right)^2&=(10)^2 \\ 6(17)-2&=100 \\ 6(17)&=102 \\ 17&=17 \end{align}

This is a true statement. Therefore, $x=17$ is a valid solution to the equation $\sqrt{6x-2}-3=7$.

Solving Radical Equations with False Solutions

Solve this equation:

$-5-\sqrt{10x-2}=5$

Isolate the radical symbol (square root):

\begin{align} -5-\sqrt{10x-2}+5&=5+5 \\ -\sqrt{10x-2}&=10 \\ (-1) \cdot (-\sqrt{10x-2})&=(-1) \cdot (10) \\ \sqrt{10x-2}&=-10 \end{align}

Now, square both sides of the equation, and solve:

\begin{align} \left( \sqrt{10x-2} \right) ^2&= (-10) ^2 \\ 10x-2&=100 \\ 10x-2+2&=100+2 \\ 10x&=102 \\ \dfrac{10x}{10}&=\dfrac{102}{10} \\ x&=10.2 \end{align}

Once we check this result, however, we discover that $\sqrt {100}=-10$. This is incorrect, because the square root is defined to be only the positive root, $10$. This means that $10.2$ is an extraneous solution. Because it is the only answer we found, the answer to this problem is “no solution.”

This problem demonstrates how important it is to check your solutions whenever you square both sides of an equation.

Equations with Absolute Value

To solve an equation with an absolute value, first isolate the absolute value, and then solve for the positive and negative cases.

Learning Objectives

Break down an absolute value equation into two equations to solve for the variable

Key Takeaways

Key Points

• Absolute values are always positive, since they represent distance.
• An absolute value equation can have one, two, or no solutions.

Key Terms

• absolute value: The magnitude (i.e., non-negative value) of a number without regard to its sign; a number’s distance from zero.

Absolute value is one of the simplest functions—and paradoxically, one of the most problematic.  At face value, nothing could be simpler: absolute value simply means the distance a number is from zero.  The absolute value of $-5$ is $5$, and the absolute value of $5$ is also $5$, since both $-5$ and $5$ are $5$ units away from $0$. Mathematically, this is represented as follows:

$\left | -5 \right |=5$ and $\left| 5 \right| =5$

The following number line also illustrates this definition: Absolute value: The absolute value of a real number may be thought of as its distance from zero. In this image, for example, $\left | -3 \right |=3$.

Types of Solutions to Absolute Value Equations

Consider the following three equations. They look very similar—only the number changes—but the solutions are completely different. These three equations demonstrate how absolute value equations can have one, two, or no solutions.

Equation 1

$\left| x \right| =10$

What value(s) will make this equation true?

$x=10$ works, as does $x=-10$. Therefore our solution is:

$x=\pm 10$

Equation 2

$\left| x \right| =-10$

Here, neither $x=10$ nor $x=-10$ works. Recall that absolute value is a measure of distance, so it can never be a negative value. This equation therefore has no solution.

Equation 3

$\left| x \right| =0$

What value(s) will make this equation true? $x=0$ is the only solution.

Solving Absolute Value Equations

The following steps describe how to solve an absolute value equation:

1. Isolate the absolute value term algebraically.
2. Set up two separate equations: For the first, keep the new equation you found in step 1, but remove the absolute value signs; for the second, keep the equation you found in step 1, remove the absolute value signs, and multiply one side by -1.
3. Solve the pair of equations.

For example, let’s solve the following equation for $x$:

$3\left| 2x+1 \right| -7=5$

Step 1

First, algebraically isolate the absolute value by adding 7 to both sides of equation and then dividing both sides by 3:

\begin{align} 3 \left| 2x+1 \right| -7 +7 &=5 +7 \\ 3 \left| 2x+1 \right| &= 12 \\ \dfrac{3 \left| 2x+1 \right|}{3} &= \dfrac{12}{3} \\ \left| 2x+1 \right| &=4 \end{align}

Step 2

Now, set up two separate equations. The first is the equation we found in Step 1, but with the absolute value signs removed:

$2x+1 =4$

The second equation is the one we found in Step 1, with the absolute value signs removed, and with the other side multiplied by -1:

$2x+1 =-4$

Step 3

Now, solve both equations. For the first:

\begin{align} 2x+1&=4 \\ 2x+1-1&=4-1 \\ 2x&=3 \\ \dfrac{2x}{2}&=\dfrac{3}{2} \\ x&=\dfrac{3}{2} \end{align}

For the second equation:

\begin{align} 2x+1&=-4 \\ 2x+1-1&=-4-1 \\ 2x&=-5 \\ \dfrac{2x}{2}&=\dfrac{-5}{2} \\ x&=\dfrac{-5}{2} \end{align}

Therefore, this problem has two answers:

$x=\dfrac{3}{2}$ and $x=\dfrac{-5}{2}$