3.1 One-Step Equations
Introduction
Imagine you are regularly setting aside \(\$75\) into a savings account every month. The total amount of money in the account can be determined by the expression \(75m\), where \(m\) represents the number of months.
Now, picture this: you have a plan to go on a trip that requires \(\$800\). The big question is, how many months do you need to save before you have enough for your adventure? The answer to this question involves solving an equation like \(75m=800\).
Whenever we want to find unknown values, solving equations will be a reliable tool for finding the answers.
In this chapter and the next we will learn to write equations, use them to represent real life scenarios and solve them for unknown values.
Vocabulary
As we begin, it will be helpful if we review some common vocabulary that was introduced back in section 2.1.
- Constant: A quantity that does not change. Numbers are constants. Letters can also be used for constants when their value is unknown.
- Variable: A quantity whose value can change. Variables are usually represented by letters like \(x\), \(y\), etc.
- Operation: A mathematical way of combining two numbers. Addition, subtraction, multiplication and division are the four basic operations.
- Expression: A collection of constants and/or variables that are combined by operations.
- Equation: A mathematical sentence stating that two things are exactly the same, or equal.
- Solution: The specific value(s) for a variable that make an equation true.
- Solving: The process of finding the solution(s) of an equation.
- Equivalent Equations: Equations that have the exact same solution(s).
- Inverse Operations: Operations that have the opposite effect and undo each other. Adding \(5\) and subtracting \(5\) would be inverse operations, for instance.
Informally, we can think of an equation as any math sentence that includes an equal sign. The equal sign \(=\) means "is the same as".
The equal sign was invented in 1557 by the Welsh mathematician Robert Recorde who was tired of having to write the words "is equal to" over and over. Up until then, equations were more like sentences with actual words mixed with numbers!
For example, the equation \(3+2 = 5\) tells us that \(2+3\) is the same as \(5\). This equation is true since \(3+2\) really does equal as \(5\). True equations are known as identities.
On the other hand, the equation \(2 + 3 = 8\) is not true. It is a false statement or a contradiction.
Equations with variable quantities are intriguing. For example, \(x + 7 = 12\) is true when \(x\) is \(5\) but false for every other value. Equations like this, that could be true or could be false depending on the value of the variable, are called conditional equations.
Since the equation \(x + 7 = 12\) is true when \(x\) is replaced with \(5\), we refer to \(5\) as the solution of the equation. Solutions are typically written in equation form, such as \(x = 5\)
It is important to note that not every equation that contains a variable is conditional or has a single solution. For example, \(x -4 = x - 4\) is always true (it's an identity) and any value for \(x\) is a solution. But \(x+7 = x+8\) is always false (it's a contradiction) and no value will ever be a solution.
One of the most important things we will learn in Algebra is how to find the solution(s) of an equation when they exist. We turn our attention for the remainder of this section to finding solutions to equations where there is a single operation. That will lay the foundation for solving more complex equations in the following sections.
Mental Math vs. Algebraic Solving
Some conditional equations contain simple numbers and operations where we might be able to spot the solution simply by looking at it and applying our own number sense.
For example, consider this equation: \(3x = 6\). It's a basic fact of multiplication that \(3(2) = 6\) so the solution must be \(x=2\).
While we were able to solve that equation mentally, it certainly would be more challenging to mentally solve an equation like \(3.835x = 62.894\). Because of that, it's important that we develop a systematic process for solving equations algebraically.
That process will allow us to transform a given equation where an uncertain solution into an equivalent equation that explicitly states the solution.
Since the goal is to get an equivalent equation, whatever we do to solve an equation must maintain the equality of both sides. There are a number of properties that do not impact the equality of an equation. We list several of them below.
Generally speaking, these rules can be summarized as "whatever you do to one side of an equation you must do to the other side".
Inverse Operations
Now that we've seen the importance of performing operations on both sides of an equation, we want to make sure we always know which operation to perform when solving. Since our goal is to end up with an equivalent equation in which the variable is isolated, we need to be able to get rid of any number that appears with the variable. This is done by using inverse operations.
You are likely already familiar with some inverse operations. Addition and subtraction, for instance, are usually thought of as inverse operations, as are multiplication and division. Powers and roots are also inverse operations. In more advanced algebra courses, you will make use of another pair of inverse operations: exponentials and logarithms.
One method of testing if two operations are inverses is by picking a number, applying the first operation, then applying the second operation and checking to see if it returns the initial value. Here are a few examples.
adding and subtracting \(6\) are inverse operations | multiplying and dividing by \(-3\) are inverse operations | cubing a number and taking the cube root are inverse operations |
There are, however, some finer points that need to be discussed. The first is that addition/subtraction and multiplication/division are only inverse operations when using the same number. Subtracting \(6\) is not the inverse of adding \(200\), even though one is addition and the other is subtraction.
The second point of interest is that division by \(0\) is not defined, so multiplying and dividing by \(0\) are not inverse operations.
The third point concerns powers and roots. Exponents and roots are fairly straightforward if the exponent and/or root is odd. For example, an exponent of \(3\) and a \(3\)rd root are inverse operations. As a demonstration of that, a quick calcultion shows \((-4)^3 = -64\) and \(\sqrt[3]{-64} = -4\), exactly the behavior we expect from inverse operations.
Things are more complicated with even exponents and roots, however. For instance, \((-3)^2 = 9\) but \(\sqrt{9} = 3\), which is not the number we started with.
Does this mean that an exponent of \(2\) and a \(2\)nd root are never inverse operations? Not exactly. If we had started with positive \(3\), then \((3)^2 = 9\) and \(\sqrt{9} = 3\) and this time the operations do indeed undo each other.
So what went wrong with \(-3\)? Well, the issue is that there are two numbers that equal \(9\) when you square them, \(3\) and \(-3\), but the square root will only return the positive (or principle) one. This is true for all even roots and is something we will need to take into consideration later when solving equations with even powers.
Give the inverse of each operation:
- Adding \(81\)
- Dividing by \(-5\)
- Applying a power of \(4\)
- Subtracting \(-23\)
- Applying the \(5\)th root
Solution
- Subtracting \(81\)
- Multiplying by \(-5\)
- Applying the \(4\)th root
- Adding \(-23\)
- Applying a power of \(5\)
Solving with Inverse Operations
We now have the foundational for solving equations: identify the operation applied to the variable and then apply the inverse operation.
To see this process in action, let's revisit the equation \(3x = 6\). How can we solve it using inverse operations? Since \(x\) is being multiplied by \(3\), we should undo that multiplication by dividing by \(3\). If we divide both sides of the equation by \(3\), the resulting equation will show us the solution:
\begin{align} 3x &= 6 && \small \color{#5fa2ce} {\text{Original equation}} \\ \frac{3x}{3} &= \frac{6}{3} && \small \color{#5fa2ce} {\text{Divide both sides by }3} \\ x &= 2 && \small \color{#5fa2ce} {\text{Simplify}} \end{align}Notice that we had to divide both sides of the equation by \(3\) in order to get an equivalent equation. It's critical to remember to do the same thing to both sides to maintain the equality of the equation.
(a) Solve \(x - 13 = 35\)
(b) Solve \(\frac{n}{4} = 24\)
(c) Solve \(75m=800\)
Solution
(a) Notice that that \(13\) is being subtracted from \(x\), so we need to add \(13\) to both sides.
\begin{align} x - 13 &= 35 && \small \color{#5fa2ce} {\text{Notice } 13 \text{ is being subtracted from } x} \\ x - 13 + 13 &= 35 + 13 && \small \color{#5fa2ce} {\text{Add }13 \text{ to both sides}} \\ x &= 48 && \small \color{#5fa2ce} {\text{Simplify}} \\ \end{align}(b) Since \(n\) is being divided by \(4\), we will multiply both sides by \(4\).
\begin{align} \frac{n}{4} &= 24 && \small \color{#5fa2ce} {\text{Notice } n \text{ is being divided by } 4} \\ 4 \cdot \frac{n}{4} &= 4 \cdot 24 && \small \color{#5fa2ce} {\text{Multiply both sides by } 4} \\ n &= 96 && \small \color{#5fa2ce} {\text{Simplify}} \\ \end{align}(b) Since \(m\) is being multiplied by \(75\), we will divide both sides by \(75\).
\begin{align} 75m &= 800 && \small \color{#5fa2ce} {\text{Notice } m \text{ is being multiplied by } 75} \\ \frac{75m}{75} &= \frac{800}{75} && \small \color{#5fa2ce} {\text{Divide by } 75} \\ m &\approx 10.667 && \small \color{#5fa2ce} {\text{Evaluate}} \\ \end{align}This is the equation from our introduction. By solving it we now know you would need to save for almost \(11\) months to have enough for the \(\$800\) trip.
The final two examples involve powers and roots, so we will proceed with caution. If an equation has an even power then the opposite operation would be the corresponding even root, and there are three possible outcomes.
If \(x^n = A\) and \(n\) is an even number then:
- If\(A\) is positive, then there are two solutions: the positive (principle) root \(x=\sqrt[n]{A}\) and the negative (secondary) root \(x=-\sqrt[n]{A}\).
- If\(A = 0\) then there is one solution: \(x=0\)
- If\(A\) is negative, then there is no real number solution.
The "plus/minus" symbol \(\pm\) can be used to indicate the two solutions in the case where \(A\) is positive.
Also, remember that for a square root (or \(2\)nd root) we don't write the root index \(2\) with the radical symbol. So rather than writing \(\sqrt[2]{25}\) just write \(\sqrt{25}\).
(a) Solve \(x^2 = 49\)
(b) Solve \(n^5 = -5.37824\)
(c) Solve \(p^6 = -729\)
Solution
(a) In the equation \(x^2 = 49\), \(x\) is raised to an even power and the result is positive, so there are two solutions. Since the power is \(2\) we will use the square root to solve.
\begin{align} x^2 &= 49 \\ x &= \pm \sqrt{49} \\ x &= \pm 7 \end{align}We can either list the solutions separately as \(x = 7, x = -7\) or, as we've done here, use the \(\pm\) symbol to indicate both values.
(b) In the equation \(n^5 = -5.37824\) the power is \(5\) which is odd, so there is a single solution and we will a use \(5\)th root to find it.
\begin{align} n^5 &= -5.37824 \\ n &= \sqrt[5]{-5.37824} \\ n &= -1.4 \end{align}Tip: Evaluating Roots on a Calculator
In example 3 part (b), it was necessary to calculate a 5th root. Every calculator should have an easy-to-find square root button, but a different button must be used for higher roots. If you are using a TI-83 or TI-84, here are the steps for calculating a 5th root:
- Begin your calculation by entering the number 5, since you want to do a 5th root.
- Press the [MATH] button
- Select \(\sqrt[x]{ \, }\)
- Enter the number you're finding the 5th root of. For the example above, that would be -5.37824.
- Press [ENTER] to get the result.
Roots can also be entered as fractional exponents. For instance, the square root is the same as a power of \(1/2\) and a cube root is the same as a power of \(1/3\). If you use fractional exponents, be sure they are enclosed in parenthesis.
(c) In the equation \(p^6 = -729\), \(p\) is raised to an even power and the result is negative. Therefore, the equation has no real number solution.
We'll conclude by looking at equations with roots. As discussed earlier, odd roots aren't particularly complicated. However, we need to be careful with equations involving even roots. When we are given an even root of a number, it is understood to be the positive (or principal) root. Because of that, it can't be set equal to a negative number.
If \(\sqrt[n]{x} = A\) and \(n\) is an even number then:
- If\(A\) is a non-negative number, then there is one solution: \(x = A^n\)
- If\(A\) is negative, then there is no solution.
(a) Solve \(\sqrt[3]{x} = -5\)
(b) Solve \(\sqrt[4]{w} = 3.5\)
(c) Solve \(\sqrt{n} = -7\)
Solution
(a) Since the equation \(\sqrt[3]{x} = -5\) inovles and odd root we can solve it by raising both sides to the \(3\)rd power
\begin{align} \sqrt[3]{x} &= -5 \\ x &= (-5)^3 \\ x &= -125 \end{align}(b) The equation \(\sqrt[4]{w} = 3.5\) involves an even root and is equal to a positive number, so we can solve it by raising both sides to the \(4\)th power.
\begin{align} \sqrt[4]{x} &= 3.5 \\ x &= (3.5)^4 \\ x &= 150.0625 \end{align}(c) The equation \(\sqrt{n} = -7\) has a square root equal to a negative number, so there is no real number solution to this equation.
Conclusion
In this section we've learned to Solve Six Types of Equations using inverse operations.
In the next stage of our journey we will explore equations like \(3x-12 = 4\) that involve more than one operation.