Section 4.1 What Are Equations?
This might seem like a silly question, but it’s important to address clearly. An equation is a mathematical statement involving equality “\(=\)” that evaluates as either TRUE or FALSE. The following equation
\begin{equation*}
2 + 4 \times (2 - 2) - 2 = 0
\end{equation*}
is TRUE because the values on either side of the equal sign are the same. On the other hand, the equation
\begin{equation*}
2+2 = 5
\end{equation*}
is FALSE. An algebraic equation is an equation involving one or more variables. For instance,
\begin{equation*}
4x - 3y + 2z = 6
\end{equation*}
is an algebraic equation in three variables. The equation is TRUE or FALSE depending on what value the variables take. If \(x=1, y=1, z=1\) then it is FALSE, but if \(x=2, y=2, z=2\text{,}\) then it is TRUE.
To solve an equation, means to find all possible values of the variable(s) that make the equation TRUE. These values are called solutions of the equation. For instance, the equation
\begin{equation*}
x^2 = 4
\end{equation*}
has two solutions \(x = 2\) AND \(x=-2\text{.}\) The solution set is the set of all solutions of the equation. In this case, the solution set is \(\{-2,2\}\text{,}\) where it is common to enclose the solutions in curly braces. On the other hand, the equation
\begin{equation*}
x^2 = -4
\end{equation*}
has no (real number) solutions, so that its solution set is the empty set, \(\emptyset\text{.}\)
When solving an equation, we usually do so by replacing it with an equivalent equation, that is, with an equation with the same solution set. Performing operations of addition or multiplication by the same value to both sides of an equation generally results in an equivalent equation. The following example illustrates this in the case of a linear equation.
Example 4.1. Solving a Linear Equation.
To solve \(2(x-3)+5\,=\,9\text{,}\) we replace it with a sequence of equivalent equations.
\begin{align*}
2(x-3)+5\,\amp=\,9\\
2(x-3)+5\,{\color{blue}-5}\,\amp=\,9\,{\color{blue}-5}\\
2(x-3)\,\amp=\,4\\
2(x-3)\,{\color{blue}\div 2}\,\amp=\,4 \,{\color{blue}{\div 2}}\\
x-3 \,\amp=\, 2\\
x-3 \,{\color{blue}+3}\,\amp=\, 2 \,{\color{blue}+3}\\
x \,\amp=\, 5
\end{align*}
showing that \(x=5\) is the unique solution to the original equation.
It is not necessary to write as much as the previous example might suggest. The following would be suitable in the college classroom.
Example 4.2. Solving a Linear Equation.
Solve the linear equation \(\frac{1}{3} x \,=\, \frac{2}{3}(x+2)\text{.}\)
\begin{align*}
\frac{1}{3} x \,\amp=\, \frac{2}{3}(x+2)\\
x \,\amp=\, 2(x+2)\\
x \,\amp=\, 2x+4\\
0 \,\amp=\, x+4\\
x \,\amp=\, -4
\end{align*}
Example 4.4. Solving a Linear Equation.
Solve for \(z\text{.}\)
\begin{equation*}
\frac{2z-3}{4} = 2 - \frac{3z+1}{2}
\end{equation*}
Solution.
If we multiply both sides by 4, we will clear the fractions.
\begin{gather*}
\frac{2z-3}{4} = 2 - \frac{3z+1}{2}\\
4\cdot \frac{2z-3}{4} = 4\cdot \left(2 - \frac{3z+1}{2}\right)\\
4\cdot \frac{2z-3}{4} = 4\cdot 2 - 4\cdot \frac{3z+1}{2}\\
2z-3 = 8 - 2(3z+1)\\
2z-3 = 8 - 6z-2\\
2z-3 = 6 - 6z\\
8z = 9\\
z = 9/8
\end{gather*}
Checkpoint 4.6.
Are the following pairs of equations equivalent? That is, do they have the same solution set? Why or why not.
\(x = 2\) and \(x^2 = 4\)
\(\sqrt{x^2} = \sqrt{4}\) and \(x = 2\)
