An equation is a mathematical statement that two expressions are equal, such as \(3x12=0\text{.}\) A solution to an equation is any set of values that can replace the variables to produce a true statement. The variable in the equation \(3x12=0\) is \(x\) and the solution is \(x=4\text{.}\) To verify this, substitute the value 4 in for \(x\) and check that you get a true statement:
\begin{align*}
3x12 \amp =0\\
3(\alert{4})12 \amp=0\\
1212\amp =0\\
0\amp =0
\end{align*}
Equations relating two variables are particularly useful. If we know the value of one of the variables, we can find the corresponding value of the other variable by solving the equation.
Example29
The equation \(w = 6h\) gives Emily's wages, \(w\text{,}\) in terms of the number of hours she works, \(h\text{.}\) How much in wages will Emily make next week if she works 38 hours??
SolutionWe know that \(h = 38\text{,}\) and we would like to know the value of \(w\text{.}\) We substitute the value for \(h\) into our equation and then simplify.
\begin{align*}
w \amp= 6h\amp\amp \text{Substitute 38 for }h.\\
w \amp= 6(\alert{38})\amp\amp \text{Simplify the right side}.\\
w \amp= 228
\end{align*}
If Emily works \(38\) hours she will earn \(\$228\text{.}\)
To solve an equation we can generate simpler equations that have the same solutions. Equations that have identical solutions are called equivalent equations. For example,
\begin{equation*}
3x  5 = x + 3
\end{equation*}
and
\begin{equation*}
2x = 8
\end{equation*}
are equivalent equations because the solution of each equation is \(4\text{.}\) Often we can find simpler equivalent equations by undoing in reverse order the operations performed on the variable.
Linear, or firstdegree, equations can be written so that every term is either a constant or a constant times the variable. The equations above are examples of linear equations, as are the following:
\begin{equation*}
y=5,\: \: \: y=4x,\: \: \: 2x+4=13,\: \: \: y=142x,\: \: \: y=13x+93x
\end{equation*}
Notice that the last equation above is linear, even though it is not in its simplified form. All the terms \(13x, 9, 3x\) are either constants or constants times the variable \(x\text{.}\) If we simplify the equation, we have \(y=13x3x+9=10x+9\text{,}\) and again, the terms \(10x\) and \(9\) are a constant times the variable and a constant, satisfying the conditions for a linear equation.
Some examples of equations that are NOT linear are:
\begin{equation*}
y=x^3,\: \: \: y=4x^2+3,\: \: \: y=3x17+5x^3,\: \: \: y=\sqrt{x4},\: \: \: y=\frac{1}{x}+2
\end{equation*}
SubsectionSolving Linear Equations
While considering linear equations in more than one variable is a useful task, in the remainder of this section we only consider linear equations in a single variable.
The following list provides important rules for solving linear equations.
To Generate Equivalent Equations
We can add or subtract the same number on both sides of an equation.
We can multiply or divide \(both\) sides of an equation by the \(same\) number (except zero).
Applying either of these rules produces a new equation equivalent to the old one and thus preserves the solution. We use the rules to isolate the variable on one side of the equation.
Example30
Solve the equation \(3x  5 = x + 3\text{.}\)
SolutionWe first collect all the variable terms on one side of the equation, and the constant terms on the other side.
\begin{align*}
3x  5 \alert{{} x} \amp = x + 3 \alert{{} x}\amp\amp \text{Subtract }x \text{ from both sides.}\\
2x  5 \amp = 3\amp\amp \text{Simplify.}\\
2x  5 \alert{{}+ 5} \amp = 3 \alert{{}+ 5}\amp\amp \text{Add 5 to both sides.}\\
2x \amp = 8 \amp\amp\text{Simplify.}\\
\frac{2x}{\alert{2}}\amp =\frac{8}{\alert{2}}\amp\amp\text{Divide both sides by 2.}\\
x\amp = 4\amp\amp\text{Simplify.}
\end{align*}
The solution is \(4\text{.}\) (You can check the solution by substituting \(4\) into the original equation to show that you get a true statement as the result.)
The following steps should enable you to solve any linear equation. Of course, you may not need all the steps for a particular equation.
To Solve a Linear Equation

Simplify each side of the equation separately.
Apply the distributive law to remove parentheses.
Combine like terms.
By adding appropriate terms to both sides of the equation or subtracting appropriate terms from both sides of the equation, get all the variable terms on one side and all the constant terms on the other.
Divide both sides of the equation by the coefficient of the variable.
Example31
Solve \(3(2x  5)  4x = 2x  (6  3x)\text{.}\)
SolutionWe begin by simplifying each side of the equation.
\begin{align*}
3(2x  5)  4x \amp = 2x  (6  3x) \amp\amp\text{Apply the distributive law.}\\
6x  15  4x \amp= 2x  6 + 3x \amp\amp\text{Combine like terms on each side.}\\
2x  15 \amp= 5x  6
\end{align*}
Next, we collect all the variable terms on the left side of the equation, and all the constant terms on the right side.
\begin{align*}
2x  15  5x \amp= 5x  6  5x \amp\amp\text{Subtract }5x \text{ from both sides.}\\
3x  15 + 15 \amp=  6 + 15 \amp\amp\text{Add } 15 \text{ to both sides.}\\
3x \amp= 9
\end{align*}
Finally, we divide both sides of the equation by the coefficient of the variable.
\begin{align*}
3x \amp= 9 \amp\amp\text{Divide both sides by }3.\\
x \amp =3
\end{align*}
The solution is \(3\text{.}\)
We will often come across the need to solve linear equations with fraction or decimal coefficients. In this case, we should "clear" the denominators by multiplying both sides of the equation by the least common multiple of the denominators, as shown in the following example.
Example32
Solve \(\frac{p2}{3}+\frac{p}{4}=\frac{1}{2}\text{.}\)
SolutionWe begin by finding the least common multiple of the three denominators: LCM\((3,4,2)=12\text{.}\) Then we multiply both sides of the equation by \(12\text{,}\) and continue solving as normal.
\begin{equation*}
\begin{aligned}
\frac{p2}{3}+\frac{p}{4}\amp =\frac{1}{2}\\
(\alert{12})\left(\frac{p2}{3}+\frac{p}{4}\right)\amp =(\alert{12})\frac{1}{2}\\
4(p2)+3p\amp =6\\
4p8+3p\amp =6\\
7p8\amp = 6\\
7p\amp =14\\
p\amp =2
\end{aligned}
\end{equation*}
The solution is \(p=2\text{.}\)
When we have decimal coefficients, it is sometimes easier to multiply through by \(10,100\text{,}\) etc. before solving the equation.
Example33
Solve \(0.04x+0.08x=0.1+0.17x\text{.}\)
SolutionSince the decimal coefficients have up to \(2\) decimal places, we will multiply each side of the equation by \(100\) before solving the equation.
\begin{equation*}
\begin{aligned}
0.04x+0.08x \amp =0.1+0.17x\\
(\alert{100})(0.04x+0.08x) \amp =(\alert{100})(0.1+0.17x)\\
4x+8x \amp =10+17x\\
12x\amp =10+17x\\
5x\amp =10\\
x\amp =2
\end{aligned}
\end{equation*}
The solution is \(x=2\text{.}\)
SubsectionNumber of Solutions to Linear Equations of One Variable.
Not all linear equations have one solution. Although all of the examples above of linear equations with one variable have one solution, there are two other possible scenarios. Firstly, an equation of one variable which has equivalent expressions on both sides of the equals sign has infinitely many solutions. For example,
\begin{equation*}
3x2 = 3x2, \hspace{1cm} x=x, \hspace{.5cm}\text{and}\hspace{.5cm} 2(x+1) = 2x+2
\end{equation*}
all have infinitely many solutions. In this situation, any values we substitute for the variables will result in a true statement. Secondly, an equation which has two expressions that are not equal for any values of the variables has no solution. For example,
\begin{equation*}
7x+2 = 7x+10, \hspace{1cm} 17 = 0, \hspace{.5cm}\text{and}\hspace{.5cm} 2(8x+3) = 16x 5
\end{equation*}
all have no solutions. Here, no matter what values we substitute for the variables we will receive a false statement.
The number of solutions to a linear equation may not be immediately obvious. To determine the number of solutions, follow the procedures above for solving a linear equation. If at any point we have a statement which we recognize as always true or always false, then we know the equation has infinitely many or zero solutions, respectively. Else, we will finish solving the equation to determine the single solution.
Example34
Solve each equation. Determine whether there is one solution, infinitely many solutions, or no solutions.
\(5(72x) +1 = 10x + 25\)
\(6x = 12  (x7)\)
\(3612x = 3(12+4x)\)
Solution

\begin{align*}
5 (7  2x) +1 \amp= 10x + 25 \amp\amp \text{Apply the distributive property.}\\
35  10x +1 \amp= 10x + 25 \amp\amp \text{Combine like terms.}\\
10x+36 \amp= 10x + 25
\end{align*}
Since \(10x+36\) is not equal to \(10x+25\) for any value of \(x\text{,}\) has no solution.

\begin{align*}
6x\amp= 12  (x7) \amp\amp \text{Apply the distributive property.}\\
6x \amp= 12x+7 \amp\amp \text{Combine like terms.}\\
6x \amp= x+19 \amp\amp \text{Move variable terms to one side.}\\
7x \amp= 19 \amp\amp \text{Divide by coefficient on variable.}\\
x \amp=\frac{19}{7}
\end{align*}
This equation has exctly one solution, \(x=\frac{19}{7}\text{.}\)

\begin{align*}
3612x \amp= 3(12+4x) \amp\amp \text{Apply the distributive property.}\\
3612x \amp= 3612x
\end{align*}
It is always the case that \(3612x = 3612x\text{,}\) regardless of the value of \(x\text{.}\) In other words, this equation has infinitely many solutions.
Exercise35
Solve \(0.3x+1.4=0.6x2.2\text{.}\)