1.2: Absolute Value Equations
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When solving equations with absolute value, the solution may result in more than one possible answer because, recall, absolute value is just distance from zero. Since the integer \(−4\) has distance \(4\) units from zero, and 4 has distance 4 units from zero, then there are two integers that have distance \(4\) from zero, \(−4,\: 4\). We extend this concept to algebraic absolute value equations. This is illustrated in the following example.
Solve for \(x\): \(|x|=7\)
Solution
\[\begin{array}{rl}|x|=7&\text{Expression in the absolute value can be positive or negative} \\ x=7\text{ or }x=-7&\text{Solution}\end{array}\nonumber\]
Let’s think about the solution set. The equation is asking for all numbers in which the distance from zero is \(7\). Well, there are two integers that have a distance \(7\) from zero, \(−7\) and \(7\). Hence, the solution set \(\{−7, 7\}\).
The first set of rules for working with negative numbers came from \(7^{\text{th}}\) century India. However, in 1758, more than a thousand years later, British mathematician Francis Maseres claimed that negatives “Darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple.”
Absolute value for linear equations in one variable is given by \[\text{If }|x|=a,\text{ then }x=a\text{ or }x=-a\nonumber\] where \(a\) is a real number.
When we have an equation with absolute value, it is important to first isolate the absolute value, then remove the absolute value by applying the definition.
Solve for \(x\): \(5+|x|=8\)
Solution
\[\begin{array}{rl}5+|x|=8&\text{Isolate the absolute value by subtracting }5\text{ from each side} \\ |x|=3&\text{Rewrite as two linear equations} \\ x=3\text{ or }x=-3&\text{Solution}\end{array}\nonumber\]
Thus, the solution set is \(\{-3,3\}\).
Solve for \(x\): \(-4|x|=-20\)
Solution
\[\begin{array}{rl}-4|x|=-20&\text{Isolate the absolute value} \\ \frac{-4|x|}{-4}=\frac{-20}{-4}&\text{Divide each side by }-4 \\ |x|=5&\text{Rewrite as two linear equations} \\ x=5\text{ or }x=-5&\text{Solution}\end{array}\nonumber\]
Thus, the solution set is \(\{-5,5\}\).
Never combine the inside of the absolute value with factors or terms from outside the absolute value. We always have to isolate the absolute value first, then apply the definition to obtain two equations without the absolute value.
Solve for \(y\): \(5|y| − 4 = 26\)
Solution
\[\begin{array}{rl}5|y|-4=26&\text{Isolate the absolute value term by adding }4\text{ to each side} \\ 5|y|=30&\text{Divide each side by }5 \\ \\ |y|=6&\text{Rewrite as two linear equations} \\ y=6\quad\text{or}\quad y=-6&\text{Solution}\end{array}\nonumber\]
Thus, the solution set is \(\{-6,6\}\).
Absolute Value Equations with Different Solutions
Often, we will have linear arguments inside the absolute value which changes the solution. Previously, all solution sets have been opposite integers, but in these cases, the solution sets contain different sized numbers.
Solve for \(t\): \(|2t − 1| = 7\)
Solution
\[\begin{array}{rl} |2t-1|=7&\text{The absolute value term is isolated. Rewrite as two linear equations.} \\ 2t-1=7\quad\text{or}\quad 2t-1=-7&\text{Solve each equation.}\end{array}\nonumber\]
Notice we have two equations to solve where each equation results in a different solution. In any case, we solve as usual.
\[\begin{array}{lll} 2t-1=7&& 2t-1=-7 \\ 2t=8&\text{or}& 2t=-6 \\ t=4&&t=-3\end{array}\nonumber\]
Thus, the solution set is \(\{-3,4\}\).
Multiple-Step Absolute Value Equations
Solve for \(x\): \(2 − 4|2x + 3| = −18\)
Solution
To isolate the absolute value, we first apply the addition rule for equations. Then apply the multiplication rule for equations.
\[\begin{array}{rl}2-4|2x+3|=-18&\text{Isolate the absolute value term by subtracting }2\text{ from each side} \\ -4|2x+3|=-20&\text{Divide each side by }-4 \\ |2x+3|=5&\text{Rewrite as two linear equations.} \\ 2x+3=5\quad\text{or}\quad 2x+3=-5\end{array}\nonumber\]
Solve each equation.
\[\begin{array}{lll}2x+3=5&&2x+3=-5 \\ 2x=2&\text{or}& 2x=-8 \\ x=1&&x=-4\end{array}\nonumber\]
We now have obtained two solutions, \(x = 1\) and \(x = −4\). Thus, the solution set is \(\{−4, 1\}\).
Equations with Two Absolute Values
In this case, we have an absolute value on each side of the equals sign. However, even though there are two absolute values, we apply the same process. Recall, methods never change, only problems.
Solve for \(m\): \(|2m-7|=|4m+6|\)
Solution
In order to apply the definition, we rewrite this equation as two linear equations, but with the left side as its positive and negative value: \[\begin{array}{rl}|2m-7|=|4m+6|&\text{Rewrite as two linear equations} \\ 2m-7=\color{blue}{4m+6}\color{black}{}\quad\text{or}\quad 2m-7=\color{blue}{-(4m+6)}\color{black}{}\end{array}\nonumber\]
Now, we can solve as usual. Be sure to distribute the negative for the equation on the right.
\[\begin{array}{lll} &&2m-7=\color{blue}{-}\color{black}{}(4m+6) \\ 2m-7=4m+6&&2m-7=\color{blue}{-}\color{black}{}4m\color{blue}{-}\color{black}{}6 \\ -13=2m&&6m-7=-6 \\ -\frac{13}{2}=m&\text{or}&6m=1 \\ &&m=\frac{1}{6}\end{array}\nonumber\]
Thus gives the solutions, \(m=-\frac{13}{2}\) or \(m=\frac{1}{6}\). Thus, the solution set is \(\left\{-\frac{13}{2},\frac{1}{6}\right\}\).
In Example \(\PageIndex{7}\), because there are absolute value expressions on both sides of the equation, we could have easily applied the definition to the left side and obtained \[\color{blue}{2m-7}\color{black}{}=4m-6\quad\text{or}\quad\color{blue}{-(2m-7)}\color{black}{}=4m-6\nonumber\]
Then solved each linear equation as usual and obtained the same results.
Special Cases
As we are solving absolute value equations, it is important to be aware of special cases. Remember, the result after evaluating absolute value must always be non-negative.
Solve for \(x\): \(7 + |2x − 5| = 4\)
Solution
\[\begin{array}{rl}7+|2x-5|=4&\text{Isolate the absolute value term by subtracting }7\text{ from each side} \\ |2x-5|=-3&X\text{ False}\end{array}\nonumber\]
Careful! Observe the absolute value of \(2x − 5\) is a negative number. This is impossible with absolute value because the result after evaluating absolute value must always be non-negative. Thus, we say this equation has no solution.
Absolute Value Equations Homework
Solve each equation.
\(|x| = 8\)
\(|b| = 1\)
\(|5 + 8a| = 53\)
\(|3k + 8| = 2\)
\(|9 + 7x| = 30\)
\(|8 + 6m| = 50\)
\(|6 − 2x| = 24\)
\(−7| − 3 − 3r| = −21\)
\(7| − 7x − 3| = 21\)
\(\frac{|−4b − 10|}{8} = 3\)
\(8|x + 7| − 3 = 5\)
\(5|3 + 7m| + 1 = 51\)
\(3 + 5|8 − 2x| = 63\)
\(|6b − 2| + 10 = 44\)
\(−7 + 8| − 7x − 3| = 73\)
\(|5x + 3| = |2x − 1|\)
\(|3x − 4| = |2x + 3|\)
\(\left|\frac{4x-2}{5}\right|=\left|\frac{6x+3}{2}\right|\)
\(|n| = 7\)
\(|x| = 2\)
\(|9n + 8| = 46\)
\(|3 − x| = 6\)
\(|5n + 7| = 23\)
\(|9p + 6| = 3\)
\(|3n − 2| = 7\)
\(|2 + 2b| + 1 = 3\)
\(\frac{|-4-3n|}{4}=2\)
\(8|5p + 8| − 5 = 11\)
\(3 − |6n + 7| = −40\)
\(4|r + 7| + 3 = 59\)
\(5 + 8| − 10n − 2| = 101\)
\(7|10v − 2| − 9 = 5\)
\(8|3 − 3n| − 5 = 91\)
\(|2 + 3x| = |4 − 2x|\)
\(\left|\frac{2x-5}{3}\right|=\left|\frac{3x+4}{2}\right|\)
\(\frac{|-n+6|}{6}=0\)