Which One Doesn't Belong?
Source: wodb.ca
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Previously in this unit, we solved linear equations. We will extend this topic to solve absolute value equations.
The definition of absolute value is the distance a value is from zero on the number line. Because absolute value is a distance, the absolute value of a number is always positive.
For example, \(\left|-3\right|=3\ \text {and} \left|3\right|=3\) because both \(-3\) and \(3\) are \(3\) units from zero on the number line as illustrated below.
When solving an absolute value equation, we are trying to determine the value or values of \(x\) so that the entire expression inside the absolute value symbol is a certain distance from zero on the number line. Since you can move in two directions from zero (left, which is negative, and right, which is positive), it is necessary to set up two different equations to solve.
Before solving absolute value equations, explore the interactive graph to determine how many solutions are possible in an equation of the form \(\left|x\right|=a\). Solutions occur at the values of \(x\) where the two graphs intersect.
From you exploration, you should have noticed that you can get exactly two solutions, exactly one solution, or no solutions to an absolute value equation.
Steps for solving an absolute value equation
1. Isolate the absolute value expression by using inverse operations.
a) Eliminate any addition or subtraction outside the absolute value symbol.
b) Eliminate any multiplication or division outside the absolute value symbol (do NOT distribute a number outside the absolute value to the terms inside the absolute value).
c) If the isolated absolute value equals a negative number, there is no solution because absolute value cannot be negative.
2. Set up two equations.
a) Drop the absolute value symbols when you set up the two equations.
b) Do NOT change the expression that is inside the absolute value symbol.
c) Set the expression equal to a negative value and equal to a positive value.
3. Solve both equations.
4. Check both solutions in the ORIGINAL problem.
Example 1: Solve the absolute value equation \(3\left|2x-1\right|-6=12\).
\(\begin{align}3\left|2x-1\right|-6&=12 \ & \ &\text{1) Rewrite the equation.}\\3\left|2x-1\right|&=18 \ & \ &\text{2) Add 6 to both sides to isolate the absolute value.}\\\left|2x-1\right|&=6 \ & \ &\text{3) Divide both sides by 3 to isolate the absolute value.}\\2x-1=-6\ &\text {or}\ 2x-1=6 \ & \ &\text{4) Set up two equations.}\\2x=-5\ &\text {or}\ 2x=7 \ & \ &\text{5) Add 1 to both sides (in each equation).}\\x=-\frac{5}{2}\ &\text {or}\ x=\frac{7}{2} \ & \ &\text{6) Divide both sides by 2 (in each equation).}\end{align}\)
Don’t forget to check your solutions in the original equation to make sure both sides are equal! The check is shown below.
\( \begin{align} 3\left|2\left(-\frac{5}{2}\right)-1\right|-6=12&\ \text {and}\ 3\left|2\left(\frac{7}{2}\right)-1\right|-6=12 \\ 3\left|-5-1\right|-6=12&\ \text {and}\ 3\left|7-1\right|-6=12\\3\left|-6\right|-6=12&\ \text {and}\ 3\left|6\right|-6=12 \\ 3\left(6\right)-6=12&\ \text {and}\ 3\left(6\right)-6=12 \\ 18-6=12&\ \text {and}\ 18-6=12 \\ 12=12&\ \text {and}\ 12=12 \ \ \end{align} \)
\(x=-\frac{5}{2}\) and \( x=\frac{7}{2}\) are both solutions!
Example 2: Solve the absolute value equation \(-\frac{3}{2}\left|x-7\right|+1=-8\).
\( \begin{align} 3\left|2\left(-\frac{5}{2}\right)-1\right|-6=12&\ \text {and}\ 3\left|2\left(\frac{7}{2}\right)-1\right|-6=12 \\ 3\left|-5-1\right|-6=12&\ \text {and}\ 3\left|7-1\right|-6=12\\3\left|-6\right|-6=12&\ \text {and}\ 3\left|6\right|-6=12 \\ 3\left(6\right)-6=12&\ \text {and}\ 3\left(6\right)-6=12 \\ 18-6=12&\ \text {and}\ 18-6=12 \\ 12=12&\ \text {and}\ 12=12 \ \ \end{align} \)
\(x=-\frac{5}{2}\) and \( x=\frac{7}{2}\) are both solutions!
Example 2: Solve the absolute value equation \(-\frac{3}{2}\left|x-7\right|+1=-8\).
We can also solve absolute value equations that have variables on both sides of the equal sign. Explore the interactive graph to determine how many solutions are possible in an equation of the form \(\left|x\right|=mx+b\). Solutions occur at the values of x where the two graphs intersect.
From the exploration, you should have noticed that there can be two solutions, one solution or no solutions to these types of problems.
When determining the number of solutions to \(|ax+b|=cx\), you can compare the slopes. Consider setting each side of the equation equal to \(y\) to create two functions so that you could visualize the intersection(s) in order to find solutions as we did in the explore above. This would give you \(y=|ax+b|\) and \(y=cx\). If \(a = c\), then the line \(y = cx\) is parallel to one branch of the absolute value graph, so there will be only one solution.
The process to solve these types of equations is the same as if the equation is set equal to a number. However, in these types of problems we often will have an extraneous solution. An extraneous solution is an algebraic solution that will not check when substituted back into the original equation.
Example 3: Solve the absolute value equation \(\left|5x-3\right|=2x-7\).
\(\begin{align}\left|5x-3\right|&=2x-7 \ & \ &\text{1) Rewrite the original equation}\\5x-3=-\left(2x-7\right)\ &\text {or}\ 5x-3=2x-7 \ & \ &\text{2) Set up two equations}\\5x-3=-2x+7\ &\text {or}\ 5x-3=2x-7 \ & \ &\text{3) Distribute the negative sign}\\7x=10\ &\text{or}\ 3x=-4 \ & \ &\text{4) Group like terms on same side of equal sign}\\x=\frac{10}{7}\ &\text{or}\ x=-\frac{4}{3} \ & \ &\text{5) Solve for x}\end{align}\)
Now it is time to check the solutions!
\(\begin{align}\left|5\left(\frac{10}{7}-3\right)\right|&=2\left(\frac{10}{7}\right)-7\\\left|\frac{50}{7}-\frac{21}{7}\right|&=\frac{20}{7}-\frac{49}{7}\\\left|\frac{29}{7}\right|&=-\frac{29}{7}\\\frac{29}{7}&\ne-\frac{29}{7}\end{align}\)
So \(x=\frac{10}{7}\) is NOT a solution. It is an extraneous solution.
\(\begin{align}\left|5\left(\frac{-4}{3}\right)-3\right|&=2\left(\frac{-4}{3}\right)-7\\\left|-\frac{20}{3}-\frac{9}{3}\right|&=-\frac{8}{3}-\frac{21}{3}\\\left|-\frac{29}{3}\right|&=-\frac{29}{3}\\\frac{29}{3}&\ne-\frac{29}{3}\end{align}\)
So, \(x=-\frac{4}{3}\) is NOT a solution. It is an extraneous solution.
There is no solution to \(\left|5x-3\right|=2x-7\) as illustrated in the graph.
\(\begin{align}\left|5\left(\frac{10}{7}-3\right)\right|&=2\left(\frac{10}{7}\right)-7\\\left|\frac{50}{7}-\frac{21}{7}\right|&=\frac{20}{7}-\frac{49}{7}\\\left|\frac{29}{7}\right|&=-\frac{29}{7}\\\frac{29}{7}&\ne-\frac{29}{7}\end{align}\)
So \(x=\frac{10}{7}\) is NOT a solution. It is an extraneous solution.
\(\begin{align}\left|5\left(\frac{-4}{3}\right)-3\right|&=2\left(\frac{-4}{3}\right)-7\\\left|-\frac{20}{3}-\frac{9}{3}\right|&=-\frac{8}{3}-\frac{21}{3}\\\left|-\frac{29}{3}\right|&=-\frac{29}{3}\\\frac{29}{3}&\ne-\frac{29}{3}\end{align}\)
So, \(x=-\frac{4}{3}\) is NOT a solution. It is an extraneous solution.
There is no solution to \(\left|5x-3\right|=2x-7\) as illustrated in the graph.
Example 4: Solve the absolute value equation \(\left|3x+1\right|=5x-4\).