Linear Equations with a Single Unknown: Variations
Equations with absolute values
We remind you that the absolute value of a real number #x# is defined by \[| x | = \begin{cases} \phantom{-}x & \mbox{ if } x \ge 0\\ -x & \mbox{ if } x\lt0\end{cases}\]
Below, the graphs of #y=|x + b|# and #y = c# are drawn. Use the sliders to explore the possible numbers of solutions to the equation #|x + b| = c# with unknown #x#. The solutions are the #x#-coordinates of the points lying on each of the two graphs.
Here we present a method to get rid of the absolute value from an equation.
Absolute values in an equation
An equation with unknown #x# in which an expression such as #|ax + b|# occurs, with real numbers #a\ne0# and #b#, can be solved in the following manner. Get rid of the absolute value by identifying two cases:
- #x\ge - \frac{b}{a}#: everywhere in the equation replace #|ax + b|# by #ax + b#.
- #x\lt - \frac{b}{a}#: everywhere in the equation replace #|ax + b|# by #-ax - b#.
Summarized in a formula, in which #\land# stands for "and", and #\lor# stands for "or":\[\begin{array}{rcl}{\Huge[}\left(x\ge - \frac{b}{a}\right)&\land& \text{equation with }|ax + b|\text{ replace by }ax + b{\Huge]}\\ &\lor&\\ {\Huge[}\left( x\lt - \frac{b}{a}\right)&\land& \text{equation with }|ax + b|\text{ replace by }-ax - b{\Huge]}\end{array}\]
For each of the two cases, solve the equation and check if the value of #x# of that solution satisfies the inequality.
The large rectangular brackets are made this big for sake of clarity. It signifies nothing more than a regular bracket.
As an example we will discuss solving the equation #|ax+b| + d = c#. First, we deal with the absolute value. The definition of #|ax+b|# gives that the equation becomes #ax+b+d= c# if #ax+b\ge0#, and #-ax-b +d=c# if #ax+b\lt 0#. This can be expressed as \[\begin{array}{c}\big[x\ge - \frac{b}{a} \text{ and } ax+b+d= c\big]\\ \text{ or }\\ \big[x\lt - \frac{b}{a}\text{ and } -ax-b+d= c\big]\end{array}\]This gives us the formula of the expression in our example if we write #\land# for "and" and #\lor# for "or".
The equation #|x + b| = c# is equivalent to \[x+b=c\lor x+b=-c\tiny.\]The two equations herein contained are linear and can hence be solved.
If #c\lt0#, then the left hand side of the equation #|x+b| = c# is non-negative and the right hand side is negative. This situation is impossible.
If #c\ge0#, then #|x+b|=c# is equivalent to\[\left(x+b=c\land x+b\ge0\right)\lor \left(x+b=-c\land x+b\lt0\right)\tiny.\]
But
- if #x+b=c#, then #x+b\ge0# is also satisfied and
- if #x+b=-c#, then #x+b\lt0# is also satisfied, unless #c=0#, a case we already dealt with because then #-c=-0=0=c#.
This means the original equation is equivalent to \[x+b=c\lor x+b=-c\tiny.\] Hence the expression is proven.
The method is even more general than the formulation: she can be used for every expression in #x# instead of #x+b# between the absolute value symbols. In the following examples we show this for expressions like #ax+b# and #\frac{ax+b}{cx+d}#.
Because getting rid of the absolute value gives \[\left(x-2=4\right) \vee \left( x-2=-4 \right)\tiny.\]
We will consider the two cases (before and after the #\vee#-sign) separately:
- The equation #x - 2= 4# has solution #x = 6#.
- The equation #x - 2= -4# has solution #x = -2#.
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