Absolute Value Equations

The absolute value of a number is defined by

$$|c| = \cases{+c & if $c \ge 0$ \cr -c & if $c < 0$ \cr}.$$

Suppose you want to solve an absolute value equation of the form

$$|(\hbox{stuff})| = (\hbox{a number}).$$

Replace $|(\hbox{stuff})|$ with $\pm (\hbox{stuff})$ , then take cases.


Example. Solve $|x| = 7$ .

Replace $|x|$ with $\pm x$ :

$$\matrix{ & & \pm x = 7 & & \cr & \swarrow & & \searrow & \cr x = 7 & & & & -x = 7 \cr x = 7 & & & & x = -7 \cr}$$

The solutions are $x =
   7$ and $x = -7$ .

It's often easier to put the "$\pm$ " on the side with the number rather than the side with the variable, this way:

$$\matrix{ & & x = \pm 7 & & \cr & \swarrow & & \searrow & \cr x = 7 & & & & x = -7 \cr}$$

This is the approach I'll use in the examples that follow.


Example. Solve $|x - 3| = 2$ .

Remove the absolute values and put a "$\pm$ " on the "2". Then take cases and solve.

$$\matrix{ & & x - 3 = \pm 2 & & \cr & \swarrow & & \searrow & \cr x - 3 = 2 & & & & x - 3 = -2 \cr x = 5 & & & & x = 1 \cr}$$

The solutions are $x =
   5$ and $x = 1$ .


Example. Solve $|x + 4| = 6$ .

Remove the absolute values and put a "$\pm$ " on the "6". Then take cases and solve.

$$\matrix{ & & x + 4 = \pm 6 & & \cr & \swarrow & & \searrow & \cr x + 4 = 6 & & & & x + 4 = -6 \cr x = 2 & & & & x = -10 \cr}$$

The solutions are $x =
   2$ and $x = -10$ .


Example. Solve $|2x - 3| = 5$ .

Remove the absolute values and put a "$\pm$ " on the "5". Then take cases and solve.

$$\matrix{ & & 2x - 3 = \pm 5 & & \cr & \swarrow & & \searrow & \cr 2x - 3 = 5 & & & & 2x - 3 = -5 \cr 2x = 8 & & & & 2x = -2 \cr x = 4 & & & & x = -1 \cr}$$

The solutions are $x =
   4$ and $x = -1$ .


Example. Solve $|x - 7| = -8$ .

If you work this problem like the others, you'll get two answers, but they won't be right.

The equation says an absolute value ($|x - 7|$ ) is negative (-8). Since an absolute value can't be negative, the equation has no solutions.


*Example. Solve $|x| = |x - 4| + 9$ .

Since I have two absolute value expressions, I'll go back to my original procedure: Remove the absolute values from an expression and put "$\pm$ " on it. Doing so, I get

$$\pm x = \pm(x - 4) + 9.$$

Now I have 4 cases:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & & & x & & $x - 4$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & case 1 & & + & & + & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & case 2 & & + & & - & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & case 3 & & - & & + & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & case 4 & & - & & - & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Case 1.

$$\eqalign{ x &= (x - 4) + 9 \cr 0 &= 5 \cr}$$

Since this is a contradiction, this case doesn't give any solutions.

Case 2.

$$\eqalign{ x &= -(x - 4) + 9 \cr x &= -x + 4 + 9 \cr 2x &= 13 \cr \noalign{\vskip2pt} x &= \dfrac{13}{2} \cr}$$

Case 3.

$$\eqalign{ -x &= (x - 4) + 9 \cr -x &= x + 5 \cr -2x &= 5 \cr \noalign{\vskip2pt} x &= -\dfrac{5}{2} \cr}$$

Case 4.

$$\eqalign{ -x &= -(x - 4) + 9 \cr x &= -x + 4 + 9 \cr 0 &= 13 \cr}$$

Since this is a contradiction, this case doesn't give any solutions.

The solutions are $x =
   \dfrac{13}{2}$ and $x = -\dfrac{5}{2}$ .


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