Definition
Absolute value
The absolute value of a number is always positive or zero. For example, $|+5| = 5$ and $|-5| = 5$. Formal definition:
When $a$ is an expression that depends on $x$, its sign changes at the points where the expression vanishes — the critical points. These points tell us where we must "break" the equation.
General strategy
Simple case $\;|E(x)| = k\;$ (with $k > 0$ constant): no critical points are needed. It is solved directly by splitting into two cases:
General case $\;|E_1(x)| \pm |E_2(x)| \pm \dots = F(x)$: we set each argument equal to $0$ to find the critical points; these points split the real line into intervals. Within each interval, the arguments have a constant sign — so we can remove the absolute values (putting $+$ or $-$ according to the sign), solve the resulting linear equation and check that the value obtained belongs to the interval we were analysing.
Mandatory check
Every solution we find in an interval is valid only if it falls inside that same interval. Otherwise, it is an extraneous solution generated by the sign change and must be discarded.
Example 1 — Simple case $|E(x)| = k$
$|2x+1| = 5$
Since the right-hand side is constant and positive, we split into two cases without needing critical points:
Check: $\;|2\cdot 2 + 1| = |5| = 5$ ✓ and $|2\cdot(-3)+1| = |-5| = 5$ ✓.
Example 2 — Critical-points method: $|E(x)| = F(x)$
$|3x+6| = x+3$
We set the argument of the absolute value equal to $0$ to find the critical point:
This point splits the line into two intervals: $x < -2$ and $x > -2$.
Case 1 — if $x < -2$: the argument $3x+6$ is negative, so $|3x+6| = -(3x+6) = -3x-6$. The equation becomes:
Membership check: $-\tfrac{9}{4} = -2.25 < -2$ ✓. Valid solution.
Case 2 — if $x > -2$: the argument is positive, so $|3x+6| = 3x+6$. The equation becomes:
Check: $-\tfrac{3}{2} = -1.5 > -2$ ✓. Valid solution.
Check in the original equation:
$x = -\tfrac{9}{4}$: $\;|3(-\tfrac{9}{4})+6| = |-\tfrac{27}{4} + \tfrac{24}{4}| = \tfrac{3}{4}\,$ and $\,-\tfrac{9}{4} + 3 = \tfrac{3}{4}$ ✓.
$x = -\tfrac{3}{2}$: $\;|3(-\tfrac{3}{2})+6| = |\tfrac{3}{2}| = \tfrac{3}{2}\,$ and $\,-\tfrac{3}{2} + 3 = \tfrac{3}{2}$ ✓.
Example 3 — Several absolute values (4 intervals)
$|2x+2| - |3x-6| = |x-4|$
We have three absolute values. Setting each argument equal to $0$ we find the critical points:
These three points split the line into four intervals: $\;x < -1,\;\; -1 < x < 2,\;\; 2 < x < 4,\;\; x > 4$. We solve each case with the appropriate signs.
Case 1 — if $x < -1$: the three arguments are negative. The equation becomes:
But $x = 6 \not< -1$ → discard.
Case 2 — if $-1 < x < 2$: the first argument is positive, the other two negative:
$\tfrac{4}{3} \approx 1.33$ belongs to $(-1, 2)$ ✓. Valid solution.
Case 3 — if $2 < x < 4$: the first two positive, the last negative:
Impossible equation → no solution in this interval.
Case 4 — if $x > 4$: the three arguments positive:
$x = 6 > 4$ ✓. Valid solution.
Summary of the four cases:
| Interval | Linear equation | Candidate | Valid? |
|---|---|---|---|
| $x < -1$ | $x - 8 = -x + 4$ | $x = 6$ | no (doesn't satisfy $x<-1$) |
| $-1 < x < 2$ | $5x - 4 = -x + 4$ | $x = \tfrac{4}{3}$ | yes |
| $2 < x < 4$ | $0 \cdot x = -4$ | — | impossible |
| $x > 4$ | $-x + 8 = x - 4$ | $x = 6$ | yes |
Check in the original equation:
$x = \tfrac{4}{3}$: $\;|\tfrac{8}{3}+2| - |4-6| = \tfrac{14}{3} - 2 = \tfrac{8}{3}\,$ and $\,|\tfrac{4}{3}-4| = \tfrac{8}{3}$ ✓.
$x = 6$: $\;|14| - |12| = 14 - 12 = 2\,$ and $\,|6-4| = 2$ ✓.