To confirm this graphically,
you can look for the intersections of y_{1} = | x^{2} – 4x – 5 | and y_{2} = 7:

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What if your equation has
two absolute values? Then you will need to be more explicit about the
cases you are checking, but afterwards, the general approach will be the
same.

Solve | x – 3 | = | 3x + 2 | – 1

First I need to find the
break-points for each of these absolute values. Where do the arguments
(the expressions between the bars) of these absolute values switch from
being positive inside the bars to being negative? I'll look at each
absolute value separately.

| x – 3 | > 0 for x> 3

| 3x + 2 | > 0 for x> –2/3

How did I arrive at these
conclusions? I know that x – 3 = 0 at x =
3, and that the line y = x – 3 has a positive
slope and thus an increasing line. So x – 3 = 0 at x =
3, and x – 3 must be positive
after x =
3. Also, I know that 3x +
2 = 0 at x = –2/3, and that the
line y = 3x + 2 has a
positive slope and thus an increasing line. So 3x + 2 = 0 at x = –2/3, and 3x + 2 must be positive
after x = –2/3.

These points, x = –2/3 and x =
3, are where the absolute-value
expressions equal zero. Since these expressions must be negative or positive
for other x-values,
then these points divide the number line into intervals (before x = –2/3, between x = –2/3 and x =
3, and after x =
3), each of which should
be considered separately.

The zeroes of the two absolute-value
expressions give me three intervals: (–infinity,
–2/3),
(–2/3, 3), and (3, infinity).
On the first interval, both absolute-value expressions will have negative
values, so I'll need to change the signs on both of them when I
take the bars off.

This tells me that the
solution to the original equation, on the interval (–infinity,
–2/3),
is x = –3. Since x = –3 is contained
within this interval, this solution is valid.

On the second interval
(where x is
between –2/3 and 3),
the absolute value on the left-hand side of the equation, | x – 3 |, has
a negative argument; I'll have to change its sign when I take off the
bars. But the absolute value on the right-hand side of the equation, | 3x + 2 |, has a positive
argument, so I can just take the bars off.

This tells me that the
solution to the original equation, on the interval (–2/3,
3), is x = 1/2. Since 1/2 is
between –2/3 and 3,
this solution is valid.

On the third interval (where x is 3 or
more), both absolute values have positive arguments, so I can just take
the bars off.

| x – 3 | = | 3x + 2 | – 1 x – 3 = 3x + 2 – 1
–4
= 2x
–2
= x

However, on this interval, x was
already fixed as being greater than 3;
the solution then cannot be "x = –2", since –2 is actually less than 3.
So "x = –2" is not actually
a valid solution!

Then the answer is:

x = –3 or
x = ^{1}/_{2}

To confirm this graphically,
look for the intersections of y_{1} = | x – 3 | and y_{2} = | 3x + 2 | – 1:

You don't often need to take
different intervals into consideration— but sometimes you do. So make
sure you understand the last exercise above.

You can use the Mathway widget below to
practice solving an absolute-value equation. Try the entered exercise,
or type in your own exercise. Then click "Answer" to compare
your answer to Mathway's. (Or skip the widget and return to the index.)

(Clicking on "View Steps"
on the widget's answer screen will take you to the Mathway site, where
you can register for a free seven-day trial of the software.)