# What does it mean to have an absolute value equal an absolute value?

I have no problem reading absolute value equations such as $|x -2| = 2$.

I know this means that the distance of some real number is $2$ away from the origin. Because the origin splits the number line into a negative side and positive side then the numbers inside the absolute value symbol will be $2$ and $-2$, since those are the only two numbers $2$ units away from the origin. Then, it's just a matter of finding the values of $x$ which will give $2$ and $-2$ inside the absolute value.

Therefore, $|x - 2| = 2$ which is

$x - 2 = 2$

or

$x - 2 = -2$

And the solutions are $\{0, 4\}$

But when I see $|3x - 1| = |x + 5|$, I have no idea know what this means. I know how to solve it, but I don't know how this relates to the distance from the origin or how to interpret this on a number line. My initial interpretation is to say, "the absolute value of some unknown number is the absolute value of some unknown number," but that doesn't tell me the distance from $0$.

My Algebra textbook gave the following definition:

If $|u| = |v|$, then $u = v$ or $u = -v$.

But I can't really tell why this is the case.

• It might help to plot the equation $y = |3x -1| = 3|x-1/{3}|$, and then plot the equation $y=|x+5|$ on the same graph. Your equation is valid at the intersection point(s) of those two plots. Sep 10, 2018 at 23:25
• Maybe this will help: $|u|=|v|$ is equivalent to $u=v$ or $u=-v$ or $-u=v$ or $-u=-v$. But of course, the last two of these are redundant being equivalent to the first two. Sep 10, 2018 at 23:43
• The algebraic $definition$ of $|x|$ is that $|x|=x$ if $x\geq 0$ and $|x|=-x$ if $x<0$. In all cases, $|x|$ is the non-negative member of $\{x,-x\}.$ So if $|x|=|y|$ then one member of $\{x,-x\}$ is equal to one member of $\{y,-y\},$ which implies $(x=y\lor x=-y\lor -x=y\lor -x=-y)$, which is equivalent to $(x=y \lor x=- y)$. Conversely , if $x=\pm y,$ then the non-negative member of $\{x,-x\},$ which is $|x|,$ must equal the non-negative member of $\{y,-y\},$ which is $|y|.$ Sep 11, 2018 at 0:10

Any value $v$ is at distance $|v|$ from origin. Sometimes we are given the distance and are asked to find original value. When something ($\in \mathbb{R}$) is at distance $|w|$ from origin, it has a value either $w$ or $-w$.

As you said, $|x-2|=2$ means that $(x-2)$ is at distance $2$ from origin.

The same goes for the example that confuses you; $$|3x-1|=|x+5|$$ means that $(3x-1)$ is at distance $|x+5|$ from origin.

And what can we conclude from this? That the value of $(3x-1)$ is either $(x+5)$ or $-(x+5)$, and that is what your textbook says using $u$ and $v$.

[Also, you can flip it and say that $(x+5)$ is at distance $|3x-1|$ from origin, and those are the redundant cases you see mentioned in other answers.]

And to directly refer to your title question; Having two absolute values equal means, in terms of distance from origin, that they are both equaly far from origin. So $$|u|=|v|$$ means that $u$ and $v$ are equaly far from origin. How far specificaly? Exactly $|u|$ (or $|v|$, because they are equal).

• Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, {0, 4}, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is {-1, 3}. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line? Sep 11, 2018 at 1:48
• @Slecker As opposed to the simpler equation of the form $|u| = 2$ that asks you “which values of $x$ cause $u$ to be 2 away from the origin”, you have an equation $|u|=|v|$ and the question is “which values of $x$ cause $u$ to be as far away from the origin as $v$ is”. $x=-1$ causes $u=-4$ and $v=4$ which fits (both are 4 away from the origin, on different sides), and $x=3$ results in $u=v=8$ (both are obviously 8 away from the origin on the same side, on the same exact point). Sep 11, 2018 at 2:48
• @Roman Aha! Thank you very much for that last bit of insight: “|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head! Sep 11, 2018 at 3:04

In real line, when do two points $u$ and $v$ have the same absolute value (the same distance away from the origin)?

If you’re having trouble imagining what the equation might represent, consider two objects moving along the number line. Let the coordinate of one of those change with time as $3t - 1$ and the other one as $t + 5$ (this would imply that at the moment of time $t=0$, presumably when you started observing the situation, one was at $-1$ and was moving at a constant speed of 3 units of distance per unit of time in the positive direction, while the second one was at the point 5 and its speed was 1).

Now you’re asked the question, at what time one was as far away from the origin as the other one? (Maybe you’re a spy trying to determine the timestamp of some event, but all you know is that the signal from two ships or planes or whatever was equally strong at that point.) That’s what your equation $$|3t - 1| = |t + 5|$$ says. Intuitively, because the first object’s velocity is greater but at $t=0$ it’s to the left of the second object, this would happen two times: once in the past, when the origin was exactly between the objects, and once in the future, when one object overtakes the other. The equation $3t - 1 = -(t + 5)$ corresponds to the former case (coordinates are $-4$ and $4$ respectively), and $3t-1 = t+5$ to the latter (both objects at $8$).

• That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful. Sep 11, 2018 at 3:19

if $|u| = |v|$ there are actually four possibilities, but they are two redundant pairs.

The four possibilities are.

A) $u \ge 0; v\ge 0$ and therefore $u = |u|; v=|v|=|u| = u$ and RESULT 1) $u = v$.

B) $u < 0; v < 0$ and therefore $u = -|u|; v = -|v|=-|u| = u$ and RESULT 1) $u = v$. (that's redudant.)

C) $u\ge 0; v < 0$ and therefore $u = |u|; v = -|v|=-|u|=-u$ and RESULT 2) $u = -v$

D) $u < 0; v \ge 0$ and therefore $u = -|u|; v = |v| = |u| = -u$ and RESULT 2) $u = -v$. (that's reducant).

Now if $|3x -1| = |x+5|$ we could solve by doing all four cases but that is unnecessary as we will get redundant and contradictory results.

Let's do it to see what happens and see if we can learn from it:

A) $3x - 1 \ge 0$ and $x + 5 \ge 0$ and therefore $3x-1 = x + 5$.

In other words $3x \ge 1$ and $x \ge -5$ and $2x = 6$

In other words $x \ge \frac 13$ and $x \ge -5$ and $x = 3$.

Or in other words $x = 3$.

B) $3x -1 < 0$ and $x + 5 < 0$ and therefore $3x -1 =x+5$.

Or $x < \frac 13$ and $x < -5$ and $x =3$.

That's a contradiction. Notice there was no reason to consider the two case whether $3x -1$ and $x + 5$ are greater or less than $0$. That was just a waste of time. It would have been just as well to only consider that $3x - 1= x+5\implies x = 3$. That's all we had to do.

C) $3x - 1< 0$ and $x + 5 \ge 0$ and therefore $3x -1 = -x -5$

Or $x < \frac 13$ and $x \ge -5$ and $4x = -4\implies x = -1$. So $x =-1$ is possible.

D) $3x -1 \ge 0$ and $x + 5 < 0$ and therefore $3x-1 = -x -5$

Or $x \ge \frac 13$ and $x < -5$ and $4x = -4\implies x = -1$. That's a contradiction.

But again we didn't have to do both C) and D). That was redundant.

It's enough to know that if $|3x -1| = |x+5|$ then either $3x - 1 = x+5$ (and we don't care if they are both positive or both negative-- we can figure that out later) or $3x - 1 = -x -5$ (ditto).

If $|u| = |v|$, then $u = v$ or $u = -v$.

An easy way to understand why this is true is to square both sides of the equation. The statement $|u|=|v|$ is the same statement as $u^2=v^2$ (since $\sqrt{x^2}=|x|$ for any $x$). Rearrange this to $u^2-v^2=0$, then factor the LHS to obtain $(u-v)(u+v)=0$. Conclude that the original statement $|u|=|v|$ is equivalent to the statement $u-v=0$ or $u+v=0$.

You can similarly solve $|3x-1|=|x+5|$ by squaring both sides.