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I am trying to solve this equation $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ (specifically, find set of all points $(x, y)$ satisfying equation).

$\lvert 2x \rvert = \lvert x - 2 + y \rvert$ is equivalent to $2x = x - 2 + y \lor 2x = -(x - 2 + y)$.

If $2x = x - 2 + y$, then I am assuming that $2x \ge 0$ and $x - 2 + y \ge 0$, but do I have to check if solutions meet those conditions?

In general, while solving any equation with absolute values you have to check conditions, but sometimes you don't have to.

For example,

$$\lvert\lvert x-1\rvert +5\rvert = 7$$

I can write it as $\lvert x - 1\rvert + 5 = 7$ or $\lvert x - 1\rvert + 5 = -7$. In this case, I don't have to check conditons.

user128409235
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  • "If 2x=x-2+y then I am assume that 2x $\ge $ 0". No, no you are not making any such assumption. If |x| = |y| there are four possible equations. x=y; x=-y;-x=y and -x=-y. However x=y <=> -x=-y and x=-y <=> -y=x, so you only have to consider two of them. So solve x=y and x=-y. Now you have no idea whether x is positive or negative. But x=y will give you the exact same answers as solving -x=-y. – fleablood Mar 27 '16 at 01:08

3 Answers3

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It is obvious that $2x = x - 2 + y$ or $2x = -(x - 2 + y)$.

If I take $2x = x - 2 + y$, then I am assuming that $2x \ge 0$ and $x - 2 + y \ge 0$, but do I have to check if solutions meet those conditions?

No, you are not assuming that.   Rather when this is so: $2x>0 \wedge x-2+y >0\textsf{, OR }\\2x<0\wedge x-2+y <0\textsf{, OR }\\2x=0\wedge x-2+y=0$.

And of course, when $2x = -(x - 2 + y)$, then too: $2x>0 \wedge x-2+y <0\textsf{, OR }\\2x<0\wedge x-2+y >0\textsf{, OR }\\2x=0\wedge x-2+y=0$.

So, $y=x+2$ or $y=-3x+2$ , are the two solutions to $\lvert 2x\rvert =\lvert x-2+y\rvert$.   Each will meet one of those conditions.   No need to verify, as they must do so.

However:

$$\lvert\lvert x-1\rvert +5\rvert = 7$$

I can write it as $\lvert x - 1\rvert + 5 = 7$ or $\lvert x - 1\rvert + 5 = -7$. In this case, I don't have to check conditons.

No, but you may need to check that both your solutions do exist (or at least have real values).   You have $\lvert x - 1\rvert = 2$ or $\color{red}{\lvert x - 1\rvert = -12}$.

Community
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Graham Kemp
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It is obvious that $2x = x - 2 + y$ or $2x = -(x - 2 + y)$.

True, but is it obvious for you that $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ is equivalent to $2x =\pm( x - 2 + y)$ ?

We have $$\begin{align}|X|=|Y|&\iff |X|^2=|Y|^2\\&\iff X^2-Y^2=0\\&\iff (X-Y)(X+Y)=0\\&\iff X=\pm Y\end{align}$$

So, $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ is equivalent to $2x = \pm (x - 2 + y)$.

Hence, we can have $$\begin{align}\lvert 2x \rvert = \lvert x - 2 + y \rvert &\iff 2x = x - 2 + y\quad \text{or}\quad 2x = -(x - 2 + y)\\&\iff y=x+2\quad\text{or}\quad y=-3x+2\end{align}$$

You don't have to check if solutions meet the conditions.

mathlove
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The absolute value $\lvert x \rvert$ is defined as $$ \lvert x \rvert = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases} $$ So one has to treat two cases. Your example involves two such expressions, so up to $2\cdot 2 = 4$ cases can occur.

  1. $-,-$: $x < 0$, $x-2+y < 0 \iff y < 2 - x$. Equality condition is $-2x = -(x-2+y) \iff y = 2 + x$. Combining them gives the condition $2 + x < 2 - x \iff x < 0$. So this case gives the line segment $y = 2 + x$ for $x < 0$.

  2. $0+,0+$: $x\ge 0$, $x-2+y \ge 0 \iff y \ge 2 - x$. Equality condition is $y = 2 + x$ again. Combination: $2+x \ge 2-x \iff x \ge 0$. So the whole line $y = 2 + x$ is part of the solution.

  3. $-,0+$: $x<0$, $y \ge 2-x$. Equality: $-2x = x-2+y \iff y = 2-3x$. Combination: $2-3x \ge 2-x \iff 0 \ge 2x \iff x \le 0$. This case gives the line segment $y=2-3x$ for $x < 0$.

  4. $0+,-$: $x\ge 0$, $y<2-x$. Equality: $y=2-3x$. Combination: $2-3x<2-x\iff 0<2x \iff x > 0$. This gives the line segment $y=2-3x$ for $x > 0$.

Thus the solution are both lines $y = 2 + x$ and $y = 2-3x$.

Do you have to consider all cases like above? The alternatives shown here are to employ properties of $\lvert . \rvert$ to operate on a slightly more abstract level (mathlove's answer) or to manipulate the logical conditions in a different order (Graham's answer). Of the above I prefer that I know I have handled all possible cases.

mvw
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