# Double sided modulus equation: $|2x-1| = |3x+2|$

Considering this equation $$|2x-1| = |3x+2|$$ My question is what is the reasoning behind taking only the positive and negative values of one side of the equation to solve the problem. I have seen many people do that and it seems to work. For e.g., $$2x-1 = 3x+2$$ $$2x-1 = -(3x+2),$$ then solve for $x$.On the other hand, I have been taught that I should be testing both positive and negative values on both sides which will eventually give 4 cases to solve instead of 2. This makes more sense since disregarding the absolute value sign give rise to the situation, a negative and positive value.

• The equations $2x-1 = 3x+2$ and $-(2x-1) = -(3x+2)$ have the same solutions. It might look like you have four cases, but two pairs of them are identical, so you're just duplicating your work. – Billy Feb 22 '18 at 23:31

You are correct that there are four possibilities with

$|a| =|b|$

So $\pm a = \pm b$

1) $a = b$.

2) $a = -b$

3) $-a = b$

4) $-a = -b$

But notice

A) $a = b \iff -a = -b$. So 1) and 4) are the exact same thing!

and

B) $a = -b$ \iff -a = b$so 2) and 3) are the exact same thing. So for$|2x-1| = |3x+2|$and$\pm(2x-1) = \pm(3x+2)$there are two options: A)$2x-1 = 3x+ 2$(which would also mean$-(2x-1) = -(3x +2)$.) or B)$2x - 1 = -(3x+2)$(which would also mean$-(2x-1) = 3x + 2$.) ...... In other words$|a| = |b| \iff \pm a = \pm b \iff a = \pm b \iff \pm a = b$. But in a comment above Yolo Inver raises a good point about inequalities:$|a| < |b| \implies -|b| < -|a| \le 0 \le |a| < |b| \implies$four possibilities:$-b< -a \le 0 \le a < b$OR$b < a \le 0 \le a < b$OR$-b < a \le 0 \le -a < b$OR$b< -a \le 0 \le a < -b$. For inequalities there are$4$cases to consider. In this example you actually can reduce this to two cases. Either the expressions inside the absolute values share the same sign, or they have opposite signs. If you assume the left is negative and the right is positive.$-(2x-1) = 3x + 2$And the case the the right is positive and the left is negative.$2x -1 = -(3x+2)$You will get the same solutions. However, for one case, the results are in contradiction to the assumption. If you have a more if you added in something not inside the absolute value brakets... e.g.$|x+1| - |2x-1| = 1$You should be testing all 4 cases. Or$x<-1, -1<x<\frac 12, \frac 12<x$will suffice. In my opinion, this workaround is much easier the case for inequalities, and would be squaring both sides to eliminate the negative. This also helps avoiding the individual$4$cases. $$|2x-1|^2 = |3x+2|^2$$ $$4x^2-4x+1=9x^2+12x+4$$ $$5x^2+16x+3=0$$ Thus, (spoiler below) $$(5x+1)(x+3)=0 \implies x=-3, -\frac 15$$ • If you are eliminating the negatives, doesnt it mean that you are also eliminating possible solutions for x? – Rayne zick Feb 22 '18 at 23:29 • I don't get what you mean @Raynezick. Take:$$-x-4=0$$ If you multiply both sides by$-1$, you get $$x+4=0$$ How are you "eliminating possible solutions"? – user532449 Feb 22 '18 at 23:31 • Ok now i see it – Rayne zick Feb 22 '18 at 23:36 • @rayne Please consider marking it correct if you think it is correct. – user532449 Feb 22 '18 at 23:38 • You're kidding, right? Who on earth would think solving a quadratic is simpler the taking cases.$2x -1 =3x+2\implies x = -3$and$2x - 1 = 2-3x\implies x = 5$. Obviously you made some error if you got an irrational solution to linear integer equations. – fleablood Feb 22 '18 at 23:40 There are actually just two cases. Case I: $$2x - 1 = 3x + 2$$ yields$x = -3$. Now for case II: $$2x - 1 = -3x - 2$$ yields$x = -1/5\$.