# When solving an equation with absolute value on both sides, how to choose the side to work with?

When solving an equation with absolute value on both sides, such as $$|2x-1|=|4x+3|$$ how to choose one side of which to use the definition of absolute value?

For example, if we apply absolute value just for right side so we have $2x-1=-(4x+3)$ and $2x-1=(4x+3)$. But why we don't solve for left hand $-(2x-1)=4x+3$ and $(2x-1)=4x+3$?

I know how to solve these but I don't understand the logic.

• You did do "both sides". More precisely $|2x-1|=|4x+3|$ if and only if $2x-1= |4x+3|$ or $2x-1=-|4x+3|$ if and only $2x-1=4x+3$ or $-(2x-1)=4x+3$ or $-(2x-1)=4x+3$ or $(2x-1)=4x+3$. However this equivalent to $2x-1=4x+3$ or $2x-1=-(4x+3)$. – Nex Nov 5 '15 at 8:06

## 2 Answers

The logic is that, faced with something like $\lvert X \rvert = \lvert Y \rvert$ where $X$ and $Y$ are any expressions, we know that $X$ and $Y$ are both the same distance from $0$.

In other words, if they have the same absolute value, then either they're the same so that $X = Y$ or they're negatives of each other (think $\lvert 3 \rvert = \lvert -3\rvert$) so that $X = -Y$, and both cases are perfectly valid.

But multiplying each of the above equations by $-1$, we have a total of four equations

\begin{align*} X = Y\quad &\text{ is equivalent to} \quad-X = -Y \\ X = -Y \quad &\text{ is equivalent to}\quad -X = Y \end{align*}

that form two pairs: As long as you pick one equation from each set of equivalent pairs, you'll get all possible solutions. With your method, whether we choose $X = -Y$ or $-X = Y$ depends on who we keep the absolute value with:

$|X| = Y$ gives $X = Y$ or $X = -Y$, while $X = |Y|$ gives $X = Y$ or $-X = Y$.

But remember, multiplying an equation by a negative doesn't change the solutions at all, so either one is perfectly good.

The argument of the LHS module changes sign at $x=\frac{1}{2}$. The argument in RHS changes sign at $x=-\frac{3}{4}$, so we have three systems: $$\begin{cases} x<-\frac{3}{4}\\ 1-2x=-(4x+3) \end{cases} \quad \lor \quad \begin{cases} -\frac{3}{4}\le x<\frac{1}{2}\\ 1-2x=4x+3 \end{cases} \quad \lor \quad \begin{cases} x\ge\frac{1}{2}\\ 2x-1=4x+3 \end{cases}$$ can you solve these?