1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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 $2x-1=-(4x+3)$. $\endgroup$ – Nex Nov 5 '15 at 8:06
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.