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Given the equation $$\lvert2x\rvert=x-1,$$ rewriting the left-hand side as $\pm2x$ results in the unique candidate solution set $\left\{-1,\frac13\right\}$ that nonetheless fails to satisfy the equation:

\begin{align}\forall x\in\mathbb R \Bigg[\quad\quad&x\in\emptyset\\\iff&\bigg(x<0 \,\text{ and}\,-2x=x-1\bigg) \:\text{ or }\: \bigg(x\geq0 \,\text{ and }\, 2x=x-1\bigg)\\\iff&\lvert2x\rvert=x-1\\\implies&\pm2x=x-1\\\iff &x\in\left\{-1,\frac13\right\}\quad\Bigg].\end{align}

Due to deductive explosion, performing a valid operation on an inconsistent equation, which has no solution, has given rise to an extraneous solution set.

Given the equation $$\lvert2x\rvert=x-1,$$ rewriting the left-hand side as $\pm2x$ results in the unique candidate solution set $\left\{-1,\frac13\right\}$ that nonetheless fails to satisfy the equation:

\begin{align}\forall x\in\mathbb R \Bigg[\quad\quad\quad&x\in\emptyset\\\iff{}&\bigg(x<0 \,\text{ and}\,-2x=x-1\bigg) \:\text{ or }\: \bigg(x\geq0 \,\text{ and }\, 2x=x-1\bigg)\\\iff{}&\lvert2x\rvert=x-1\\\implies{}&\pm2x=x-1\\\iff {}&x\in\left\{-1,\frac13\right\}\quad\Bigg].\end{align}

Due to deductive explosion, performing a valid operation on an inconsistent equation, which has no solution, has given rise to an extraneous solution set.

Given the equation $$\lvert2x\rvert=x-1,$$ rewriting the left-hand side as $\pm2x$ results in the unique candidate solution set $\{-1,\frac13\}$ that nonetheless fails to satisfy the equation:

\begin{align}&0=1\\\iff&\bigg(x<0 \,\text{ and}\,-2x=x-1\bigg) \:\text{ or }\: \bigg(x\geq0 \,\text{ and }\, 2x=x-1\bigg)\\\iff&\lvert2x\rvert=x-1\\\implies&\pm2x=x-1\\\iff &x=-1 \;\textbf{ or }\; \frac13.\end{align}

Due to deductive explosion, performing a valid operation on an inconsistent equation, which has no solution, has given rise to an extraneous solution set.

Given the equation $$\lvert2x\rvert=x-1,$$ rewriting the left-hand side as $\pm2x$ results in the unique candidate solution set $\left\{-1,\frac13\right\}$ that nonetheless fails to satisfy the equation:

\begin{align}\forall x\in\mathbb R \Bigg[\quad\quad&x\in\emptyset\\\iff&\bigg(x<0 \,\text{ and}\,-2x=x-1\bigg) \:\text{ or }\: \bigg(x\geq0 \,\text{ and }\, 2x=x-1\bigg)\\\iff&\lvert2x\rvert=x-1\\\implies&\pm2x=x-1\\\iff &x\in\left\{-1,\frac13\right\}\quad\Bigg].\end{align}

Due to deductive explosion, performing a valid operation on an inconsistent equation, which has no solution, has given rise to an extraneous solution set.

Given the equation $$\lvert2x\rvert=x-1,$$ rewriting the left-hand side as $\pm2x$ results in the unique candidate solution set $\{-1,\frac13\}$ that nonetheless fails to satisfy the equation:

\begin{align}&x\in\emptyset\\\iff&\bigg(x<0 \,\text{ and}\,-2x=x-1\bigg) \:\text{ or }\: \bigg(x\geq0 \,\text{ and }\, 2x=x-1\bigg)\\\iff&\lvert2x\rvert=x-1\\\implies&\pm2x=x-1\\\iff &x=-1 \;\textbf{ or }\; \frac13.\end{align}

Due to deductive explosion, performing a valid operation on an inconsistent equation, which has no solution, has given rise to an extraneous solution set.

Given the equation $$\lvert2x\rvert=x-1,$$ rewriting the left-hand side as $\pm2x$ results in the unique candidate solution set $\{-1,\frac13\}$ that nonetheless fails to satisfy the equation:

\begin{align}&0=1\\\iff&\bigg(x<0 \,\text{ and}\,-2x=x-1\bigg) \:\text{ or }\: \bigg(x\geq0 \,\text{ and }\, 2x=x-1\bigg)\\\iff&\lvert2x\rvert=x-1\\\implies&\pm2x=x-1\\\iff &x=-1 \;\textbf{ or }\; \frac13.\end{align}

Due to deductive explosion, performing a valid operation on an inconsistent equation, which has no solution, has given rise to an extraneous solution set.