$|4 \mathrm{x}-7|=|4 \mathrm{x}+6|$ A)$\left\{-\frac{1}{8}, \frac{1}{8}\right\}$ B)$\left\{\frac{13}{8}\right\}$ C)$\left\{\frac{1}{8}\right\}$ D)$\varnothing$

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The Answer of $|4 \mathrm{x}-7|=|4 \mathrm{x}+6|$ A)$\left\{-\frac{1}{8}, \frac{1}{8}\right\}$ B)$\left\{\frac{13}{8}\right\}$ C)$\left\{\frac{1}{8}\right\}$ D)$\varnothing$

$|4 \mathrm{x}-7|=|4 \mathrm{x}+6|$ A) $\left\{-\frac{1}{8}, \frac{1}{8}\right\}$ B) $\left\{\frac{13}{8}\right\}$ C) $\left\{\frac{1}{8}\right\}$ D) $\varnothing$

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∣4x−7∣=∣4x+6∣|4 \mathrm{x}-7|=|4 \mathrm{x}+6|∣4x−7∣=∣4x+6∣ A) {−18,18}\left\{-\frac{1}{8}, \frac{1}{8}\right\}{−81​,81​} B) {138}\left\{\frac{13}{8}\right\}{813​} C) {18}\left\{\frac{1}{8}\right\}{81​} D) ∅\varnothing∅

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$|4 \mathrm{x}-7|=|4 \mathrm{x}+6|$ A) $\left\{-\frac{1}{8}, \frac{1}{8}\right\}$ B) $\left\{\frac{13}{8}\right\}$ C) $\left\{\frac{1}{8}\right\}$ D) $\varnothing$

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