$\frac{4 \mathrm{x}-3}{2 \mathrm{x}+1}=\frac{2 \mathrm{x}-1}{\mathrm{x}+4}$ A)$\{-1\}$ B)$\left\{\frac{15}{13}\right\}$ C)$\left\{\frac{11}{13}\right\}$ D)$\varnothing$

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The Answer of $\frac{4 \mathrm{x}-3}{2 \mathrm{x}+1}=\frac{2 \mathrm{x}-1}{\mathrm{x}+4}$ A)$\{-1\}$ B)$\left\{\frac{15}{13}\right\}$ C)$\left\{\frac{11}{13}\right\}$ D)$\varnothing$

$\frac{4 \mathrm{x}-3}{2 \mathrm{x}+1}=\frac{2 \mathrm{x}-1}{\mathrm{x}+4}$ A) $\{-1\}$ B) $\left\{\frac{15}{13}\right\}$ C) $\left\{\frac{11}{13}\right\}$ D) $\varnothing$

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4x−32x+1=2x−1x+4\frac{4 \mathrm{x}-3}{2 \mathrm{x}+1}=\frac{2 \mathrm{x}-1}{\mathrm{x}+4}2x+14x−3​=x+42x−1​ A) {−1}\{-1\}{−1} B) {1513}\left\{\frac{15}{13}\right\}{1315​} C) {1113}\left\{\frac{11}{13}\right\}{1311​} D) ∅\varnothing∅

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$\frac{4 \mathrm{x}-3}{2 \mathrm{x}+1}=\frac{2 \mathrm{x}-1}{\mathrm{x}+4}$ A) $\{-1\}$ B) $\left\{\frac{15}{13}\right\}$ C) $\left\{\frac{11}{13}\right\}$ D) $\varnothing$

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