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Question Number 117027 by I want to learn more last updated on 09/Oct/20

$$\mathrm{solve}:\frac{\mathrm{x}-4}{\mathrm{x}-3}<\frac{2\mathrm{x}-1}{2}$$

Answered by bemath last updated on 09/Oct/20

$$\Rightarrow \frac{\mathrm{x}-4}{\mathrm{x}-3}-\frac{2\mathrm{x}-1}{2}<0$$ $$\Rightarrow \frac{2\mathrm{x}-8-(2\mathrm{x}-1)(\mathrm{x}-3)}{2(\mathrm{x}-3)}<0$$ $$\Rightarrow \frac{2\mathrm{x}-8-({2\mathrm{x}}^2-7\mathrm{x}+3)}{2(\mathrm{x}-3)}<0$$ $$\Rightarrow \frac{-{2\mathrm{x}}^2+9\mathrm{x}-11}{2(\mathrm{x}-3)}<0$$ $$\Rightarrow \frac{{2\mathrm{x}}^2-9\mathrm{x}+11}{2(\mathrm{x}-3)}>0$$ $$\Rightarrow \frac{{\mathrm{x}}^2-\frac{9}{2}\mathrm{x}+\frac{11}{2}}{\mathrm{x}-3}>0$$ $$\Rightarrow \frac{{(\mathrm{x}-\frac{9}{4})}^2-\frac{81}{16}+\frac{88}{16}}{\mathrm{x}-3}>0$$ $$\Rightarrow \frac{{(\mathrm{x}-\frac{9}{4})}^2+\frac{7}{16}}{\mathrm{x}-3}>0$$ $$\mathrm{the}\mathrm{solution}\mathrm{x}>3$$

Commented byI want to learn more last updated on 09/Oct/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by 1549442205PVT last updated on 09/Oct/20

$$\frac{\mathrm{x}-4}{\mathrm{x}-3}<\frac{2\mathrm{x}-1}{2}\iff \frac{\mathrm{x}-4}{\mathrm{x}-3}-\frac{2\mathrm{x}-1}{2}<0$$ $$\frac{2\mathrm{x}-8-({2\mathrm{x}}^2-7\mathrm{x}+3)}{2(\mathrm{x}-3)}<0$$ $$\iff \frac{-{2\mathrm{x}}^2+9\mathrm{x}-11}{\mathrm{x}-3}<0\iff \frac{{2\mathrm{x}}^2-9\mathrm{x}+11}{\mathrm{x}-3}>0\left(1\right)$$ $${2\mathrm{x}}^2-9\mathrm{x}+11=2{(\mathrm{x}-\frac{9}{4})}^2+\frac{7}{8}>0\mathrm{\forall }\mathrm{x}\in \mathrm{R}$$ $$\mathrm{Hence},\left(1\right)\iff \mathrm{x}-3>0\iff \mathrm{x}>3$$ $$\mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{inequality}\mathrm{has}\mathrm{set}\mathrm{of}$$ $$\mathrm{roots}\mathrm{is}\mathrm{S}=(3;+\mathrm{\infty })$$

Commented bybemath last updated on 09/Oct/20

$$\mathrm{gave}\mathrm{kudos}$$

Commented byI want to learn more last updated on 09/Oct/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by1549442205PVT last updated on 09/Oct/20

$$\mathrm{You}\mathrm{are}\mathrm{welcome}.$$

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