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Question Number 80687 by Power last updated on 05/Feb/20

Commented by Power last updated on 05/Feb/20

a)(1/6) b)(1/3) c)(1/2) d)(1/4)

$$\left.\mathrm{a}\left.\right)\left.\frac{\mathrm{1}}{\mathrm{6}}\left.\:\:\:\:\:\mathrm{b}\right)\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\mathrm{c}\right)\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\mathrm{d}\right)\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by MJS last updated on 05/Feb/20

((x−1)/(2x))=(1/6) ⇒ x=(3/2) ((x−1)/(2x))=(1/3) ⇒ x=3 ((x−1)/(2x))=(1/2) ⇒ no solution ((x−1)/(2x))=(1/4) ⇒ x=2 but for x∈{(3/2), 2, 3} the given equation is wrong the only solution is x≈1.06427 ⇒ ((x−1)/(2x))≈.0301927

$$\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow\:{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{x}=\mathrm{3} \\ $$$$\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:{x}=\mathrm{2} \\ $$$$\mathrm{but}\:\mathrm{for}\:{x}\in\left\{\frac{\mathrm{3}}{\mathrm{2}},\:\mathrm{2},\:\mathrm{3}\right\}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\:{x}\approx\mathrm{1}.\mathrm{06427}\:\Rightarrow\:\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\approx.\mathrm{0301927} \\ $$

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