Question Number 222172 by hardmath last updated on 19/Jun/25
$$\frac{\mathrm{4}}{\mathrm{5}}\:>\:\frac{\mathrm{8}}{\mathrm{3x}\:−\:\mathrm{6}}\:>\:\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$
Answered by A5T last updated on 19/Jun/25
![(4/5)>(8/(3x−6)) and 3x−6>0 (x>2)⇒ 12x−24>40 ⇒x>((16)/3) ... (i) (4/5)>(8/(3x−6)) and 3x−6<0 (x<2) ⇒ 24−12x>−40 ⇒x<((16)/3) but x<2 ⇒ x<2 ...(ii) (i)∪(ii) = x<2 ∪ x>((16)/3) (8/(3x−6))>(2/9) and 3x−6>0(x>2) ⇒ 4×3>x−2⇒x<14 but x>2 ⇒ 2<x<14...(iii) (8/(3x−6))>(2/9) and 3x−6<0 (x<2) ⇒−4×3>4−2x⇒x>14 but x<2 ⇒ x∈∅...(iv) (iii)∪(iv)= 2<x<14 [(i)∪(ii)]∩[(iii)∪(iv)] = ((16)/3)<x<14 ⇒x∈ (((16)/3),14)](https://www.tinkutara.com/question/Q222178.png)
$$\frac{\mathrm{4}}{\mathrm{5}}>\frac{\mathrm{8}}{\mathrm{3x}−\mathrm{6}}\:\mathrm{and}\:\mathrm{3x}−\mathrm{6}>\mathrm{0}\:\left(\mathrm{x}>\mathrm{2}\right)\Rightarrow\:\mathrm{12x}−\mathrm{24}>\mathrm{40} \\ $$$$\Rightarrow\mathrm{x}>\frac{\mathrm{16}}{\mathrm{3}}\:…\:\left(\mathrm{i}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{5}}>\frac{\mathrm{8}}{\mathrm{3x}−\mathrm{6}}\:\mathrm{and}\:\mathrm{3x}−\mathrm{6}<\mathrm{0}\:\left(\mathrm{x}<\mathrm{2}\right)\:\Rightarrow\:\mathrm{24}−\mathrm{12x}>−\mathrm{40} \\ $$$$\Rightarrow\mathrm{x}<\frac{\mathrm{16}}{\mathrm{3}}\:\mathrm{but}\:\mathrm{x}<\mathrm{2}\:\Rightarrow\:\mathrm{x}<\mathrm{2}\:…\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)\cup\left(\mathrm{ii}\right)\:=\:\mathrm{x}<\mathrm{2}\:\cup\:\mathrm{x}>\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\frac{\mathrm{8}}{\mathrm{3x}−\mathrm{6}}>\frac{\mathrm{2}}{\mathrm{9}}\:\mathrm{and}\:\mathrm{3x}−\mathrm{6}>\mathrm{0}\left(\mathrm{x}>\mathrm{2}\right)\: \\ $$$$\Rightarrow\:\mathrm{4}×\mathrm{3}>\mathrm{x}−\mathrm{2}\Rightarrow\mathrm{x}<\mathrm{14}\:\mathrm{but}\:\mathrm{x}>\mathrm{2}\:\:\Rightarrow\:\mathrm{2}<\mathrm{x}<\mathrm{14}…\left(\mathrm{iii}\right) \\ $$$$\frac{\mathrm{8}}{\mathrm{3x}−\mathrm{6}}>\frac{\mathrm{2}}{\mathrm{9}}\:\mathrm{and}\:\mathrm{3x}−\mathrm{6}<\mathrm{0}\:\left(\mathrm{x}<\mathrm{2}\right)\: \\ $$$$\Rightarrow−\mathrm{4}×\mathrm{3}>\mathrm{4}−\mathrm{2x}\Rightarrow\mathrm{x}>\mathrm{14}\:\mathrm{but}\:\mathrm{x}<\mathrm{2}\:\Rightarrow\:\mathrm{x}\in\emptyset…\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{iii}\right)\cup\left(\mathrm{iv}\right)=\:\mathrm{2}<\mathrm{x}<\mathrm{14} \\ $$$$ \\ $$$$\left[\left(\mathrm{i}\right)\cup\left(\mathrm{ii}\right)\right]\cap\left[\left(\mathrm{iii}\right)\cup\left(\mathrm{iv}\right)\right]\:=\:\frac{\mathrm{16}}{\mathrm{3}}<\mathrm{x}<\mathrm{14} \\ $$$$\Rightarrow\mathrm{x}\in\:\left(\frac{\mathrm{16}}{\mathrm{3}},\mathrm{14}\right) \\ $$
Answered by Frix last updated on 19/Jun/25
$${x}\neq\mathrm{2} \\ $$$$\frac{\mathrm{5}}{\mathrm{4}}<\frac{\mathrm{3}{x}−\mathrm{6}}{\mathrm{8}}<\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\mathrm{10}<\mathrm{3}{x}−\mathrm{6}<\mathrm{36} \\ $$$$\mathrm{16}<\mathrm{3}{x}<\mathrm{42} \\ $$$$\frac{\mathrm{16}}{\mathrm{3}}<{x}<\mathrm{14} \\ $$
Commented by hardmath last updated on 19/Jun/25
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$