Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

solve:    ((x  −  4)/(x  −  3))   <   ((2x  −  1)/2)

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

Answered by bemath last updated on 09/Oct/20

⇒((x−4)/(x−3)) −((2x−1)/2) < 0  ⇒ ((2x−8−(2x−1)(x−3))/(2(x−3))) < 0  ⇒ ((2x−8−(2x^2 −7x+3))/(2(x−3))) < 0  ⇒((−2x^2 +9x−11)/(2(x−3))) < 0  ⇒((2x^2 −9x+11)/(2(x−3))) > 0  ⇒((x^2 −(9/2)x+((11)/2))/(x−3)) > 0  ⇒(((x−(9/4))^2 −((81)/(16))+((88)/(16)))/(x−3)) > 0  ⇒ (((x−(9/4))^2 +(7/(16)))/(x−3)) > 0   the solution x > 3

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

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

Thanks sir

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

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

((x−4)/(x−3))<((2x−1)/2)⇔((x−4)/(x−3))−((2x−1)/2)<0  ((2x−8−(2x^2 −7x+3))/(2(x−3)))<0  ⇔((−2x^2 +9x−11)/(x−3))<0⇔((2x^2 −9x+11)/(x−3))>0(1)  2x^2 −9x+11=2(x−(9/4))^2 +(7/8)>0 ∀x∈R  Hence,(1)⇔x−3>0⇔x>3  Thus,the given inequality has set of  roots is   S=(3;+∞)

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

Commented by bemath last updated on 09/Oct/20

gave kudos

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

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

Thanks sir

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

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

You are welcome.

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com