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Question Number 120953 by bemath last updated on 04/Nov/20
solve in x∈R ∣ 3x−4 ∣ = x−5
$$\mathrm{solve}\:\mathrm{in}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mid\:\mathrm{3x}−\mathrm{4}\:\mid\:=\:\mathrm{x}−\mathrm{5}\: \\ $$
Answered by TANMAY PANACEA last updated on 04/Nov/20
when (3x−4)>0→ x>(4/3) (3x−4)=x−5 2x=−1 x=((−1)/5) but x=((−1)/5) does not follow condition x>(4/3) so x=((−1)/5) can not be a solution ★ now when (3x−4)<0 x<(4/3) −(3x−4)=x−5 −3x−x=−9 4x=9 x=(9/4) but x=(9/4) does not follow condition x<(4/3) So no solution
$${when}\:\left(\mathrm{3}{x}−\mathrm{4}\right)>\mathrm{0}\rightarrow\:\:\:\:{x}>\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\left(\mathrm{3}{x}−\mathrm{4}\right)={x}−\mathrm{5} \\ $$$$\mathrm{2}{x}=−\mathrm{1} \\ $$$${x}=\frac{−\mathrm{1}}{\mathrm{5}} \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{x}}=\frac{−\mathrm{1}}{\mathrm{5}}\:\:\boldsymbol{{does}}\:\boldsymbol{{not}}\:\boldsymbol{{follow}}\:\boldsymbol{{condition}} \\ $$$$\boldsymbol{{x}}>\frac{\mathrm{4}}{\mathrm{3}}\:\:\boldsymbol{{so}}\:\boldsymbol{{x}}=\frac{−\mathrm{1}}{\mathrm{5}}\:{can}\:{not}\:{be}\:{a}\:{solution} \\ $$$$\bigstar\:{now}\: \\ $$$${when}\:\left(\mathrm{3}{x}−\mathrm{4}\right)<\mathrm{0} \\ $$$${x}<\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$−\left(\mathrm{3}{x}−\mathrm{4}\right)={x}−\mathrm{5} \\ $$$$−\mathrm{3}{x}−{x}=−\mathrm{9} \\ $$$$\mathrm{4}{x}=\mathrm{9} \\ $$$${x}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{x}}=\frac{\mathrm{9}}{\mathrm{4}}\:\boldsymbol{{does}}\:\boldsymbol{{not}}\:\boldsymbol{{follow}}\:\:\boldsymbol{{condition}}\:\boldsymbol{{x}}<\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\boldsymbol{{S}}{o}\:{no}\:{solution} \\ $$$$ \\ $$$$ \\ $$
Answered by ebi last updated on 04/Nov/20
∣3x−4∣=x−5 Case A 3x−4≥0 3x−4=x−5 2x=−1 →x=−0.5 ∣3(−0.5)−4∣=^? (−0.5)−5 ∣−5.5∣=^? −5.5 5.5≠−5.5 ∴ x=−0.5 is not the solution. Case B 3x−4<0 3x−4=−(x−5) 4x=9 →x=2.25 ∣3(2.25)−4∣=^? 2.25−5 ∣−2.75∣=^? −2.75 2.75≠−2.75 ∴ x=2.25 is not the solution. no solution
$$\mid\mathrm{3}{x}−\mathrm{4}\mid={x}−\mathrm{5} \\ $$$$ \\ $$$${Case}\:{A} \\ $$$$\mathrm{3}{x}−\mathrm{4}\geqslant\mathrm{0} \\ $$$$\mathrm{3}{x}−\mathrm{4}={x}−\mathrm{5} \\ $$$$\mathrm{2}{x}=−\mathrm{1}\:\rightarrow{x}=−\mathrm{0}.\mathrm{5} \\ $$$$\mid\mathrm{3}\left(−\mathrm{0}.\mathrm{5}\right)−\mathrm{4}\mid\overset{?} {=}\left(−\mathrm{0}.\mathrm{5}\right)−\mathrm{5} \\ $$$$\mid−\mathrm{5}.\mathrm{5}\mid\overset{?} {=}−\mathrm{5}.\mathrm{5} \\ $$$$\mathrm{5}.\mathrm{5}\neq−\mathrm{5}.\mathrm{5} \\ $$$$\therefore\:{x}=−\mathrm{0}.\mathrm{5}\:{is}\:{not}\:{the}\:{solution}. \\ $$$$ \\ $$$${Case}\:{B} \\ $$$$\mathrm{3}{x}−\mathrm{4}<\mathrm{0} \\ $$$$\mathrm{3}{x}−\mathrm{4}=−\left({x}−\mathrm{5}\right) \\ $$$$\mathrm{4}{x}=\mathrm{9}\:\rightarrow{x}=\mathrm{2}.\mathrm{25} \\ $$$$\mid\mathrm{3}\left(\mathrm{2}.\mathrm{25}\right)−\mathrm{4}\mid\overset{?} {=}\mathrm{2}.\mathrm{25}−\mathrm{5} \\ $$$$\mid−\mathrm{2}.\mathrm{75}\mid\overset{?} {=}−\mathrm{2}.\mathrm{75} \\ $$$$\mathrm{2}.\mathrm{75}\neq−\mathrm{2}.\mathrm{75} \\ $$$$\therefore\:{x}=\mathrm{2}.\mathrm{25}\:{is}\:{not}\:{the}\:{solution}. \\ $$$${no}\:{solution}\: \\ $$
Commented by MJS_new last updated on 04/Nov/20
∣3×2.25−4∣≠2.25−5 2.75≠−2.75
$$\mid\mathrm{3}×\mathrm{2}.\mathrm{25}−\mathrm{4}\mid\neq\mathrm{2}.\mathrm{25}−\mathrm{5} \\ $$$$\mathrm{2}.\mathrm{75}\neq−\mathrm{2}.\mathrm{75} \\ $$
Commented by ebi last updated on 04/Nov/20
Yes. Thank you sir for notice the mistake.
$${Yes}.\:{Thank}\:{you}\:{sir}\:{for}\:{notice}\:{the}\:{mistake}. \\ $$$$ \\ $$
Answered by liberty last updated on 04/Nov/20
∣ 3x−4∣ = x−5 , since ∣3x−4∣ ≥ 0 for all xεR then it follow that x≥5 consider ∣3x−4 ∣ = x−5 → { ((3x−4=x−5 )),((or )),((3x−4=5−x)) :} { ((2x=−1→x=−(1/2) (rejected))),((4x=9→x=(9/4) (rejected))) :} thus the equality has no solution or x = ∅ .
$$\mid\:\mathrm{3x}−\mathrm{4}\mid\:=\:\mathrm{x}−\mathrm{5}\:,\:\mathrm{since}\:\mid\mathrm{3x}−\mathrm{4}\mid\:\geqslant\:\mathrm{0}\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\epsilon\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{it}\:\mathrm{follow}\:\mathrm{that}\:\mathrm{x}\geqslant\mathrm{5} \\ $$$$\mathrm{consider}\:\mid\mathrm{3x}−\mathrm{4}\:\mid\:=\:\mathrm{x}−\mathrm{5}\:\rightarrow\begin{cases}{\mathrm{3x}−\mathrm{4}=\mathrm{x}−\mathrm{5}\:}\\{\mathrm{or}\:}\\{\mathrm{3x}−\mathrm{4}=\mathrm{5}−\mathrm{x}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2x}=−\mathrm{1}\rightarrow\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{rejected}\right)}\\{\mathrm{4x}=\mathrm{9}\rightarrow\mathrm{x}=\frac{\mathrm{9}}{\mathrm{4}}\:\left(\mathrm{rejected}\right)}\end{cases} \\ $$$$\mathrm{thus}\:\mathrm{the}\:\mathrm{equality}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}\: \\ $$$$\mathrm{or}\:\mathrm{x}\:=\:\varnothing\:. \\ $$
Answered by mathmax by abdo last updated on 04/Nov/20
∣3x−4∣=x−5 ⇒∣3x−4∣−x+5 =0 ⇒f(x)=0 x −∞ (4/3) +∞ ∣3x−4∣ −3x+4 0 3x−4 f(x) −4x+9 2x+1 3x−4−x+5 =2x+1 ⇒f(x)= { ((−4x+9 if x≤(4/3))),((2x+1 if x≥(4/3))) :} if x≤(4/3) f(x)=0 ⇒−4x+9 =0 ⇒x =(9/4) but(9/4)>(4/3) its not solution if x≥(4/3) f(x)=0 ⇒2x+1 =0 ⇒x=−(1/2)<(4/3) →not solution so no solution for this equation
$$\mid\mathrm{3x}−\mathrm{4}\mid=\mathrm{x}−\mathrm{5}\:\Rightarrow\mid\mathrm{3x}−\mathrm{4}\mid−\mathrm{x}+\mathrm{5}\:=\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$$\mid\mathrm{3x}−\mathrm{4}\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3x}+\mathrm{4}\:\:\:\:\:\:\mathrm{0}\:\:\:\:\mathrm{3x}−\mathrm{4} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{4x}+\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2x}+\mathrm{1} \\ $$$$\mathrm{3x}−\mathrm{4}−\mathrm{x}+\mathrm{5}\:=\mathrm{2x}+\mathrm{1}\:\:\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\:\:\begin{cases}{−\mathrm{4x}+\mathrm{9}\:\:\mathrm{if}\:\mathrm{x}\leqslant\frac{\mathrm{4}}{\mathrm{3}}}\\{\mathrm{2x}+\mathrm{1}\:\:\mathrm{if}\:\:\:\mathrm{x}\geqslant\frac{\mathrm{4}}{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{if}\:\mathrm{x}\leqslant\frac{\mathrm{4}}{\mathrm{3}}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0}\:\Rightarrow−\mathrm{4x}+\mathrm{9}\:=\mathrm{0}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{9}}{\mathrm{4}}\:\:\mathrm{but}\frac{\mathrm{9}}{\mathrm{4}}>\frac{\mathrm{4}}{\mathrm{3}}\:\:\mathrm{its}\:\mathrm{not}\:\mathrm{solution} \\ $$$$\mathrm{if}\:\mathrm{x}\geqslant\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0}\:\Rightarrow\mathrm{2x}+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{2}}<\frac{\mathrm{4}}{\mathrm{3}}\:\rightarrow\mathrm{not}\:\mathrm{solution} \\ $$$$\mathrm{so}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{this}\:\mathrm{equation} \\ $$

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