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Question Number 49857 by afachri last updated on 11/Dec/18

Please guide me Sir. I was trying to solve this eq for searching possible values of x. eq is : ∣x − 2∣ < 3∣x + 7∣ the range of x whom i got : −((23)/2) < x < −((19)/4) but the result do not satisfy the eq, instead i put x > −4 , they satisfy the eq. please help me out of this pickle. Not because i didn′t try, yet i always stuck in this type of function.

$$\mathrm{Please}\:\mathrm{guide}\:\mathrm{me}\:\mathrm{Sir}.\:\mathrm{I}\:\mathrm{was}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{solve}\: \\ $$ $$\mathrm{this}\:\mathrm{eq}\:\mathrm{for}\:\mathrm{searching}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:{x}. \\ $$ $$\mathrm{eq}\:\mathrm{is}\:: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{x}\:−\:\mathrm{2}\mid\:<\:\mathrm{3}\mid{x}\:+\:\mathrm{7}\mid \\ $$ $$\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{x}\:\mathrm{whom}\:\mathrm{i}\:\mathrm{got}\::\:\:\:−\frac{\mathrm{23}}{\mathrm{2}}\:<\:{x}\:<\:−\frac{\mathrm{19}}{\mathrm{4}} \\ $$ $$\mathrm{but}\:\mathrm{the}\:\mathrm{result}\:\mathrm{do}\:\mathrm{not}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{eq},\:\mathrm{instead} \\ $$ $$\mathrm{i}\:\mathrm{put}\:{x}\:>\:−\mathrm{4}\:,\:\mathrm{they}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{eq}.\: \\ $$ $$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{out}\:\mathrm{of}\:\mathrm{this}\:\mathrm{pickle}. \\ $$ $$\mathrm{Not}\:\mathrm{because}\:\mathrm{i}\:\mathrm{didn}'\mathrm{t}\:\mathrm{try},\:\mathrm{yet}\:\mathrm{i}\:\mathrm{always} \\ $$ $$\mathrm{stuck}\:\mathrm{in}\:\mathrm{this}\:\mathrm{type}\:\mathrm{of}\:\mathrm{function}. \\ $$

Commented bymaxmathsup by imad last updated on 11/Dec/18

the best way is this let A(x)=∣x−2∣−3∣x+7∣ (e) ⇔ A(x)<0 we eradicate the abslutevalue x −∞ −7 2 +∞ ∣x−2∣ −x+2 −x+2 0 x−2 ∣x+7∣ −x−7 0 x+7 x+7 A(x) 2x+23 −4x−19 −2x−23 case 1 x≤−7 (e) ⇔ 2x+23 <0 ⇔ x<−((23)/2) ⇒ S_1 =]−∞ ,−((23)/2)[ case 2 −7≤x≤2 (e) ⇔ −4x −19 <0 ⇔ −4x<19 ⇔4x >−19 ⇔x>−((19)/4) ⇒ S_2 =]−((19)/4) ,2] case 3 x≥2 ⇒ (e)⇔−2x −23 <0 ⇔ −2x<23 ⇔ 2x>−23 ⇔x>−((23)/2) ⇒S_3 =[2,+∞[ and we take ∪S_i S=S_1 ∪S_2 ∪ S_3 =]−∞,−((23)/2)[∪]−((19)/4),+∞[ .

$${the}\:{best}\:{way}\:{is}\:{this}\: \\ $$ $${let}\:{A}\left({x}\right)=\mid{x}−\mathrm{2}\mid−\mathrm{3}\mid{x}+\mathrm{7}\mid\:\:\:\left({e}\right)\:\Leftrightarrow\:{A}\left({x}\right)<\mathrm{0}\:\:{we}\:{eradicate}\:{the}\:{abslutevalue} \\ $$ $${x}\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:−\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$ $$\mid{x}−\mathrm{2}\mid\:\:\:\:\:−{x}+\mathrm{2}\:\:\:−{x}+\mathrm{2}\:\:\mathrm{0}\:\:\:{x}−\mathrm{2} \\ $$ $$\mid{x}+\mathrm{7}\mid\:\:−{x}−\mathrm{7}\:\:\mathrm{0}\:\:\:\:{x}+\mathrm{7}\:\:\:\:\:\:\:\:\:\:{x}+\mathrm{7} \\ $$ $${A}\left({x}\right)\:\:\:\:\mathrm{2}{x}+\mathrm{23}\:\:\:\:\:\:\:−\mathrm{4}{x}−\mathrm{19}\:\:\:\:−\mathrm{2}{x}−\mathrm{23} \\ $$ $$\left.{case}\:\mathrm{1}\:\:\:{x}\leqslant−\mathrm{7}\:\:\:\left({e}\right)\:\Leftrightarrow\:\mathrm{2}{x}+\mathrm{23}\:<\mathrm{0}\:\Leftrightarrow\:{x}<−\frac{\mathrm{23}}{\mathrm{2}}\:\Rightarrow\:{S}_{\mathrm{1}} =\right]−\infty\:,−\frac{\mathrm{23}}{\mathrm{2}}\left[\right. \\ $$ $${case}\:\mathrm{2}\:\:\:\:−\mathrm{7}\leqslant{x}\leqslant\mathrm{2}\:\:\:\:\left({e}\right)\:\Leftrightarrow\:−\mathrm{4}{x}\:−\mathrm{19}\:<\mathrm{0}\:\:\Leftrightarrow\:−\mathrm{4}{x}<\mathrm{19}\:\:\Leftrightarrow\mathrm{4}{x}\:>−\mathrm{19}\:\Leftrightarrow{x}>−\frac{\mathrm{19}}{\mathrm{4}} \\ $$ $$\left.\:\left.\Rightarrow\:{S}_{\mathrm{2}} =\right]−\frac{\mathrm{19}}{\mathrm{4}}\:,\mathrm{2}\right] \\ $$ $${case}\:\mathrm{3}\:\:\:\:\:{x}\geqslant\mathrm{2}\:\Rightarrow\:\left({e}\right)\Leftrightarrow−\mathrm{2}{x}\:−\mathrm{23}\:<\mathrm{0}\:\Leftrightarrow\:−\mathrm{2}{x}<\mathrm{23}\:\Leftrightarrow\:\mathrm{2}{x}>−\mathrm{23}\:\Leftrightarrow{x}>−\frac{\mathrm{23}}{\mathrm{2}} \\ $$ $$\Rightarrow{S}_{\mathrm{3}} =\left[\mathrm{2},+\infty\left[\:\:{and}\:\:{we}\:{take}\:\cup{S}_{{i}} \right.\right. \\ $$ $$\left.{S}={S}_{\mathrm{1}} \cup{S}_{\mathrm{2}} \cup\:{S}_{\mathrm{3}} \:=\right]−\infty,−\frac{\mathrm{23}}{\mathrm{2}}\left[\cup\right]−\frac{\mathrm{19}}{\mathrm{4}},+\infty\left[\:.\right. \\ $$

Commented byafachri last updated on 11/Dec/18

yes it is Sir. Seems this be my another references. thank you Mr Max. soon after Mr Hassen was giving me explanation, i re−tried and get the answer −((23)/2)> x >−((19)/4). and i found match answer to both of yours Sir.

$$\mathrm{yes}\:\mathrm{it}\:\mathrm{is}\:\mathrm{Sir}.\:\mathrm{Seems}\:\mathrm{this}\:\mathrm{be}\:\mathrm{my}\:\mathrm{another}\:\mathrm{references}.\: \\ $$ $$\mathrm{thank}\:\mathrm{you}\:\mathrm{Mr}\:\mathrm{Max}.\:\:\mathrm{soon}\:\mathrm{after}\:\mathrm{Mr}\:\mathrm{Hassen}\:\mathrm{was} \\ $$ $$\mathrm{giving}\:\mathrm{me}\:\mathrm{explanation},\:\mathrm{i}\:\mathrm{re}−\mathrm{tried}\:\mathrm{and}\:\mathrm{get} \\ $$ $$\mathrm{the}\:\mathrm{answer}\:\:−\frac{\mathrm{23}}{\mathrm{2}}>\:{x}\:>−\frac{\mathrm{19}}{\mathrm{4}}.\:\mathrm{and}\:\mathrm{i}\:\mathrm{found} \\ $$ $$\mathrm{match}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{both}\:\mathrm{of}\:\mathrm{yours}\:\mathrm{Sir}.\: \\ $$

Commented byafachri last updated on 11/Dec/18

i′m glad to be part of this forum. Hail Math !

$$\mathrm{i}'\mathrm{m}\:\mathrm{glad}\:\mathrm{to}\:\mathrm{be}\:\mathrm{part}\:\mathrm{of}\:\mathrm{this}\:\mathrm{forum}. \\ $$ $$\mathrm{Hail}\:\mathrm{Math}\:! \\ $$

Commented byAbdo msup. last updated on 12/Dec/18

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Answered by hassentimol last updated on 11/Dec/18

Well... We write it as : ∣x−2∣ − 3∣x+7∣ < 0 We have : { ((x−2 > 0 ⇔ x > 2)),((x+7 > 0 ⇔ x > −7)) :} Therefore we define I_1 , I_2 and I_3 such as : I_1 = ]−∞,−7] : −x+2+3x+21 = 2x+23 I_2 =[−7,2] : −x+2−3x−21 = −4x−19 I_3 =[2,+∞[ : x−2−3x−21 = −2x−23 We have : Let S_n be the set of solutions for the I_n interval. For I_1 : 2x+23<0 ⇔ x<((−23)/2) : S_1 =]−∞,((−23)/2)[ For I_2 : −4x−19<0 ⇔ x>((−19)/4) : S_2 =[((−19)/4),2[ For I_3 : −2x−23<0 ⇔ x>((−23)/2) : S_3 =[2,+∞[ Finally : We can consider S the set of solutions : S = { x ∣ x∈]−∞, ((−19)/4)[ ∪ ]2_ ^ , +∞[ } In other terms : Either x<((−19)/4) , either x>2. Thanks T.H.

$$\mathrm{Well}... \\ $$ $$\mathrm{We}\:\mathrm{write}\:\mathrm{it}\:\mathrm{as}\::\:\:\:\mid{x}−\mathrm{2}\mid\:−\:\mathrm{3}\mid{x}+\mathrm{7}\mid\:<\:\mathrm{0} \\ $$ $$\mathrm{We}\:\mathrm{have}\::\:\begin{cases}{{x}−\mathrm{2}\:>\:\mathrm{0}\:\:\:\Leftrightarrow\:\:\:{x}\:>\:\mathrm{2}}\\{{x}+\mathrm{7}\:>\:\mathrm{0}\:\:\:\Leftrightarrow\:\:\:{x}\:>\:−\mathrm{7}}\end{cases} \\ $$ $$ \\ $$ $$\mathrm{Therefore}\:\mathrm{we}\:\mathrm{define}\:\mathrm{I}_{\mathrm{1}} ,\:\mathrm{I}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{I}_{\mathrm{3}} \:\mathrm{such}\:\mathrm{as}\:: \\ $$ $$\left.\mathrm{I}_{\mathrm{1}} \left.=\:\right]−\infty,−\mathrm{7}\right]\::\:−{x}+\mathrm{2}+\mathrm{3}{x}+\mathrm{21}\:=\:\:\:\mathrm{2}{x}+\mathrm{23} \\ $$ $$\mathrm{I}_{\mathrm{2}} =\left[−\mathrm{7},\mathrm{2}\right]\::\:−{x}+\mathrm{2}−\mathrm{3}{x}−\mathrm{21}\:=\:\:\:−\mathrm{4}{x}−\mathrm{19} \\ $$ $$\mathrm{I}_{\mathrm{3}} =\left[\mathrm{2},+\infty\left[\::\:{x}−\mathrm{2}−\mathrm{3}{x}−\mathrm{21}\:=\:\:\:−\mathrm{2}{x}−\mathrm{23}\right.\right. \\ $$ $$ \\ $$ $$\mathrm{We}\:\mathrm{have}\:: \\ $$ $${Let}\:{S}_{{n}} \:{be}\:{the}\:{set}\:{of}\:{solutions}\:{for}\:{the}\:{I}_{{n}} \:{interval}. \\ $$ $$\left.\mathrm{For}\:\mathrm{I}_{\mathrm{1}} \::\:\mathrm{2}{x}+\mathrm{23}<\mathrm{0}\:\Leftrightarrow\:{x}<\frac{−\mathrm{23}}{\mathrm{2}}\::\:\mathrm{S}_{\mathrm{1}} =\right]−\infty,\frac{−\mathrm{23}}{\mathrm{2}}\left[\right. \\ $$ $$\mathrm{For}\:\mathrm{I}_{\mathrm{2}} \::\:−\mathrm{4}{x}−\mathrm{19}<\mathrm{0}\:\Leftrightarrow\:{x}>\frac{−\mathrm{19}}{\mathrm{4}}\::\:\mathrm{S}_{\mathrm{2}} =\left[\frac{−\mathrm{19}}{\mathrm{4}},\mathrm{2}\left[\right.\right. \\ $$ $$\mathrm{For}\:\mathrm{I}_{\mathrm{3}} \::\:−\mathrm{2}{x}−\mathrm{23}<\mathrm{0}\:\Leftrightarrow\:{x}>\frac{−\mathrm{23}}{\mathrm{2}}\::\:\mathrm{S}_{\mathrm{3}} =\left[\mathrm{2},+\infty\left[\right.\right. \\ $$ $$ \\ $$ $$\mathrm{Finally}\:: \\ $$ $$ \\ $$ $$\mathrm{We}\:\mathrm{can}\:\mathrm{consider}\:\mathbb{S}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{solutions}\:: \\ $$ $$\:\:\:\:\mathbb{S}\:=\:\left\{\:{x}\:\mid\:{x}\in\right]−\infty,\:\frac{−\mathrm{19}}{\mathrm{4}}\left[\:\cup\:\right]\underset{} {\overset{} {\mathrm{2}}},\:+\infty\left[\:\right\} \\ $$ $$ \\ $$ $$\mathrm{In}\:\mathrm{other}\:\mathrm{terms}\:: \\ $$ $$\mathrm{Either}\:{x}<\frac{−\mathrm{19}}{\mathrm{4}}\:,\:\mathrm{either}\:{x}>\mathrm{2}. \\ $$ $$ \\ $$ $$\boldsymbol{{Thanks}} \\ $$ $$\boldsymbol{{T}}.\boldsymbol{{H}}. \\ $$

Commented byafachri last updated on 11/Dec/18

Thanks for your explanation Mr. Hassen. I will learn from this. But Sir, is x < −((19)/4) satisfy the eq ???

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{explanation}\:\mathrm{Mr}.\:\mathrm{Hassen}. \\ $$ $$\mathrm{I}\:\mathrm{will}\:\mathrm{learn}\:\mathrm{from}\:\mathrm{this}. \\ $$ $$\mathrm{But}\:\mathrm{Sir},\:\mathrm{is}\:{x}\:<\:−\frac{\mathrm{19}}{\mathrm{4}}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{eq}\:??? \\ $$

Commented byhassentimol last updated on 11/Dec/18

You are welcome sir. It was a pleasure.

$$\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir}.\:\mathrm{It}\:\mathrm{was}\:\mathrm{a}\:\mathrm{pleasure}. \\ $$

Commented byhassentimol last updated on 11/Dec/18

Well...any number smaller than −19/4 will verify the equation. But a number which would be greater than 2 would also satisfy the inequation...

$$\mathrm{Well}...\mathrm{any}\:\mathrm{number}\:\mathrm{smaller}\:\mathrm{than}\:−\mathrm{19}/\mathrm{4}\:\mathrm{will} \\ $$ $$\mathrm{verify}\:\mathrm{the}\:\mathrm{equation}. \\ $$ $$\mathrm{But}\:\mathrm{a}\:\mathrm{number}\:\mathrm{which}\:\mathrm{would}\:\mathrm{be}\:\mathrm{greater}\:\mathrm{than} \\ $$ $$\mathrm{2}\:\mathrm{would}\:\mathrm{also}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{inequation}... \\ $$

Commented byhassentimol last updated on 11/Dec/18

Commented byhassentimol last updated on 11/Dec/18

This is the curve of the function when all is put on one side of the inequation. The solutions are when the curve is negative...

$$\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{when}\:\mathrm{all}\:\mathrm{is}\:\mathrm{put} \\ $$ $$\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{inequation}. \\ $$ $$\mathrm{The}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{when}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{is}\:\mathrm{negative}... \\ $$

Commented byafachri last updated on 11/Dec/18

i put x = (−5) for trial, ∣x − 2∣ − 3∣x + 7∣ < 0 ∣−5 − 2∣ − 3∣−5+7∣ < 0 7 − 3(2) < 0 why x = (−5) doesn′t satisfy the eq Sir ? correct me Sir, please.

$$\:\mathrm{i}\:\mathrm{put}\:{x}\:=\:\left(−\mathrm{5}\right)\:\mathrm{for}\:\mathrm{trial},\: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{x}\:−\:\mathrm{2}\mid\:−\:\mathrm{3}\mid{x}\:+\:\mathrm{7}\mid\:\:<\:\mathrm{0} \\ $$ $$\:\:\:\:\:\mid−\mathrm{5}\:−\:\mathrm{2}\mid\:\:−\:\:\:\mathrm{3}\mid−\mathrm{5}+\mathrm{7}\mid\:\:<\:\:\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}\:\:−\:\mathrm{3}\left(\mathrm{2}\right)\:\:<\:\mathrm{0}\: \\ $$ $$\mathrm{why}\:{x}\:=\:\left(−\mathrm{5}\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{eq}\:\mathrm{Sir}\:? \\ $$ $$\mathrm{correct}\:\mathrm{me}\:\mathrm{Sir},\:\mathrm{please}. \\ $$

Commented byafachri last updated on 11/Dec/18

Commented byafachri last updated on 11/Dec/18

thank you very much Sir. i′ve looked in my graphic too and understand now. :)

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}.\:\mathrm{i}'\mathrm{ve}\:\mathrm{looked}\:\mathrm{in} \\ $$ $$\mathrm{my}\:\mathrm{graphic}\:\mathrm{too}\:\mathrm{and}\:\mathrm{understand}\:\mathrm{now}. \\ $$ $$\left.:\right) \\ $$

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