Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 49809 by Abdo msup. last updated on 10/Dec/18

solve the system { ((((4(√(1+x^2 )))/x) =((5(√(1+y^2 )))/y)=((6(√(1+z^2 )))/z))),((x+y+z=xyz.)) :} {: (),() }

$${solve}\:{the}\:{system}\:\:\:\begin{cases}{\frac{\mathrm{4}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\:=\frac{\mathrm{5}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}}=\frac{\mathrm{6}\sqrt{\mathrm{1}+{z}^{\mathrm{2}} }}{{z}}}\\{{x}+{y}+{z}={xyz}.}\end{cases} \\ $$$$\left.\begin{matrix}{}\\{}\end{matrix}\right\} \\ $$

Commented by MJS last updated on 12/Dec/18

((a(√(1+x^2 )))/x)=((b(√(1+y^2 )))/y)=((c(√(1+z^2 )))/z) x+y+z=xyz leads to x=±((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/(a^2 −b^2 −c^2 )) y=±((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/(b^2 −a^2 −c^2 )) z=±((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/(c^2 −a^2 −b^2 )) ⇒ a, b, c must form a triangle but not a rectangular one...

$$\frac{{a}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}=\frac{{b}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}}=\frac{{c}\sqrt{\mathrm{1}+{z}^{\mathrm{2}} }}{{z}} \\ $$$${x}+{y}+{z}={xyz} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$${x}=\pm\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$${y}=\pm\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$${z}=\pm\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{a},\:{b},\:{c}\:\mathrm{must}\:\mathrm{form}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{but}\:\mathrm{not}\:\mathrm{a} \\ $$$$\mathrm{rectangular}\:\mathrm{one}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

x=tana y=tanb z=tanc ((4(√(1+tan^2 a)) )/(tana))=((5(√(1+tan^2 b)))/(tanb))=((6(√(1+tan^2 c)))/(tanc)) (4/(cosa×((sina)/(cosa))))=(5/(sinb))=(6/(sinc))=(1/k) sina=4k sinb=5k sinc=6k x+y+z=xyz x+y=xyz−z x+y=z(xy−1) x+y=−z(1−xy) ((x+y)/(1−xy))=−z ((tana+tanb)/(1−tanatanb))=−tanc tan(a+b)=−tanc tan(a+b)=tan(π−c) a+b+c=π wait... sin(a+b)=sin(π−c) sinacosb+cosasinb=sinc 4k×(√(1−25b^2 )) +(√(1−16k^2 )) ×5k=6k 4(√(1−25k^2 )) +5(√(1−16k^2 )) =6 16(1−25k^2 )=36+25(1−16k^2 )−60(√(1−16k^2 )) 16−400k^2 =36+25−400k^2 −60(√(1−16k^2 )) −45=−60(√(1−16k^2 )) 9=16(1−16k^2 ) 9−16=−16×16k^2 256k^2 =7 k=±((√7)/(16)) sina=((±4(√7))/(16)) x=tana=((±4(√7))/(√(256−112)))=((±4(√7))/(12)) sinb=((±5(√7))/(16)) y=tanb=((±4(√7))/(√(256−175)))=((±5(√7))/9) sinc=((±6(√7))/(16)) z=tanc=((±4(√7))/(√(256−252)))=((±4(√7))/2)←silly error in place of 6 i put 4 so corrected.. z=((±6(√7))/2) pls check...

$${x}={tana}\:\:\:{y}={tanb}\:\:\:\:{z}={tanc} \\ $$$$\frac{\mathrm{4}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {a}}\:}{{tana}}=\frac{\mathrm{5}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {b}}}{{tanb}}=\frac{\mathrm{6}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {c}}}{{tanc}} \\ $$$$\frac{\mathrm{4}}{{cosa}×\frac{{sina}}{{cosa}}}=\frac{\mathrm{5}}{{sinb}}=\frac{\mathrm{6}}{{sinc}}=\frac{\mathrm{1}}{{k}} \\ $$$${sina}=\mathrm{4}{k} \\ $$$${sinb}=\mathrm{5}{k} \\ $$$${sinc}=\mathrm{6}{k} \\ $$$${x}+{y}+{z}={xyz} \\ $$$${x}+{y}={xyz}−{z} \\ $$$${x}+{y}={z}\left({xy}−\mathrm{1}\right) \\ $$$${x}+{y}=−{z}\left(\mathrm{1}−{xy}\right) \\ $$$$\frac{{x}+{y}}{\mathrm{1}−{xy}}=−{z} \\ $$$$\frac{{tana}+{tanb}}{\mathrm{1}−{tanatanb}}=−{tanc} \\ $$$${tan}\left({a}+{b}\right)=−{tanc} \\ $$$${tan}\left({a}+{b}\right)={tan}\left(\pi−{c}\right) \\ $$$${a}+{b}+{c}=\pi \\ $$$${wait}... \\ $$$${sin}\left({a}+{b}\right)={sin}\left(\pi−{c}\right) \\ $$$${sinacosb}+{cosasinb}={sinc} \\ $$$$\mathrm{4}{k}×\sqrt{\mathrm{1}−\mathrm{25}{b}^{\mathrm{2}} }\:+\sqrt{\mathrm{1}−\mathrm{16}{k}^{\mathrm{2}} }\:×\mathrm{5}{k}=\mathrm{6}{k} \\ $$$$\mathrm{4}\sqrt{\mathrm{1}−\mathrm{25}{k}^{\mathrm{2}} }\:+\mathrm{5}\sqrt{\mathrm{1}−\mathrm{16}{k}^{\mathrm{2}} }\:=\mathrm{6} \\ $$$$\mathrm{16}\left(\mathrm{1}−\mathrm{25}{k}^{\mathrm{2}} \right)=\mathrm{36}+\mathrm{25}\left(\mathrm{1}−\mathrm{16}{k}^{\mathrm{2}} \right)−\mathrm{60}\sqrt{\mathrm{1}−\mathrm{16}{k}^{\mathrm{2}} }\: \\ $$$$\mathrm{16}−\mathrm{400}{k}^{\mathrm{2}} =\mathrm{36}+\mathrm{25}−\mathrm{400}{k}^{\mathrm{2}} −\mathrm{60}\sqrt{\mathrm{1}−\mathrm{16}{k}^{\mathrm{2}} }\: \\ $$$$−\mathrm{45}=−\mathrm{60}\sqrt{\mathrm{1}−\mathrm{16}{k}^{\mathrm{2}} }\: \\ $$$$\mathrm{9}=\mathrm{16}\left(\mathrm{1}−\mathrm{16}{k}^{\mathrm{2}} \right) \\ $$$$\mathrm{9}−\mathrm{16}=−\mathrm{16}×\mathrm{16}{k}^{\mathrm{2}} \\ $$$$\mathrm{256}{k}^{\mathrm{2}} =\mathrm{7} \\ $$$${k}=\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{16}} \\ $$$${sina}=\frac{\pm\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{16}}\:\:{x}={tana}=\frac{\pm\mathrm{4}\sqrt{\mathrm{7}}}{\sqrt{\mathrm{256}−\mathrm{112}}}=\frac{\pm\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{12}} \\ $$$${sinb}=\frac{\pm\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{16}}\:\:\:{y}={tanb}=\frac{\pm\mathrm{4}\sqrt{\mathrm{7}}}{\sqrt{\mathrm{256}−\mathrm{175}}}=\frac{\pm\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{9}} \\ $$$${sinc}=\frac{\pm\mathrm{6}\sqrt{\mathrm{7}}}{\mathrm{16}}\:\:\:{z}={tanc}=\frac{\pm\mathrm{4}\sqrt{\mathrm{7}}}{\sqrt{\mathrm{256}−\mathrm{252}}}=\frac{\pm\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{2}}\leftarrow{silly}\:{error} \\ $$$${in}\:{place}\:{of}\:\mathrm{6}\:{i}\:{put}\:\mathrm{4} \\ $$$${so}\:{corrected}.. \\ $$$${z}=\frac{\pm\mathrm{6}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$$$ \\ $$

Answered by MJS last updated on 12/Dec/18

x>0∧y>0∧z>0 ∨ x<0∧y<0∧z<0 ((4(√(1+x^2 )))/x)=((5(√(1+y^2 )))/y) ⇒ y=±((5x)/(√(16−9x^2 ))) ((4(√(1+x^2 )))/x)=((6(√(1+z^2 )))/z) ⇒ z=±((3x)/(√(4−5x^2 ))) let x, y, x greater than zero, due to squaring we′ll get the other option anyway x+y+x−xyz=0 x+((5x)/(√(16−9x^2 )))+((3x)/(√(4−5x^2 )))−x×((5x)/(√(16−9x^2 )))×((3x)/(√(4−5x^2 )))=0 (√(4−5x^2 ))=p; (√(16−9x^2 ))=q ⇒ x((5p+pq+3q−15x^2 )/(pq))=0 p=3(((5x^2 −q))/(q+5)) p^2 =3((25x^4 −10qx^2 +q^2 )/(q^2 +10q+25)) p^2 =4−5x^2 ; q^2 =16−9x^2 ⇒ q=((9x^2 −1)/2) q^2 =((81x^4 −18x^2 +1)/4) q^2 =16−9x^2 ⇒ x^4 +(2/9)x^2 −(7/9)=0 ⇒ x^2 =−1 ∨ x^2 =(7/9) ⇒ x=±((√7)/3) ⇒ y=±((5(√7))/9); z=±3(√7) solutions ((x),(y),(z) ) = (((−((√7)/3))),((−((5(√7))/9))),((−3(√7))) ) ∨ ((x),(y),(z) ) = ((((√7)/3)),(((5(√7))/9)),((3(√7))) )

$${x}>\mathrm{0}\wedge{y}>\mathrm{0}\wedge{z}>\mathrm{0}\:\vee\:{x}<\mathrm{0}\wedge{y}<\mathrm{0}\wedge{z}<\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{4}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}=\frac{\mathrm{5}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}}\:\Rightarrow\:{y}=\pm\frac{\mathrm{5}{x}}{\sqrt{\mathrm{16}−\mathrm{9}{x}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{4}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}=\frac{\mathrm{6}\sqrt{\mathrm{1}+{z}^{\mathrm{2}} }}{{z}}\:\Rightarrow\:{z}=\pm\frac{\mathrm{3}{x}}{\sqrt{\mathrm{4}−\mathrm{5}{x}^{\mathrm{2}} }} \\ $$$$\mathrm{let}\:{x},\:{y},\:{x}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{zero},\:\mathrm{due}\:\mathrm{to}\:\mathrm{squaring} \\ $$$$\mathrm{we}'\mathrm{ll}\:\mathrm{get}\:\mathrm{the}\:\mathrm{other}\:\mathrm{option}\:\mathrm{anyway} \\ $$$${x}+{y}+{x}−{xyz}=\mathrm{0} \\ $$$${x}+\frac{\mathrm{5}{x}}{\sqrt{\mathrm{16}−\mathrm{9}{x}^{\mathrm{2}} }}+\frac{\mathrm{3}{x}}{\sqrt{\mathrm{4}−\mathrm{5}{x}^{\mathrm{2}} }}−{x}×\frac{\mathrm{5}{x}}{\sqrt{\mathrm{16}−\mathrm{9}{x}^{\mathrm{2}} }}×\frac{\mathrm{3}{x}}{\sqrt{\mathrm{4}−\mathrm{5}{x}^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\sqrt{\mathrm{4}−\mathrm{5}{x}^{\mathrm{2}} }={p};\:\sqrt{\mathrm{16}−\mathrm{9}{x}^{\mathrm{2}} }={q} \\ $$$$\Rightarrow\:{x}\frac{\mathrm{5}{p}+{pq}+\mathrm{3}{q}−\mathrm{15}{x}^{\mathrm{2}} }{{pq}}=\mathrm{0} \\ $$$${p}=\mathrm{3}\frac{\left(\mathrm{5}{x}^{\mathrm{2}} −{q}\right)}{{q}+\mathrm{5}} \\ $$$${p}^{\mathrm{2}} =\mathrm{3}\frac{\mathrm{25}{x}^{\mathrm{4}} −\mathrm{10}{qx}^{\mathrm{2}} +{q}^{\mathrm{2}} }{{q}^{\mathrm{2}} +\mathrm{10}{q}+\mathrm{25}} \\ $$$${p}^{\mathrm{2}} =\mathrm{4}−\mathrm{5}{x}^{\mathrm{2}} ;\:{q}^{\mathrm{2}} =\mathrm{16}−\mathrm{9}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{q}=\frac{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}} \\ $$$${q}^{\mathrm{2}} =\frac{\mathrm{81}{x}^{\mathrm{4}} −\mathrm{18}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}} \\ $$$${q}^{\mathrm{2}} =\mathrm{16}−\mathrm{9}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}^{\mathrm{4}} +\frac{\mathrm{2}}{\mathrm{9}}{x}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{9}}=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} =−\mathrm{1}\:\vee\:{x}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{9}} \\ $$$$\Rightarrow\:{x}=\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\:\Rightarrow\:{y}=\pm\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{9}};\:{z}=\pm\mathrm{3}\sqrt{\mathrm{7}} \\ $$$$\mathrm{solutions}\:\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\begin{pmatrix}{−\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}}\\{−\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{9}}}\\{−\mathrm{3}\sqrt{\mathrm{7}}}\end{pmatrix}\:\vee\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\begin{pmatrix}{\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}}\\{\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{9}}}\\{\mathrm{3}\sqrt{\mathrm{7}}}\end{pmatrix} \\ $$

Commented by Abdo msup. last updated on 12/Dec/18

thankyou sir

$${thankyou}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com