Menu Close

Question-110299

Tinku Tara 0 Comments
Pin




Question Number 110299 by Lekhraj last updated on 28/Aug/20
Answered by Her_Majesty last updated on 28/Aug/20
5x+5y=2xy ⇒ 5∣xy ⇒ 5∣x∨5∣y let x=5k 25k+5y=10ky y=((5k)/(2k−1)) k=1 ⇒ y=5∧x=5 but x≠y k=3 ⇒ y=3∧x=15 ⇒ (√(x+y))=3(√2) only solution because ((5k)/(2k−1))>2∀k>0 of course x=3∧y=15 is the 2^(nd) possibility but it also leads to (√(x+y))=3(√2)
$$\mathrm{5}{x}+\mathrm{5}{y}=\mathrm{2}{xy} \\ $$$$\Rightarrow\:\mathrm{5}\mid{xy}\:\Rightarrow\:\mathrm{5}\mid{x}\vee\mathrm{5}\mid{y} \\ $$$${let}\:{x}=\mathrm{5}{k} \\ $$$$\mathrm{25}{k}+\mathrm{5}{y}=\mathrm{10}{ky} \\ $$$${y}=\frac{\mathrm{5}{k}}{\mathrm{2}{k}−\mathrm{1}} \\ $$$${k}=\mathrm{1}\:\Rightarrow\:{y}=\mathrm{5}\wedge{x}=\mathrm{5}\:{but}\:{x}\neq{y} \\ $$$${k}=\mathrm{3}\:\Rightarrow\:{y}=\mathrm{3}\wedge{x}=\mathrm{15}\:\Rightarrow\:\sqrt{{x}+{y}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${only}\:{solution}\:{because}\:\frac{\mathrm{5}{k}}{\mathrm{2}{k}−\mathrm{1}}>\mathrm{2}\forall{k}>\mathrm{0} \\ $$$${of}\:{course}\:{x}=\mathrm{3}\wedge{y}=\mathrm{15}\:{is}\:{the}\:\mathrm{2}^{{nd}} \:{possibility} \\ $$$${but}\:{it}\:{also}\:{leads}\:{to}\:\sqrt{{x}+{y}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$
Commented by Lekhraj last updated on 28/Aug/20
Thanks
$$\mathrm{Thanks} \\ $$
Commented by udaythool last updated on 28/Aug/20
x+y=2xy/5⇒5k+y=2ky for some k∈Z\{0}⇒5+m=2km⇒m=±1,±5 m=1⇒k=3⇒x=15,y=3⇒(√(x+y))=±3(√2) m=−1⇒k=−2⇒x=−10,y=2⇒(√(x+y))=±2(√2)i m=5⇒k=1⇒x=5,y=5 But x≠y⇒m≠5 m=−5⇒k=0 which is not possible.
$${x}+{y}=\mathrm{2}{xy}/\mathrm{5}\Rightarrow\mathrm{5}{k}+{y}=\mathrm{2}{ky}\:\mathrm{for} \\ $$$$\mathrm{some}\:{k}\in\mathbb{Z}\backslash\left\{\mathrm{0}\right\}\Rightarrow\mathrm{5}+{m}=\mathrm{2}{km}\Rightarrow{m}=\pm\mathrm{1},\pm\mathrm{5} \\ $$$${m}=\mathrm{1}\Rightarrow{k}=\mathrm{3}\Rightarrow{x}=\mathrm{15},{y}=\mathrm{3}\Rightarrow\sqrt{{x}+{y}}=\pm\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${m}=−\mathrm{1}\Rightarrow{k}=−\mathrm{2}\Rightarrow{x}=−\mathrm{10},{y}=\mathrm{2}\Rightarrow\sqrt{{x}+{y}}=\pm\mathrm{2}\sqrt{\mathrm{2}}{i} \\ $$$${m}=\mathrm{5}\Rightarrow{k}=\mathrm{1}\Rightarrow{x}=\mathrm{5},{y}=\mathrm{5}\:\mathrm{But}\:{x}\neq{y}\Rightarrow{m}\neq\mathrm{5} \\ $$$${m}=−\mathrm{5}\Rightarrow{k}=\mathrm{0}\:\mathrm{which}\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}. \\ $$$$ \\ $$
Commented by Her_Majesty last updated on 28/Aug/20
the question says “distinct positive integers” so as I stated, there′s only one solution
$${the}\:{question}\:{says}\:“{distinct}\:{positive}\:{integers}'' \\ $$$${so}\:{as}\:{I}\:{stated},\:{there}'{s}\:{only}\:{one}\:{solution} \\ $$
Commented by Rasheed.Sindhi last updated on 28/Aug/20
(√(x+y))=±3(√2) is not correct,I think. Because the symbol “(√( )) ”is used for positive square root only. For both roots “±(√( ))”is used.(And for negative root “−(√( ))”is used.)
$$\sqrt{{x}+{y}}=\pm\mathrm{3}\sqrt{\mathrm{2}}\:\:{is}\:{not}\:{correct},{I}\:{think}. \\ $$$${Because}\:{the}\:{symbol}\:“\sqrt{\:\:\:\:}\:''{is} \\ $$$${used}\:{for}\:{positive}\:{square}\:{root} \\ $$$${only}.\:{For}\:{both}\:{roots}\:“\pm\sqrt{\:\:\:\:}''{is} \\ $$$${used}.\left({And}\:{for}\:{negative}\:{root}\:\right. \\ $$$$\left.“−\sqrt{\:\:\:\:}''{is}\:{used}.\right) \\ $$
Commented by Her_Majesty last updated on 28/Aug/20
(√x) is a unique number ∀x∈C if we ask for the solution of x^2 =y we want all possible values for x and get x_1 =−(√y) and x_2 =(√y). we write x=±(√y) but that obviously cannot mean x=−(√y) ∧ x=+(√y). we would have to write x=−(√y) xor x=+(√y) (different symbols for xor in use) but we are too lazy and write x=−(√x) ∨ x=+(√x) which is wrong... it would make absolutely no sense to think (√y)=±(√y) because then x_1 =±(−(√y)) and x_2 =±(+(√y)) ⇒ x_1 =∓(√y) and x_2 =±(√y)...
$$\sqrt{{x}}\:{is}\:{a}\:{unique}\:{number}\:\forall{x}\in\mathbb{C} \\ $$$${if}\:{we}\:{ask}\:{for}\:{the}\:{solution}\:{of}\:{x}^{\mathrm{2}} ={y}\:{we} \\ $$$${want}\:{all}\:{possible}\:{values}\:{for}\:{x}\:{and}\:{get} \\ $$$${x}_{\mathrm{1}} =−\sqrt{{y}}\:{and}\:{x}_{\mathrm{2}} =\sqrt{{y}}.\:{we}\:{write}\:{x}=\pm\sqrt{{y}} \\ $$$${but}\:{that}\:{obviously}\:{cannot}\:{mean} \\ $$$${x}=−\sqrt{{y}}\:\wedge\:{x}=+\sqrt{{y}}.\:{we}\:{would}\:{have}\:{to}\:{write} \\ $$$${x}=−\sqrt{{y}}\:{xor}\:{x}=+\sqrt{{y}}\:\left({different}\:{symbols}\:{for}\right. \\ $$$$\left.{xor}\:{in}\:{use}\right)\:{but}\:{we}\:{are}\:{too}\:{lazy}\:{and}\:{write} \\ $$$${x}=−\sqrt{{x}}\:\vee\:{x}=+\sqrt{{x}}\:{which}\:{is}\:{wrong}… \\ $$$${it}\:{would}\:{make}\:{absolutely}\:{no}\:{sense}\:{to}\:{think} \\ $$$$\sqrt{{y}}=\pm\sqrt{{y}}\:{because}\:{then}\:{x}_{\mathrm{1}} =\pm\left(−\sqrt{{y}}\right)\:{and} \\ $$$${x}_{\mathrm{2}} =\pm\left(+\sqrt{{y}}\right)\:\Rightarrow\:{x}_{\mathrm{1}} =\mp\sqrt{{y}}\:{and}\:{x}_{\mathrm{2}} =\pm\sqrt{{y}}… \\ $$
Commented by udaythool last updated on 28/Aug/20
Oh! Yes, I haven′t noticed...
$$\mathrm{Oh}! \\ $$$$\mathrm{Yes},\:\mathrm{I}\:\mathrm{haven}'\mathrm{t}\:\mathrm{noticed}… \\ $$
Answered by Rasheed.Sindhi last updated on 28/Aug/20
(1/x)=(2/5)−(1/y)=((2y−5)/(5y)) x=((5y)/(2y−5)) As x,y∈Z^+ , y≥3 For y=3,x=15 Also for symmetry x=3,y=15 (√(x+y))=(√(3+15))=3(√2) Continue
$$\:\:\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{1}}{{y}}=\frac{\mathrm{2}{y}−\mathrm{5}}{\mathrm{5}{y}} \\ $$$$\:\:{x}=\frac{\mathrm{5}{y}}{\mathrm{2}{y}−\mathrm{5}} \\ $$$${As}\:{x},{y}\in\mathbb{Z}^{+} ,\:{y}\geqslant\mathrm{3} \\ $$$${For}\:{y}=\mathrm{3},{x}=\mathrm{15} \\ $$$${Also}\:{for}\:{symmetry} \\ $$$$\:\:\:{x}=\mathrm{3},{y}=\mathrm{15} \\ $$$$\sqrt{{x}+{y}}=\sqrt{\mathrm{3}+\mathrm{15}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${Continue} \\ $$
Commented by Lekhraj last updated on 28/Aug/20
Thanks
$$\mathrm{Thanks} \\ $$
Answered by 1549442205PVT last updated on 29/Aug/20
(1/x)+(1/y)=(2/5)⇔((x+y)/(xy))=(2/5)⇔2xy=5(x+y) ⇔4xy−10(x+y)+25=25 ⇔(2x−5)(2y−5)=25 (1)⇒(2x−5 )is an odd divisor of 25 ⇒2x−5∈{1,−1,5,−5,−25,25} i)2x−5=1⇒x=3 .Replace into (1) we get y=15 ii)2x−5=−1⇒x=2.Replace into (1)we get y=−10 ∉Z^+ ⇒rejected iii)2x−5=5⇒x=5.Replace into (1) we get y=5 iv)2x−5=−25⇒x=−10⇒rejected v)2x−5=25⇒x=15⇒y=3 Combinating all above cases we get (x,y)∈{(3,15),(5,5),(15,3)} ⇒(√(x+y)) ∈{3(√2) ,(√(10))}
$$\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}=\frac{\mathrm{2}}{\mathrm{5}}\Leftrightarrow\frac{\mathrm{x}+\mathrm{y}}{\mathrm{xy}}=\frac{\mathrm{2}}{\mathrm{5}}\Leftrightarrow\mathrm{2xy}=\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\Leftrightarrow\mathrm{4xy}−\mathrm{10}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{25}=\mathrm{25} \\ $$$$\Leftrightarrow\left(\mathrm{2x}−\mathrm{5}\right)\left(\mathrm{2y}−\mathrm{5}\right)=\mathrm{25}\:\:\left(\mathrm{1}\right)\Rightarrow\left(\mathrm{2x}−\mathrm{5}\:\right)\mathrm{is} \\ $$$$\mathrm{an}\:\mathrm{odd}\:\mathrm{divisor}\:\mathrm{of}\:\mathrm{25} \\ $$$$\Rightarrow\mathrm{2x}−\mathrm{5}\in\left\{\mathrm{1},−\mathrm{1},\mathrm{5},−\mathrm{5},−\mathrm{25},\mathrm{25}\right\} \\ $$$$\left.\mathrm{i}\right)\mathrm{2x}−\mathrm{5}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{3}\:.\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\:\mathrm{we} \\ $$$$\mathrm{get}\:\mathrm{y}=\mathrm{15} \\ $$$$\left.\mathrm{ii}\right)\mathrm{2x}−\mathrm{5}=−\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{2}.\mathrm{Replace}\:\mathrm{into} \\ $$$$\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{y}=−\mathrm{10}\:\notin\mathbb{Z}^{+} \Rightarrow\mathrm{rejected} \\ $$$$\left.\mathrm{iii}\right)\mathrm{2x}−\mathrm{5}=\mathrm{5}\Rightarrow\mathrm{x}=\mathrm{5}.\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{y}=\mathrm{5} \\ $$$$\left.\mathrm{iv}\right)\mathrm{2x}−\mathrm{5}=−\mathrm{25}\Rightarrow\mathrm{x}=−\mathrm{10}\Rightarrow\mathrm{rejected} \\ $$$$\left.\mathrm{v}\right)\mathrm{2x}−\mathrm{5}=\mathrm{25}\Rightarrow\mathrm{x}=\mathrm{15}\Rightarrow\mathrm{y}=\mathrm{3} \\ $$$$\mathrm{Combinating}\:\mathrm{all}\:\mathrm{above}\:\mathrm{cases}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)\in\left\{\left(\mathrm{3},\mathrm{15}\right),\left(\mathrm{5},\mathrm{5}\right),\left(\mathrm{15},\mathrm{3}\right)\right\} \\ $$$$\Rightarrow\sqrt{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}\:\in\left\{\mathrm{3}\sqrt{\mathrm{2}}\:,\sqrt{\mathrm{10}}\right\} \\ $$
Commented by Her_Majesty last updated on 29/Aug/20
saying it again, we′re searching for distinct integers ⇒ x≠y
$${saying}\:{it}\:{again},\:{we}'{re}\:{searching}\:{for}\:{distinct} \\ $$$${integers}\:\Rightarrow\:{x}\neq{y} \\ $$
Commented by Rasheed.Sindhi last updated on 29/Aug/20
Nice Sir! Only exclude (5,5) because they are not distinct. You′ve given the problem usefull form: (2x−5)(2y−5)=25 This form offers limited trials!
$${Nice}\:{Sir}! \\ $$$${Only}\:{exclude}\:\left(\mathrm{5},\mathrm{5}\right)\:{because}\:{they} \\ $$$${are}\:{not}\:\boldsymbol{{distinct}}. \\ $$$${You}'{ve}\:{given}\:{the}\:{problem}\:{usefull} \\ $$$${form}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2x}−\mathrm{5}\right)\left(\mathrm{2y}−\mathrm{5}\right)=\mathrm{25} \\ $$$$\mathcal{T}{his}\:{form}\:{offers}\:{limited}\:{trials}! \\ $$
Commented by 1549442205PVT last updated on 30/Aug/20
Thank Sir.I don′t note that hypothesis .
$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{note}\:\mathrm{that} \\ $$$$\mathrm{hypothesis}\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *