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Question Number 138673 by KwesiDerek last updated on 16/Apr/21
x^2 −x=72 y^2 −y=30 x+y=4 x−y=?
$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}=\mathrm{72} \\ $$$$\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}=\mathrm{30} \\ $$$$\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}=\mathrm{4} \\ $$$$\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}=? \\ $$
Commented by Rasheed.Sindhi last updated on 16/Apr/21
^• x^2 −x−72=0 (x−9)(x+8)=0 x=9 ∨ x=−8 ^• y^2 −y−30=0 (y−6)(y+5)=0 y=6 ∨ y=−5 ^• x+y=4 9+(−5)=4 (x,y)=(9,−5) satisfying all the three equations. x−y=9−(−5)=14
$$\:^{\bullet} {x}^{\mathrm{2}} −{x}−\mathrm{72}=\mathrm{0} \\ $$$$\left({x}−\mathrm{9}\right)\left({x}+\mathrm{8}\right)=\mathrm{0} \\ $$$${x}=\mathrm{9}\:\vee\:{x}=−\mathrm{8} \\ $$$$\:^{\bullet} {y}^{\mathrm{2}} −{y}−\mathrm{30}=\mathrm{0} \\ $$$$\:\:\left({y}−\mathrm{6}\right)\left({y}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\:\:\:{y}=\mathrm{6}\:\vee\:{y}=−\mathrm{5} \\ $$$$\:^{\bullet} {x}+{y}=\mathrm{4} \\ $$$$\:\:\mathrm{9}+\left(−\mathrm{5}\right)=\mathrm{4} \\ $$$$\left({x},{y}\right)=\left(\mathrm{9},−\mathrm{5}\right)\:{satisfying}\:{all}\:{the}\:{three} \\ $$$${equations}. \\ $$$${x}−{y}=\mathrm{9}−\left(−\mathrm{5}\right)=\mathrm{14} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Apr/21
So I think nimnim is right here.
$$\mathcal{S}{o}\:{I}\:{think}\:{nimnim}\:{is}\:{right}\:{here}. \\ $$
Commented by MJS_new last updated on 17/Apr/21
sorry for my mistake. I′m the Typo Master!!!
$$\mathrm{sorry}\:\mathrm{for}\:\mathrm{my}\:\mathrm{mistake}.\:\mathrm{I}'\mathrm{m}\:\mathrm{the}\:\mathrm{Typo}\:\mathrm{Master}!!! \\ $$
Answered by MJS_new last updated on 16/Apr/21
x=−9∨x=8 y=−5∨y=6 ⇒ x+y≠4
$${x}=−\mathrm{9}\vee{x}=\mathrm{8} \\ $$$${y}=−\mathrm{5}\vee{y}=\mathrm{6} \\ $$$$\Rightarrow\:{x}+{y}\neq\mathrm{4} \\ $$
Commented by Rasheed.Sindhi last updated on 17/Apr/21
Typo sir x=+9 ∨ x=−8
$${Typo}\:{sir}\:{x}=+\mathrm{9}\:\vee\:{x}=−\mathrm{8} \\ $$
Answered by nimnim last updated on 16/Apr/21
x^2 −x=72......(i) y^2 −y=30.......(ii) x+y=4..........(iii) (i)+(ii)⇒x^2 +y^2 −x−y=102 ⇒(x+y)^2 −2xy−(x+y)=102 ⇒(4)^2 −2xy−4=102 {using (iii)} ⇒xy=−45 (iii)→(x+y)^2 =4^2 ⇒(x−y)^2 +4xy=16 ⇒(x−y)^2 +4(−45)=16 ⇒(x−y)^2 =16+180 ⇒(x−y)=±(√(196))=±14 ⇒(x−y)=14 or (x−y)=−14 (neglected) ∴ (x−y)=14★
$${x}^{\mathrm{2}} −{x}=\mathrm{72}……\left({i}\right) \\ $$$${y}^{\mathrm{2}} −{y}=\mathrm{30}…….\left({ii}\right) \\ $$$${x}+{y}=\mathrm{4}……….\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{x}−{y}=\mathrm{102} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}−\left({x}+{y}\right)=\mathrm{102} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{4}\right)^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{4}=\mathrm{102}\:\:\:\:\left\{{using}\:\left({iii}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{xy}=−\mathrm{45} \\ $$$$\left({iii}\right)\rightarrow\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$$\:\:\:\Rightarrow\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{4}{xy}=\mathrm{16} \\ $$$$\:\:\:\Rightarrow\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{4}\left(−\mathrm{45}\right)=\mathrm{16} \\ $$$$\:\:\:\Rightarrow\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{16}+\mathrm{180} \\ $$$$\:\:\:\Rightarrow\left({x}−{y}\right)=\pm\sqrt{\mathrm{196}}=\pm\mathrm{14} \\ $$$$\:\:\:\Rightarrow\left({x}−{y}\right)=\mathrm{14}\:{or}\:\left({x}−{y}\right)=−\mathrm{14}\:\left({neglected}\right) \\ $$$$\therefore\:\left({x}−{y}\right)=\mathrm{14}\bigstar \\ $$
Commented by MJS_new last updated on 16/Apr/21
this method is wrong. a pair (x, y) is a solution of the system { ((x^2 −x=30)),((y^2 −y=72)),((x+y=4)) :} only if it solves each of the equations of this system.
$$\mathrm{this}\:\mathrm{method}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{a}\:\mathrm{pair}\:\left({x},\:{y}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{system}\:\begin{cases}{{x}^{\mathrm{2}} −{x}=\mathrm{30}}\\{{y}^{\mathrm{2}} −{y}=\mathrm{72}}\\{{x}+{y}=\mathrm{4}}\end{cases}\:\mathrm{only}\:\mathrm{if}\:\mathrm{it}\:\mathrm{solves}\:\mathrm{each} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{this}\:\mathrm{system}. \\ $$
Commented by nimnim last updated on 16/Apr/21
I think x=9 and y=−5
$${I}\:{think}\:{x}=\mathrm{9}\:{and}\:{y}=−\mathrm{5} \\ $$
Commented by MJS_new last updated on 16/Apr/21
your method: (1) 25+5=17 (2) 49+7=69 (1)+(2) 25+49+5+7=86 true
$$\:\mathrm{your}\:\mathrm{method}: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{25}+\mathrm{5}=\mathrm{17} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{49}+\mathrm{7}=\mathrm{69} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\mathrm{25}+\mathrm{49}+\mathrm{5}+\mathrm{7}=\mathrm{86}\:\mathrm{true} \\ $$
Commented by MJS_new last updated on 16/Apr/21
if x=9 the first equation is wrong if y=−5 the second equation is wrong
$$\mathrm{if}\:{x}=\mathrm{9}\:\mathrm{the}\:\mathrm{first}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\mathrm{if}\:{y}=−\mathrm{5}\:\mathrm{the}\:\mathrm{second}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Commented by nimnim last updated on 16/Apr/21
If x=9, 1st equation=9^2 −9=81−9=72 true If y=−5, 2nd equation=(−5)^2 −(−5)=25+5=30 true 3rd equation=9+(−5)=4 true But if you say so, it might be wrong. I am just a beginner, reading in 8 standard. I dont know much about maths like you do. Thanks for your help, Sir.
$${If}\:{x}=\mathrm{9},\:\mathrm{1}{st}\:{equation}=\mathrm{9}^{\mathrm{2}} −\mathrm{9}=\mathrm{81}−\mathrm{9}=\mathrm{72}\:{true} \\ $$$${If}\:{y}=−\mathrm{5},\:\mathrm{2}{nd}\:{equation}=\left(−\mathrm{5}\right)^{\mathrm{2}} −\left(−\mathrm{5}\right)=\mathrm{25}+\mathrm{5}=\mathrm{30}\:{true} \\ $$$$\mathrm{3}{rd}\:{equation}=\mathrm{9}+\left(−\mathrm{5}\right)=\mathrm{4}\:{true} \\ $$$$ \\ $$$${But}\:{if}\:{you}\:{say}\:{so},\:{it}\:{might}\:{be}\:{wrong}.\:{I}\:{am}\: \\ $$$${just}\:{a}\:{beginner},\:{reading}\:{in}\:\mathrm{8}\:{standard}.\:{I}\:{dont} \\ $$$${know}\:{much}\:{about}\:{maths}\:{like}\:{you}\:{do}. \\ $$$${Thanks}\:{for}\:{your}\:{help},\:{Sir}. \\ $$$$\:\: \\ $$
Answered by ajfour last updated on 16/Apr/21
x+y=4 x−y=2s x=s+2 y=2−s s^2 +4s+4−s−2=72 s^2 +3s−70=0 ...(i) & 4−4s+s^2 −2+s=30 s^2 −3s−28=0 ..(ii) from (i) s=((−3±(√(9+280)))/2) s=7, −10 from (ii) s=((3±(√(9+112)))/2) s=7, −4 Hence s=7 x−y=2s=14
$${x}+{y}=\mathrm{4} \\ $$$${x}−{y}=\mathrm{2}{s} \\ $$$${x}={s}+\mathrm{2} \\ $$$${y}=\mathrm{2}−{s} \\ $$$${s}^{\mathrm{2}} +\mathrm{4}{s}+\mathrm{4}−{s}−\mathrm{2}=\mathrm{72} \\ $$$${s}^{\mathrm{2}} +\mathrm{3}{s}−\mathrm{70}=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\&\:\:\mathrm{4}−\mathrm{4}{s}+{s}^{\mathrm{2}} −\mathrm{2}+{s}=\mathrm{30} \\ $$$${s}^{\mathrm{2}} −\mathrm{3}{s}−\mathrm{28}=\mathrm{0}\:\:\:..\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:\:{s}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}+\mathrm{280}}}{\mathrm{2}} \\ $$$${s}=\mathrm{7},\:−\mathrm{10} \\ $$$${from}\:\left({ii}\right) \\ $$$${s}=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}+\mathrm{112}}}{\mathrm{2}} \\ $$$${s}=\mathrm{7},\:−\mathrm{4} \\ $$$${Hence} \\ $$$${s}=\mathrm{7} \\ $$$${x}−{y}=\mathrm{2}{s}=\mathrm{14} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Apr/21
x^2 −x=72 ∧ y^2 −y=30 ∧ x+y=4 y=4−x : y^2 −y=30⇒(4−x)^2 −(4−x)=30 16−8x+x^2 −4+x=30 x^2 −7x−18=0 ( x−9)(x+2)=0 x=9 ∨ x=−2 x=9 also satisfy x^2 −x=72 Hence x=9 y=4−x=4−9=−5 (x,y)=(9,−5) x−y=9−(−5)=14
$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}=\mathrm{72}\:\:\wedge\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}=\mathrm{30}\:\wedge\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}=\mathrm{4} \\ $$$${y}=\mathrm{4}−{x}\:: \\ $$$${y}^{\mathrm{2}} −{y}=\mathrm{30}\Rightarrow\left(\mathrm{4}−{x}\right)^{\mathrm{2}} −\left(\mathrm{4}−{x}\right)=\mathrm{30} \\ $$$$\:\:\:\mathrm{16}−\mathrm{8}{x}+{x}^{\mathrm{2}} −\mathrm{4}+{x}=\mathrm{30} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{7}{x}−\mathrm{18}=\mathrm{0} \\ $$$$\:\:\:\:\:\left(\:{x}−\mathrm{9}\right)\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:{x}=\mathrm{9}\:\vee\:{x}=−\mathrm{2}\: \\ $$$${x}=\mathrm{9}\:{also}\:{satisfy}\:{x}^{\mathrm{2}} −{x}=\mathrm{72} \\ $$$${Hence}\:{x}=\mathrm{9} \\ $$$${y}=\mathrm{4}−{x}=\mathrm{4}−\mathrm{9}=−\mathrm{5} \\ $$$$\left({x},{y}\right)=\left(\mathrm{9},−\mathrm{5}\right) \\ $$$${x}−{y}=\mathrm{9}−\left(−\mathrm{5}\right)=\mathrm{14} \\ $$

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