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Question Number 16600 by tawa tawa last updated on 24/Jun/17

Solve simultaneously  x^2  + y^2  = 61         .............. equation (i)  x^3  − y^3  = 91         .............. equation (ii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{61}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$

Commented by prakash jain last updated on 24/Jun/17

x=(√(61))cos u,y=(√(61))sin u  61^(3/2) cos^3 u−61^(3/2) sin^3 u=91  (cos u−sin u)(cos^2 u+sin^2 u−cos usin u)         =((91)/(61^(3/2) ))  (cos u−sin u)(1−cos usin u)=.191  cos  ((π/4)−u)(2−sin 2u)=.191×2(√2)=.54  2cos ((π/4)−u)−cos ((π/4)−u)sin 2u=.54  continue

$${x}=\sqrt{\mathrm{61}}\mathrm{cos}\:{u},{y}=\sqrt{\mathrm{61}}\mathrm{sin}\:{u} \\ $$$$\mathrm{61}^{\mathrm{3}/\mathrm{2}} \mathrm{cos}^{\mathrm{3}} {u}−\mathrm{61}^{\mathrm{3}/\mathrm{2}} \mathrm{sin}^{\mathrm{3}} {u}=\mathrm{91} \\ $$$$\left(\mathrm{cos}\:{u}−\mathrm{sin}\:{u}\right)\left(\mathrm{cos}^{\mathrm{2}} {u}+\mathrm{sin}^{\mathrm{2}} {u}−\mathrm{cos}\:{u}\mathrm{sin}\:{u}\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{91}}{\mathrm{61}^{\mathrm{3}/\mathrm{2}} } \\ $$$$\left(\mathrm{cos}\:{u}−\mathrm{sin}\:{u}\right)\left(\mathrm{1}−\mathrm{cos}\:{u}\mathrm{sin}\:{u}\right)=.\mathrm{191} \\ $$$$\mathrm{cos}\:\:\left(\frac{\pi}{\mathrm{4}}−{u}\right)\left(\mathrm{2}−\mathrm{sin}\:\mathrm{2}{u}\right)=.\mathrm{191}×\mathrm{2}\sqrt{\mathrm{2}}=.\mathrm{54} \\ $$$$\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{4}}−{u}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−{u}\right)\mathrm{sin}\:\mathrm{2}{u}=.\mathrm{54} \\ $$$${continue} \\ $$

Commented by tawa tawa last updated on 24/Jun/17

Am with you sir. God bless you sir.

$$\mathrm{Am}\:\mathrm{with}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tinkutara last updated on 24/Jun/17

Hit and trial method:  y^2  = 61 − x^2   Let x, y ∈ N.  Then y^2  = 60, 57, 52, 45, 36, 25, 12.  But x > y. So x = 6 and y = 5.

$$\mathrm{Hit}\:\mathrm{and}\:\mathrm{trial}\:\mathrm{method}: \\ $$$${y}^{\mathrm{2}} \:=\:\mathrm{61}\:−\:{x}^{\mathrm{2}} \\ $$$$\mathrm{Let}\:{x},\:{y}\:\in\:{N}. \\ $$$$\mathrm{Then}\:{y}^{\mathrm{2}} \:=\:\mathrm{60},\:\mathrm{57},\:\mathrm{52},\:\mathrm{45},\:\mathrm{36},\:\mathrm{25},\:\mathrm{12}. \\ $$$$\mathrm{But}\:{x}\:>\:{y}.\:\mathrm{So}\:{x}\:=\:\mathrm{6}\:\mathrm{and}\:{y}\:=\:\mathrm{5}. \\ $$

Commented by tawa tawa last updated on 24/Jun/17

Am with you sir. God bless you sir.

$$\mathrm{Am}\:\mathrm{with}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17

Commented by mrW1 last updated on 26/Jun/17

geogebra is very powerful!

$$\mathrm{geogebra}\:\mathrm{is}\:\mathrm{very}\:\mathrm{powerful}! \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17

please correct your final answer  from :(6,−5)⇒(6,5).cxcuse me.

$${please}\:{correct}\:{your}\:{final}\:{answer} \\ $$$${from}\::\left(\mathrm{6},−\mathrm{5}\right)\Rightarrow\left(\mathrm{6},\mathrm{5}\right).{cxcuse}\:{me}. \\ $$

Commented by mrW1 last updated on 26/Jun/17

certainly you are right. thanks!

$$\mathrm{certainly}\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}.\:\mathrm{thanks}! \\ $$

Answered by mrW1 last updated on 26/Jun/17

x^2 +y^2 −2xy+2xy=61  (x−y)^2 +2xy=61  x^3 −y^3 =(x−y)(x^2 +xy+y^2 )=91  (x−y)(61+xy)=91    let u=x−y and v=xy  u^2 +2v=61   ...(1)  u(61+v)=91   ...(2)  v=((61−u^2 )/2)  u(61+((61−u^2 )/2))=91  u(((183−u^2 )/2))=91  u^3 −183u+182=0  (u−1)(u−13)(u+14)=0  ⇒u=(−14, 1, 13)  ⇒v=((61−(−14)^2 )/2)=−((135)/2)  ⇒v=((61−(1)^2 )/2)=30  ⇒v=((61−(13)^2 )/2)=−54  ⇒v=(−((135)/2), 30, −54)    x−y=u   ...(3)  xy=v  (x+y)^2 =(x−y)^2 +4xy=u^2 +4v=61+2v≥0  v≥−((61)/2)  ⇒for real roots the only solution is   u=1  v=30    x+y=±(√(61+2v))     ...(4)    from (3) and (4):  ⇒x=((u±(√(61+2v)))/2)  ⇒y=((−u±(√(61+2v)))/2)    with (u,v)=(1,30)  x=((1±(√(61+2×30)))/2)=((1±11)/2)=6,−5  y=((−1±(√(61+2×30)))/2)=((−1±11)/2)=5,−6    ⇒real solution is:  (x,y)=(6,5) or (−5,−6)

$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{2xy}=\mathrm{61} \\ $$$$\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{2xy}=\mathrm{61} \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} =\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{xy}+\mathrm{y}^{\mathrm{2}} \right)=\mathrm{91} \\ $$$$\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{61}+\mathrm{xy}\right)=\mathrm{91} \\ $$$$ \\ $$$$\mathrm{let}\:\mathrm{u}=\mathrm{x}−\mathrm{y}\:\mathrm{and}\:\mathrm{v}=\mathrm{xy} \\ $$$$\mathrm{u}^{\mathrm{2}} +\mathrm{2v}=\mathrm{61}\:\:\:...\left(\mathrm{1}\right) \\ $$$$\mathrm{u}\left(\mathrm{61}+\mathrm{v}\right)=\mathrm{91}\:\:\:...\left(\mathrm{2}\right) \\ $$$$\mathrm{v}=\frac{\mathrm{61}−\mathrm{u}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{u}\left(\mathrm{61}+\frac{\mathrm{61}−\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}\right)=\mathrm{91} \\ $$$$\mathrm{u}\left(\frac{\mathrm{183}−\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}\right)=\mathrm{91} \\ $$$$\mathrm{u}^{\mathrm{3}} −\mathrm{183u}+\mathrm{182}=\mathrm{0} \\ $$$$\left(\mathrm{u}−\mathrm{1}\right)\left(\mathrm{u}−\mathrm{13}\right)\left(\mathrm{u}+\mathrm{14}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{u}=\left(−\mathrm{14},\:\mathrm{1},\:\mathrm{13}\right) \\ $$$$\Rightarrow\mathrm{v}=\frac{\mathrm{61}−\left(−\mathrm{14}\right)^{\mathrm{2}} }{\mathrm{2}}=−\frac{\mathrm{135}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{v}=\frac{\mathrm{61}−\left(\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{30} \\ $$$$\Rightarrow\mathrm{v}=\frac{\mathrm{61}−\left(\mathrm{13}\right)^{\mathrm{2}} }{\mathrm{2}}=−\mathrm{54} \\ $$$$\Rightarrow\mathrm{v}=\left(−\frac{\mathrm{135}}{\mathrm{2}},\:\mathrm{30},\:−\mathrm{54}\right) \\ $$$$ \\ $$$$\mathrm{x}−\mathrm{y}=\mathrm{u}\:\:\:...\left(\mathrm{3}\right) \\ $$$$\mathrm{xy}=\mathrm{v} \\ $$$$\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} =\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{4xy}=\mathrm{u}^{\mathrm{2}} +\mathrm{4v}=\mathrm{61}+\mathrm{2v}\geqslant\mathrm{0} \\ $$$$\mathrm{v}\geqslant−\frac{\mathrm{61}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{for}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\: \\ $$$$\mathrm{u}=\mathrm{1} \\ $$$$\mathrm{v}=\mathrm{30} \\ $$$$ \\ $$$$\mathrm{x}+\mathrm{y}=\pm\sqrt{\mathrm{61}+\mathrm{2v}}\:\:\:\:\:...\left(\mathrm{4}\right) \\ $$$$ \\ $$$$\mathrm{from}\:\left(\mathrm{3}\right)\:\mathrm{and}\:\left(\mathrm{4}\right): \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{u}\pm\sqrt{\mathrm{61}+\mathrm{2v}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{y}=\frac{−\mathrm{u}\pm\sqrt{\mathrm{61}+\mathrm{2v}}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{with}\:\left(\mathrm{u},\mathrm{v}\right)=\left(\mathrm{1},\mathrm{30}\right) \\ $$$$\mathrm{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{61}+\mathrm{2}×\mathrm{30}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\mathrm{11}}{\mathrm{2}}=\mathrm{6},−\mathrm{5} \\ $$$$\mathrm{y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{61}+\mathrm{2}×\mathrm{30}}}{\mathrm{2}}=\frac{−\mathrm{1}\pm\mathrm{11}}{\mathrm{2}}=\mathrm{5},−\mathrm{6} \\ $$$$ \\ $$$$\Rightarrow\mathrm{real}\:\mathrm{solution}\:\mathrm{is}: \\ $$$$\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{6},\mathrm{5}\right)\:\mathrm{or}\:\left(−\mathrm{5},−\mathrm{6}\right) \\ $$

Commented by tawa tawa last updated on 24/Jun/17

God bless you sir. i really appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}.\: \\ $$

Commented by tawa tawa last updated on 24/Jun/17

God bless you sir. i really appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}.\: \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/Jun/17

perfect! & cxcellent.

$${perfect}!\:\&\:{cxcellent}. \\ $$

Commented by RasheedSoomro last updated on 25/Jun/17

e^x cellent!

$$\mathrm{e}^{\mathrm{x}} \mathrm{cellent}! \\ $$

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