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Question Number 143832 by mathdanisur last updated on 18/Jun/21

x^2 + y^2 = (((56)/(13)))^2 and x + ((5y)/(12)) = ((56)/(12)) find x+y=?

$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\left(\frac{\mathrm{56}}{\mathrm{13}}\right)^{\mathrm{2}} \:{and}\:\:{x}\:+\:\frac{\mathrm{5}{y}}{\mathrm{12}}\:=\:\frac{\mathrm{56}}{\mathrm{12}} \\ $$$${find}\:\:{x}+{y}=? \\ $$

Answered by MJS_new last updated on 18/Jun/21

x^2 +y^2 =p x+qy=r ⇒ y=((r−x)/q) insert in the 1^(st) equation leads to x^2 −((2r)/(q^2 +1))x−((pq^2 −r^2 )/(q^2 +1))=0 solve this for x and you′re done

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={p} \\ $$$${x}+{qy}={r}\:\Rightarrow\:{y}=\frac{{r}−{x}}{{q}}\:\mathrm{insert}\:\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{equation} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{2}{r}}{{q}^{\mathrm{2}} +\mathrm{1}}{x}−\frac{{pq}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{q}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0} \\ $$$$\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:{x}\:\mathrm{and}\:\mathrm{you}'\mathrm{re}\:\mathrm{done} \\ $$

Commented by mathdanisur last updated on 19/Jun/21

thanks Sir..

$${thanks}\:{Sir}.. \\ $$

Answered by mitica last updated on 18/Jun/21

(1∙x+(5/(12))∙y)^2 ≤(1^2 +((5/(12)))^2 )(x^2 +y^2 )=(((56)/(12)))^2 ⇒ (x/1)=(y/(5/(12)))⇒y=((5x)/(12))⇒x+((25x)/(144))=((56)/(12))⇒x=((56∙12)/(169))=((48)/(13))⇒y=((20)/(13))

$$\left(\mathrm{1}\centerdot{x}+\frac{\mathrm{5}}{\mathrm{12}}\centerdot{y}\right)^{\mathrm{2}} \leqslant\left(\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{12}}\right)^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\left(\frac{\mathrm{56}}{\mathrm{12}}\right)^{\mathrm{2}} \Rightarrow \\ $$$$\frac{{x}}{\mathrm{1}}=\frac{{y}}{\frac{\mathrm{5}}{\mathrm{12}}}\Rightarrow{y}=\frac{\mathrm{5}{x}}{\mathrm{12}}\Rightarrow{x}+\frac{\mathrm{25}{x}}{\mathrm{144}}=\frac{\mathrm{56}}{\mathrm{12}}\Rightarrow{x}=\frac{\mathrm{56}\centerdot\mathrm{12}}{\mathrm{169}}=\frac{\mathrm{48}}{\mathrm{13}}\Rightarrow{y}=\frac{\mathrm{20}}{\mathrm{13}} \\ $$

Commented by mathdanisur last updated on 19/Jun/21

thanks sir, but answer: x+y=((952)/(169))

$${thanks}\:{sir},\:{but}\:{answer}:\:{x}+{y}=\frac{\mathrm{952}}{\mathrm{169}} \\ $$

Answered by mr W last updated on 19/Jun/21

x+y=s s−((7y)/(12))=((56)/(12))⇒y=((12s−56)/7) x=s−y=((−5s+56)/7) (((−5s+56)/7))^2 +(((12s−56)/7))^2 =(((56)/(13)))^2 s^2 −((1904)/(13^2 ))s+((906304)/(13^4 ))=0 (s−((952)/(13^2 )))^2 =0 ⇒s=((952)/(169))=x+y

$${x}+{y}={s} \\ $$$${s}−\frac{\mathrm{7}{y}}{\mathrm{12}}=\frac{\mathrm{56}}{\mathrm{12}}\Rightarrow{y}=\frac{\mathrm{12}{s}−\mathrm{56}}{\mathrm{7}} \\ $$$${x}={s}−{y}=\frac{−\mathrm{5}{s}+\mathrm{56}}{\mathrm{7}} \\ $$$$\left(\frac{−\mathrm{5}{s}+\mathrm{56}}{\mathrm{7}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{12}{s}−\mathrm{56}}{\mathrm{7}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{56}}{\mathrm{13}}\right)^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} −\frac{\mathrm{1904}}{\mathrm{13}^{\mathrm{2}} }{s}+\frac{\mathrm{906304}}{\mathrm{13}^{\mathrm{4}} }=\mathrm{0} \\ $$$$\left({s}−\frac{\mathrm{952}}{\mathrm{13}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{s}=\frac{\mathrm{952}}{\mathrm{169}}={x}+{y} \\ $$

Commented by mathdanisur last updated on 19/Jun/21

cool dear Sir thank you..

$${cool}\:{dear}\:{Sir}\:{thank}\:{you}.. \\ $$

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