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

$$x^2+y^2={\left(\frac{56}{13}\right)}^{2}andx+\frac{5y}{12}=\frac{56}{12}$$$$findx+y=?$$

Answered by MJS_new last updated on 18/Jun/21

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

Commented by mathdanisur last updated on 19/Jun/21

$$thanksSir..$$

Answered by mitica last updated on 18/Jun/21

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

Commented by mathdanisur last updated on 19/Jun/21

$$thankssir,butanswer:x+y=\frac{952}{169}$$

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

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

Commented by mathdanisur last updated on 19/Jun/21

$$cooldearSirthankyou..$$

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