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Question Number 166592 by Ari last updated on 22/Feb/22

Answered by nurtani last updated on 23/Feb/22

$$x^2=17x+y...\left(i\right)$$$$y^2=x+17y...\left(ii\right)$$$$\left(i\right)-\left(ii\right):$$$$◼x^2-y^2=16x-16y$$$$\iff (x+y)\cancel{(x-y)}=16\cancel{(x-y)}$$$$\iff (x+y)=16$$$$◼x^2+y^2=18x+18y$$$$\iff x^2+y^2=18(x+y)$$$$\iff x^2+y^2=18\cdot \left(16\right)$$$$\iff x^2+y^2=288$$$$\iff x^2+y^2+1=288+1$$$$\iff x^2+y^2+1=289$$$$\therefore \sqrt{x^2+y^2+1}=\sqrt{289}=17$$

Commented by Ari last updated on 23/Feb/22

and -17

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