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# Find all pairs $(x,y)$ of real numbers $(x,y)$ such that $x + y = 10$ and $x^2 + y^2 = 56$.

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Find all pairs $(x,y)$ of real numbers $(x,y)$ such that $x + y = 10$ and $x^2 + y^2 = 56$.

May 5, 2020

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Given $$x+y=10$$ it follows, $$x=10-y$$ thus, $$x^2=(10-y)^2=(100-20y+y^2)$$ (1)

$$x^2+y^2=56$$  (2)

substitute (1) into (2)

$$100-20y+y^2+y^2=56$$ simplify and subtract 56 (Setting the quadratic equation equals zero)

$$2y^2-20y+44=0$$ Divide by 2

$$y^2-10y+22=0$$

Use the Quadratic formula

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$            (a=1, b=-10 , c=22)

$$x = {10 \pm \sqrt{(-10)^2-4*1*22} \over 2*1}=5+\sqrt{3}$$  and $$5-\sqrt{3}$$

Now using x+y=10 again, substitute the x value to find y

$$5+\sqrt{3}+y=10$$

$$y=10-(5+\sqrt{3})$$

$$y=5-\sqrt{3}$$

Substitute the other x value

$$5-\sqrt{3}+y=10$$

$$y=10-(5-\sqrt{3})$$

$$y=5+\sqrt{3}$$

Thus the pairs of real numbers that satisfy this:

$$(5+\sqrt{3},5-\sqrt{3})$$ & $$(5-\sqrt{3} ,5+\sqrt{3})$$

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May 5, 2020