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How do you solve $x^{2} + y^{2} = 32$ $x^2 + y^2 = 32$ and $y - x = 0$ $y - x = 0$ ?

1 Answer

Sep 17, 2016

$x_{1} = 4 : y_{1} = 4; x_{2} = - 4; y_{2} = - 4$ $x_1=4: y_1=4; x_2=-4; y_2=-4$

Explanation:

$y - x = 0 \Rightarrow y = x$ $y-x=0 rArr y=x$

$x^{2} + x^{2} = 32; 2 x^{2} = 32 : x^{2} = 16; x_{1} = 4; x_{2} = - 4$ $x^2+x^2=32; 2x^2=32: x^2=16; x_1=4; x_2=-4$

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