How do you solve x² + y² = 20 and x + y = 6?

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Answer 1 · 2018-05-05T09:13:22

When $y=2, x=4$

When $y=4,x=2$

$x + y = 6$ $"...................Eq1"$

$x^2 + y^2 = 20$ $"..............Eq2"$

$"Using the identity":$ $color(red)(x^2+y^2=(x+y)^2-2xy$

$x^2+y^2=(x+y)^2-2xy$

Substituting : $x + y = 6 and x^2 + y^2 = 20$

$20=(6)^2-2xy$

$36-2xy=20$

$-2xy=-16$

$xy=8$

$color(blue)(x=8/y$

Substituting $x=8/y$ in $"Eq1"$

$8/y + y = 6$

$8+y^2=6y$

$y^2-6y+8=0$

$(y-2)(y-4)=0$

So, $color(darkred)(y=2 or y=4$

When $color(magenta)(y =2$

$x+2=6$

$color(magenta)(x=4$

When $color(darkorange)(y =4$

$x+4=6$

$color(darkorange)(x=2$

~Hope this helps! :)