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

1 Answer
May 5, 2018

When y=2, x=4y=2,x=4

When y=4,x=2y=4,x=2

Explanation:

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

x^2 + y^2 = 20x2+y2=20"..............Eq2"..............Eq2

"Using the identity":Using the identity:color(red)(x^2+y^2=(x+y)^2-2xyx2+y2=(x+y)22xy

x^2+y^2=(x+y)^2-2xyx2+y2=(x+y)22xy

Substituting : x + y = 6 and x^2 + y^2 = 20x+y=6andx2+y2=20

20=(6)^2-2xy20=(6)22xy

36-2xy=20362xy=20

-2xy=-162xy=16

xy=8xy=8

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

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

8/y + y = 68y+y=6

8+y^2=6y8+y2=6y

y^2-6y+8=0y26y+8=0

(y-2)(y-4)=0(y2)(y4)=0

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

When color(magenta)(y =2y=2

x+2=6x+2=6

color(magenta)(x=4x=4

When color(darkorange)(y =4y=4

x+4=6x+4=6

color(darkorange)(x=2x=2

~Hope this helps! :)