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

1 Answer
May 5, 2018

Answer:

When #y=2, x=4#

When #y=4,x=2#

Explanation:

#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! :)