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

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Answer 1 · 2016-07-07T14:33:35

#y = 6 - x#

By substitution

#:.x^2 + (6 - x)^2 = 20#

#x^2 + 36 - 12x + x^2 = 20#

#2x^2 - 12x + 16 = 0#

#2(x^2 - 6x + 8) = 0#

#2(x - 4)(x - 2) = 0#

#x = 4 and 2#

#:.x + y = 6#

#4 + y = 6" and " 2 + y = 6#

#y = 2" and " y = 4#

Hence, the solution set is #{4, 2}# and #{2, 4}#.

Hopefully this helps!