How do you solve #x^2+2y=22# and #x^2-y=13#?

Question

1735 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-07T14:37:47

#-y = 13 - x^2#

#y = -13 + x^2#

#x^2 + 2(-13 + x^2) = 22#

#x^2 - 26 + 2x^2 = 22#

#3x^2 = 22 + 26#

#3x^2 = 48#

#x^2 = 16#

#x = +-4#

#4^2 - y = 13" and " -4^2 - y = 13#

#16 - y = 13" and "16 - y = 13#

#y = 3" and y = 3#

The solution set is therefore #{4, 3}# and #{-4, 3}#.

Hopefully this helps!

Answer 2 · 2016-07-07T14:43:14

#x=+-4#
#y=3#

#x^2+2y=22#
#x^2-y=13#
#3y=9#
#y=3#
#x^2=16#
#x=+-4#