How do you solve #x+y=3# and #x = 4-(y-1)^2#?

1 Answer
Alan P.
Apr 9, 2016

#(x,y)=(0,3)#
or
#(x,y)=(3,0)#

Explanation:

If #x+y=3#
then #y=3-x#

and we can substitute #(color(red)(3-x))# in place of #y# in the second equation: #x=4-(y-1)^2#

This gives
#color(white)("XXX")x=4-(color(red)(3-x)-1)^2#

#color(white)("XXX")x=4-(2-x)^2#

#color(white)("XXX")x=4-(4-4x+x^2)#

#color(white)("XXX")x= 4x-x^2

#color(white)("XXX")3x-x^2=0#

#color(white)("XXX")x(3-x)=0#

#color(white)("XXX")rarr x=0color(white)("XX")orcolor(white)("XX")x=3#

If #x=0# then #x+y=3 rarr y=3#

If #x=3# then #x+y=3 rarr y=0#

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