How do you solve #Y=2x^2-3# and #Y=3x-1#?

1 Answer
Sep 10, 2015

Answer:

#(x,y) = (-1/2, -2 1/2) or (2,5)#

Explanation:

If
[1]#color(white)("XXX")y = 2x^2-3#
and
[2]#color(white)("XXX")y=3x-1#
then
[3]#color(white)("XXX")2x^2-3=3x-1#

Simplifying
[4]#color(white)("XXX")2x^2-3x-2=0#
Factoring
[5]#color(white)("XXX")(2x+1)(x-2)=0#

[6]#color(white)("XXX")x=-1/2color(white)("XXX")or [7]color(white)("XXX")x=2#

#{: (color(white)("XXX"),x=-1/2,color(white)("XXXXXXX"),x=2), (color(white)("XXX"),y=2x^2-3,color(white)("XXXXXXX"),y=2x^2-3), (color(white)("XXX"),color(white)("X")=1/2-3,color(white)("XXXXXXX"),color(white)("X")=8-3), (color(white)("XXX"),color(white)("X")=-2 1/2,color(white)("XXXXXXX"),color(white)("X")=5) :}#