How do you solve the system #x^2+y^2=20# and #y=x-4#?
1 Answer
Nov 22, 2016
Explanation:
Since we are given y in terms of x, we can substitute directly into the other equation.
#rArrx^2+(x-4)^2=20larr" equation in one variable"#
#rArrx^2+x^2-8x+16-20=0larr" quadratic equation"#
#rArr2x^2-8x-4=0# divide both sides by 2
#x^2-4x-2=0# To solve use the
#color(blue)"quadratic formula"# with
#a=1,b=-4,c=-2#
#rArrx=(4+-sqrt24)/2=(4+-2sqrt6)/2=2+-sqrt6#
