What are the solution(s) to the following system of equations #y = -x^2# and y = x?
1 Answer
Aug 8, 2016
x = 0 , x = - 1
Explanation:
Since we are given 2 values that y equates to we can equate the right sides.
#rArrx=-x^2rArrx^2+x=0# factorising :
#x(x+1)=0#
#rArrx=0" or " x+1=0rArrx=-1# The solutions are
#x=0,x=-1#
