How do you solve the system #Y=x+7#, #x^2+y^2=25#?
1 Answer
Jan 1, 2016
Substitute
#(x, y) = (-3, 4) or (-4, 3)#
Explanation:
Substitute
#25= x^2+y^2#
#=x^2+(x+7)^2#
#=x^2+x^2+14x+49#
#=2x^2+14x+49#
Subtract
#2x^2+14x+24 = 0#
Divide through by
#x^2+7x+12 = 0#
We can factor this by finding a pair of factors of
Hence:
#0 = x^2+7x+12 = (x+3)(x+4)#
This has roots