How do you solve the system #16x^2-y^2+16y-128=0# and #y^2-48x-16y-32=0#?
1 Answer
Jul 27, 2016
Explanation:
If you add the two equations together, you get:
#16x^2-48x-160 = 0#
Divide through by
#x^2-3x-10 = 0#
Find a pair of factors of
#0 = x^2-3x-10 = (x-5)(x+2)#
Hence
Substituting
#0 = y^2-16y-272#
#= (y-8)^2-336#
#= (y-8)^2-(4sqrt(21))^2#
#= (y-8-4sqrt(21))(y-8+4sqrt(21))#
So
Substituting
#0 = y^2-16+64 = (y-8)^2#
So
Thus the solutions are:
#(x, y) = (5, 8+-4sqrt(21))# and#(x, y) = (-2, 8)#