How do you solve the system #16x^2-y^2+16y-128=0# and #y^2-48x-16y-32=0#?

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Answer 1 · 2016-07-27T13:25:07

#(x, y) = (5, 8+-4sqrt(21))# or #(x, y) = (-2, 8)#

If you add the two equations together, you get:

#16x^2-48x-160 = 0#

Divide through by #16# to get:

#x^2-3x-10 = 0#

Find a pair of factors of #10# which differ by #3#. The pair #5, 2# works, so:

#0 = x^2-3x-10 = (x-5)(x+2)#

Hence #x=5# or #x=-2#

#color(white)()#
Substituting #x=5# in the second equation we get:

#0 = y^2-16y-272#

#= (y-8)^2-336#

#= (y-8)^2-(4sqrt(21))^2#

#= (y-8-4sqrt(21))(y-8+4sqrt(21))#

So #y = 8 +-4sqrt(21)#

#color(white)()#
Substituting #x=-2# in the second equation we get:

#0 = y^2-16+64 = (y-8)^2#

So #y = 8#

Thus the solutions are:

#(x, y) = (5, 8+-4sqrt(21))# and #(x, y) = (-2, 8)#