How do you solve the system #(x+y)^2+3(x-y) = 30# and #xy+3(x-y) = 11# ?

1 Answer
Sep 18, 2016

Answer:

#(x, y)# is one of:

#(1+sqrt(6), -1+sqrt(6))#

#(1-sqrt(6), -1-sqrt(6))#

#(5, -2)#

#(2, -5)#

Explanation:

Notice that:

#(x+y)^2-4xy = (x-y)^2#

So we can get a quadratic in #(x-y)# by subtracting #4# times the second equation from the first...

#-14 = 30 - 4*11#

#color(white)(-14) = ((x+y)^2+3(x-y))-4(xy + 3(x-y))#

#color(white)(-14) = ((x+y)^2-4xy)+(3-12)(x-y)#

#color(white)(-14) = (x-y)^2-9(x-y)#

Add #14# to both ends to get:

#0 = (x-y)^2-9(x-y)+14#

#color(white)(0) = ((x-y)-2)((x-y)-7)#

So #x-y = 2# or #x-y = 7#

#color(white)()#
Case #x-y = 2#

From the first given equation, we have:

#30 = (x+y)^2+3(x-y)#

#color(white)(30) = (2y+2)^2+6 = 4(y+1)^2 + 6#

Subtract #6# from both ends and transpose to get:

#4(y+1)^2 = 24#

Hence:

#(y+1)^2 = 6#

So:

#y + 1 = +-sqrt(6)#

So

#y = -1+-sqrt(6)#

with corresponding values of #x# given by #x = y+2#

So solutions:

#(x, y) = (1+sqrt(6), -1+sqrt(6))#

#(x, y) = (1-sqrt(6), -1-sqrt(6))#

#color(white)()#
Case #x-y = 7#

From the first given equation, we have:

#30 = (x+y)^2+3(x-y)#

#color(white)(30) = (2y+7)^2+21#

Subtract #21# from both ends and transpose to get:

#(2y+7)^2 = 9#

Hence:

#2y+7 = +-sqrt(9) = +-3#

So:

#2y = -7+3 = -4#

or:

#2y = -7-3 = -10#

Hence #y = -2# or #y = -5#

Then we have corresponding values for #x# using #x = y+7#

Hence solutions:

#(x, y) = (5, -2)#

#(x, y) = (2, -5)#

graph{((x+y)^2+3(x-y)-30)(xy+3(x-y)-11) = 0 [-7.22, 12.78, -6.36, 3.64]}