How do you solve the system #(x+y)^2+3(x-y) = 30# and #xy+3(x-y) = 11# ?
1 Answer
#(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
#-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
#0 = (x-y)^2-9(x-y)+14#
#color(white)(0) = ((x-y)-2)((x-y)-7)#
So
Case
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
#4(y+1)^2 = 24#
Hence:
#(y+1)^2 = 6#
So:
#y + 1 = +-sqrt(6)#
So
#y = -1+-sqrt(6)#
with corresponding values of
So solutions:
#(x, y) = (1+sqrt(6), -1+sqrt(6))#
#(x, y) = (1-sqrt(6), -1-sqrt(6))#
Case
From the first given equation, we have:
#30 = (x+y)^2+3(x-y)#
#color(white)(30) = (2y+7)^2+21#
Subtract
#(2y+7)^2 = 9#
Hence:
#2y+7 = +-sqrt(9) = +-3#
So:
#2y = -7+3 = -4#
or:
#2y = -7-3 = -10#
Hence
Then we have corresponding values for
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]}