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

Question

Savannah Anderson · Accepted Answer

#(x, y)# is one of:

#(1+sqrt(6), -1+sqrt(6))#
#(1-sqrt(6), -1-sqrt(6))#
#(5, -2)#
#(2, -5)#

Take note of this: #(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# Using the first provided equation, we obtain: #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# Thus: #(y+1)^2 = 6# Thus: #y + 1 = +-sqrt(6)# Thus #y = -1+-sqrt(6)# with corresponding values of #x# given by #x = y+2# Thus, remedies: #(x, y) = (1+sqrt(6), -1+sqrt(6))# #(x, y) = (1-sqrt(6), -1-sqrt(6))# #color(white)()# Case #x-y = 7# Using the first provided equation, we obtain: #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# Thus: #2y+7 = +-sqrt(9) = +-3# Thus: #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# Thus, the solutions: #(x, y) = (5, -2)# #(x, y) = (2, -5)# graph{(xy+3(x-y)-11) = 0 [-7.22, 12.78, -6.36, 3.64]}

Nathan Axel · Answer

undefined To solve the system of equations:
1. Expand the first equation: $ (x+y)^2 + 3(x-y) = 30 $ becomes $ x^2 + 2xy + y^2 + 3x - 3y = 30 $.
2. Expand the second equation: $ xy + 3(x-y) = 11 $ becomes $ xy + 3x - 3y = 11 $.
3. Rearrange both equations to isolate terms: $ x^2 + 2xy + y^2 + 3x - 3y - 30 = 0 $ and $ xy + 3x - 3y - 11 = 0 $.
4. Use substitution or elimination method to solve for one variable in terms of the other.
5. Once one variable is solved for, substitute its value back into one of the original equations to solve for the other variable.
6. Once both variables are found, verify the solutions by substituting them back into both original equations to ensure they satisfy both equations.