How do you solve the system of equations: #(x+y)^(1/2)+3(x-y)=30# and #xy+3(x-y) = 11# ?

1 Answer
Sep 20, 2016

Answer:

If we try to solve this algebraically then this one (unavoidably) turns into a quartic with rather messy solutions...

Explanation:

Given:

#{ ((x+y)^(1/2)+3(x-y)=30), (xy+3(x-y) = 11) :}#

We can isolate the terms in #(x+y)# and #xy# on the left hand side to get:

#{ ((x+y)^(1/2) = 3(10-(x-y))), (xy = 11 - 3(x-y)) :}#

Raising the first equation to the #4#th power and multiplying the second by #4# we get:

#{ ((x+y)^2 = 81(10-(x-y))^4), (4xy = 44 - 12(x-y)) :}#

Using the identity:

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

subtract the second equation from the first to get:

#(x-y)^2 = 81(10-(x-y))^4+12(x-y)-44#

Letting #t = x-y-10#, we have:

#(t+10)^2 = 81t^4+12(t+10)-44#

which expands to:

#t^2+20t+100 = 81t^4+12t+120-44#

Hence:

#81t^4-t^2-8t-24 = 0#

This quartic has two irrational Real roots and two non-Real Complex roots. The solution is way too messy to present here.

The Real roots are approximately:

#t_1 ~~ -0.69521#

#t_2 ~~ 0.78593#

For each of these, look at:

#xy + 3(x-y) = 11#

Using #x-y = t_n + 10#, we have:

#x(x-(t_n+10)) + 3(t_n+10) = 11#

Hence:

#x^2-(t_n+10)x+(3t_n+19) = 0#

Then you can use the quadratic formula to find possible values for #x# and hence for #y = x-(t_n+10)#

These #(x, y)# pairs then need to be checked against the equation:

#(x+y)^(1/2) + 3(x-y) = 30#

since we raised both sides of this equation to the #4#th power and thus introduced a spurious solution.

The valid solution of the original equation is approximately:

#(x, y) ~~ (6.82734, -2.47744)#

#color(white)()#
Conclusion

Attempting to solve this problem algebraically gets too messy. A numerical approach would probably be better.