Question

Solve the following simultaneous equations for $x\,and\,y$ :
$m(x + y) + n(x - y) - ({m^2} + mn + {n^2}) = 0$
$n(x + y) + m(x - y) - ({m^2} - mn + {n^2}) = 0$
(1) Let $x + y = a,x - y = b$ .
(2) Transfer the terms not containing $x\,and\,y$ to RHS.
(3) Add equations, substitute the value of $a\,and\,b$ and solve.

Answer 1

Hint: In order to solve this question, to solve for $x\,and\,y$ , for the given two equations, we will first let the given facts and then substitute the given values of variables to calculate further. And then we will solve for $x\,and\,y$ separately.

Complete step-by-step answer:
The given equation are:
$m(x + y) + n(x - y) - ({m^2} + mn + {n^2}) = 0$...........(1)
$n(x + y) + m(x - y) - ({m^2} - mn + {n^2}) = 0$............(2)
So, according to the (1) of the question;
Let $x + y = a,x - y = b$ :
From (1)
$
  \therefore ma + nb\, = {m^2} + mn + {n^2}\, \;
$
From (2)
$
  na + mb\, = {m^2} - mn + {n^2} \;
 $
Adding above two equations
$
 ma + nb + na + mb\, = \,2{m^2} + 2{n^2} \;
$
Grouping terms of m and n and taking common
$
   \Rightarrow ma + mb + na + nb = 2{m^2} + 2{n^2} \\
   \Rightarrow m(a + b) + n(a + b) = 2({m^2} + {n^2}) \;
$
now taking (a+b) common from the term
$
   \Rightarrow (a + b)(m + n) = 2({m^2} + {n^2}) \;
$
substituting a = x+y , b=x-y
$
   \Rightarrow (x + y + x - y)(m + n) = 2({m^2} + {n^2}) \\
   \Rightarrow 2x = \dfrac{{2({m^2} + {n^2})}}{{m + n}} \\
   \Rightarrow x = \dfrac{{{m^2} + {n^2}}}{{m + n}} \;
 $
Now, in equation (2):
$
  \therefore nx + ny + mx - my = {m^2} - mn + {n^2} \;
$
grouping common terms of x and y and taking common
$
   \Rightarrow mx + nx + ny - my = {m^2} - mn + {n^2} \\
   \Rightarrow x(m + n) + y(n - m) = {m^2} - mn + {n^2} \;
$
substituting the value of x
$
   \Rightarrow \dfrac{{{m^2} + {n^2}}}{{m + n}}(m + n) + y(n - m) = {m^2} + {n^2} - nm \;
 $
$
   \Rightarrow y(n - m) = - mn + {m^2} + {n^2} - {m^2} - {n^2} \\
   \Rightarrow y = \dfrac{{mn}}{{m - n}} \;
 $
Hence, the required values of $x\,and\,y$ are $\dfrac{{{m^2} + {n^2}}}{{m + n}}\,and\,\dfrac{{mn}}{{m - n}}$ respectively.

Note: When there are more linear equations than unknowns, finding a solution that solves all of the equations is almost impossible. Then one frequently seeks a solution that approximates all of the equations.