Question

Solve the following simultaneous equations:-
\[m(x + y) + n(x - y) - ({m^2} + mn + {n^2}) = 0\]
\[n(x + y) + m(x - y) - ({m^2} - mn + {n^2}) = 0\]
Hence, m and n are constants.

Answer 1

Hint: We will first rearrange both the equation and write in terms of x and y. After that, we will solve these equations by adding or subtracting them (basically elimination method) and then solve that to find the required solution.

Complete step-by-step answer:
Let us consider the first equation initially:
\[m(x + y) + n(x - y) - ({m^2} + mn + {n^2}) = 0\]
We can rewrite this as:-
\[ \Rightarrow mx + my + nx - ny = {m^2} + mn + {n^2}\]
Taking x and y common to write this as follows:-
\[ \Rightarrow (m + n)x + (m - n)y = {m^2} + mn + {n^2}\] ……….(1)
Now, considering the second equation that is:-
\[n(x + y) + m(x - y) - ({m^2} - mn + {n^2}) = 0\]
We can rewrite this as:-
\[ \Rightarrow nx + ny + mx - my = {m^2} - mn + {n^2}\]
Taking x and y common to write this as follows:-
\[ \Rightarrow (m + n)x - (m - n)y = {m^2} - mn + {n^2}\] ……….(2)
Now, adding (1) and (2) to get:-
$ \Rightarrow 2(m + n)x = {m^2} + mn + {n^2} + {m^2} - mn + {n^2}$
Simplifying the RHS by clubbing the like terms, we will get:-
$ \Rightarrow 2(m + n)x = 2({m^2} + {n^2})$
Simplifying it to get as follows:-
$ \Rightarrow (m + n)x = {m^2} + {n^2}$
Taking (m + n) from LHS to RHS:-
$ \Rightarrow x = \dfrac{{{m^2} + {n^2}}}{{m + n}}$ ……………(3)
Now, subtracting (1) and (2) that is equation (1) – equation (2) to get:-
$ \Rightarrow 2(m - n)y = {m^2} + mn + {n^2} - {m^2} + mn - {n^2}$
Simplifying the RHS by clubbing the like terms, we will get:-
$ \Rightarrow 2(m - n)u = 2mn$
Simplifying it to get as follows:-
$ \Rightarrow (m - n)y = mn$
Taking (m – n) from LHS to RHS:-
$ \Rightarrow y = \dfrac{{mn}}{{m - n}}$ ……………(4)
Now, using the equation (3) and (4), we have the required answer that is:-

$ \Rightarrow x = \dfrac{{{m^2} + {n^2}}}{{m + n}}$ and $y = \dfrac{{mn}}{{m - n}}$

Additional Information: Simultaneous equations are a system of equations that are all true together. You must find an answer or answers that work for all the equations at the same time. For example, if you’re working with two simultaneous equations, even though there may be a solution that makes one of the equations true, you must find the solution that makes both equations true. Simultaneous equations can be used to solve everyday problems, especially those that are more difficult to think through without writing anything down.

Note: The students must note that after you get the answer, you must notice the denominators in both x and y, and you will get one condition for both of them to be true that will be ${m^2} \ne {n^2}$ because if ${m^2} = {n^2}$, then either $m = n$ or $m = - n$ from which x and y will come to be undefined. So, always pay attention to these little details.