Question

m(x+y) + n(x-y)-(m2+ mn +n2)=0,n(x+y) + m(x-y)- (m2-mn+n2).1)let x+y=a,x-y=b 2)transfer the terms not containing x &y to RHS. 3) Add equations,substitute the values of a and b and solve.​

Answer 1

The solved equation is $2[x(m+n)-(m^2-n^2)]=0$

Step-by-step explanation:

Given equations are $m(x+y) + n(x-y)-(m^2+ mn +n^2)=0$

and ,$n(x+y) + m(x-y)- (m^2-mn+n^2)=0$

To find for the given conditions :

1) Let x+y=a,x-y=b

2) Transfer the terms not containing x &y to RHS.

3) Add equations,substitute the values of a and b and solve.​

• 1) Put x+y=a,x-y=b  in the given equations we get

$ma+nb-(m^2+ mn +n^2)=0$ and

,$na + mb- (m^2-mn+n^2)=0$

• 2) Transfer the terms not containing x &y to RHS is

$ma+nb-(m^2+ mn +n^2)=0$ and

,$na + mb- (m^2-mn+n^2)=0$

• 3) Add equations,substitute the values of a and b and solve.​

$ma+nb-(m^2+ mn +n^2)+na + mb- (m^2-mn+n^2)=0$

$ma+nb-m^2-mn-n^2+na +mb-m^2+mn-n^2=0$

$ma+nb-2m^2-2n^2+na +mb=0$

$(m+n)a+(m+n)b-2(m^2-n^2)=0$

$(m+n)(a+b)-2(m^2-n^2)=0$

Now substitute the values of a and b in the above equation we get

$(m+n)(x+y+x-y)-2(m^2-n^2)=0$

$(m+n)(2x)-2(m^2-n^2)=0$

$2x(m+n)-2(m^2-n^2)=0$

$2[x(m+n)-(m^2-n^2)]=0$

Therefore the solved equation is $2[x(m+n)-(m^2-n^2)]=0$