A War with Modulus

The second equation can be written as :

$(2|x|-1)(a|y|-1)=0$ . We have , $|x|=\frac{1}{2},|y|=\frac{1}{a}$ .

This in turn implies $a>0$ as $|y|>0$ .

We have from first equation , $|\frac{1}{2}-\frac{1}{a}|=1$ . Since $\frac{1}{a}>1$ Which implies $\frac{1}{2}-\frac{1}{a}<0$ .

$\frac{1}{a}-\frac{1}{2}=1$ .

$a=\frac{2}{3}\implies \boxed{3a=2}$

$2a|x||y|+1=2|x|+a|y|,~~~\implies~~(1-2|x|)=a|y|(1-2|x|).\\ \therefore~~(1-2|x|)*(1 - a|y|)=0.\\ So~|x|=\dfrac 1 2 ....(A) ~~or/and~~a|y|=1....(B)\\ Substituting~(A)~in~\Bigg |~|x| - |y|~\Bigg |=1,~~we~have~\Bigg |~\dfrac 1 2 - |y|~\Bigg |=1.~\implies~|y|=\dfrac 3 2.\\ If~ both~(A)~and~(B)~are~true,~~a|y|=1,~\implies~a*\dfrac 3 2=1,~that ~is~a=\dfrac 2 3< 1....(C)\\ Let~ us~ check.~LHS~of~equation~two=2*\dfrac 2 3*\dfrac 1 2*\dfrac 2 3 +1=2.~~RHS=2*\dfrac 1 2 + \dfrac 2 3 *\dfrac 3 2=2.\\ So~LHS=RHS, ~so~assumption~is~true.~~So~3a=3*\dfrac 2 3 =\large ~~\color{#D61F06}{2}.$