A Beautiful Equation

$\begin{cases} 2x + 3y = 7 \\ a(x+y)-b(x-y) = 3a+b-2 \end{cases} \\ \Rightarrow \begin{cases} 2x + 3y = 7 \\ (a-b)x+(a+b)y = 3a+b-2 \end{cases}$

For the two equations to have infinite solutions, they must be identical.

$\Rightarrow \begin{cases} \dfrac {a+b}{a-b} = \dfrac{3}{2} & \Rightarrow 2a+2b = 3a-3b & \Rightarrow a = 5b \\ \dfrac {3a+b-2}{a-b} = \dfrac {7}{2} & \Rightarrow 32b - 4 = 28 b & \Rightarrow b = 1 & \Rightarrow a = 5 \end{cases}$

We note that substituting $a=5$ and $b=1$ in the second equation we have:

$(a-b)x+(a+b)y = 3a+b-2$

$\Rightarrow 4x + 6y = 14$ identical to $2x+3y = 7$

Therefore, $a^2 +2ab + b^2 = (a+b)^2 = (5+1)^2 = \boxed{36}$

We can rewrite the system as:

$\begin{aligned} 2x +3y &=7\\ (a-b)x +(a+b)y &=3a+b-2. \end{aligned}$

This system will always have 1 solution, unless the two equations on the left hand side are proportional (i.e. the matrix of coefficients has determinant zero). Hence we require:

$2\times(a+b)=3\times(a-b)\Rightarrow a=5b.$

If we substitute this and multiply the first equation by $2b$ , we obtain:

$\begin{aligned} b(4x+6y)&=14b\\ b(4x+6y)&=16b-2 \end{aligned}$ .

This only has solutions if $14b=16b-2$ and so $b=1$ and $a=5$ . Hence we get $a^2+2ab+b^2=5^2+2\cdot 5\cdot 1+1^2=\boxed{36}$ .

In this case, both equations are equivalent and we see that any pair $(x,y)$ on the line $y=\frac13(7-2x)$ is a solution.