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From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B.  Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.

Then the line is y = -x/3 - 19/3.  Then 3y = -x - 19, so 3y + x = -19.  We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3.  Therefore, B - A = -19/3 - 3 = -28/3.

Putting the equations into slope-intercept form,

$y = -\frac{Ax}{B} + \frac{3}{B}$

$y = -\frac{x}{3} - \frac{5}{3}$

Both lines are parallel, so $-\frac{x}{3} = -\frac{Ax}{B}$. Simplify to find that $\frac{A}{B} = \frac{1}{3}$, meaning that the slope is $-\frac{1}{3}$.

$y = -\frac{x}{3} + \frac{3}{B}$

Plug in (-7, 2).

$2 = -\frac{-7}{3} + \frac{3}{B}$

$6B = 7B + 9$

$B = -9$

Because A and B are in ratio $\frac{A}{B} = \frac{1}{3}$, B = -9 and A = -3. So B - A = -9 - (-3) = -6

x + 3y = -5               re arrange to y = mx+b from:

y = -1/3 x - 5/3         slope = -1/3     ( parallel slope is the same)

y = -1/3 x -  b           sub in the given point to calculate 'b'

2 = -1/3 (-7) + b       shows b = - 1/3

y = -1/3 x - 1/3         multiply by 3 to get rid of fraction

3y = 3x - 3

3x -3y = 3                 then   B - A =  -3 - 3 = - 6