Given

$\rm \frac{a}{2x - 3} - \frac{b}{3x + 4} = \frac{x + 7}{6x^2 - x - 12}$

Solving the Left-Hand Side of the equation first, we get,

$\rm \frac{a}{2x- 3} - \frac{b}{3x + 4}$

$\rm = \frac{a}{2x - 3} \cdot \frac{3x + 4}{3x + 4} - \frac{b}{3x + 4} \cdot \frac{2x - 3}{2x - 3}$

Simplifying the expression

$\rm = \frac{a (3x + 4)}{(2x-3)(3x + 4)} - \frac{ b(2x - 3)}{(3x + 4)(2x - 3)}$

$\rm = \frac{ a(3x + 4) - b(2x - 3)}{ (2x -3)(3x + 4) }$

$\rm = \frac{ 3ax + 4a - 2bx + 3b }{ (2x - 3)(3x  + 4)}$

$\rm = \frac{ 3ax - 2bx + 4a + 3b }{ 2x (3x + 4) - 3(3x + 4)}$

$\rm = \frac{ (3a - 2b)x + (4a + 3b) }{ 6x^2 + 8x - 9x - 12}$

$\rm = \frac{ (3a - 2b)x + (4a + 3b)}{6x^2 - x - 12}$

From the given, the expression obtained above for the LHS is equal to the RHS. So,

$\rm or, \frac{ (3a - 2b)x + (4a + 3b)}{6x^2  - x - 12} = \frac{x + 7}{6x^2 - x - 12}$

The denominators on both sides of the equation are the same, so we equate their numerators.

$\rm or, (3a - 2b) x + (4a + 3b) = x + 7$

For the above equation to hold, the coefficient of like terms on both sides of the equation must be the same.

We equate the coefficients of x, we get,

$\rm or, 3a - 2b = 1$

$\rm or, 3a = 1 + 2b$

$\rm or, a = \frac{1 + 2b}{3}$ – (1)

We equate the coefficients of constant terms, and we get,

$\rm or, 4a + 3b = 7$

$\rm or, 4a = 7 - 3b$

$\rm or, a = \frac{7 - 3b}{4}$ – (2)

From equations (1) and (2), we get,

$\rm or, \frac{1 + 2b}{3} = \frac{7 - 3b}{4}$

Multiplying both sides of the equation by 12, we get,

$\rm or, \frac{1 + 2b}{3} \cdot 12 = \frac{7 - 3b}{4} \cdot 12$

$\rm or, (1 + 2b) \cdot 4 = (7 - 3b) \cdot 3$

$\rm or, 4 + 8b = 21 - 9b$

$\rm or, 8b + 9b = 21 - 4$

$\rm or, 17b = 17$

$\rm \therefore b = 1$

We put the value of b = 1 in equation (1) to find the value of a, we get,

$\rm a = \frac{1 + 2 \cdot 1}{3} = \frac{1 + 2}{3} = \frac{3}{3}$

$\rm \therefore a = 1$

Hence, the required solution is (a,b) = (1,1).