Solve the equations by substitution method and partial fraction decomposition.

The given expression can be written as:

$$\frac{4x+3}{(x+2)(x+4)} = \frac{A}{x+2} + \frac{B}{x+4}$$

Multiplying both sides by $(x+2)(x+4)$ gives:

$$4x+3 = A(x+4) + B(x+2)$$

Expanding the right side:

$$4x+3 = Ax + 4A + Bx + 2B$$

Grouping terms by powers of $x$:

$$4x+3 = (A+B)x + (4A+2B)$$

Equating the coefficients of $x$ and the constant terms on both sides, we get a system of linear equations:

$$A+B = 4 \quad \text{(i)}$$ $$4A+2B = 3 \quad \text{(ii)}$$

From equation (i), express $A$ in terms of $B$:

$$A = 4-B \quad \text{(iii)}$$

Substitute equation (iii) into equation (ii):

$$4(4-B) + 2B = 3$$ $$16 - 4B + 2B = 3$$ $$16 - 2B = 3$$ $$-2B = 3 - 16$$ $$-2B = -13$$ $$B = \frac{-13}{-2}$$ $$B = \frac{13}{2}$$

Substitute the value of $B$ back into equation (iii) to find $A$:

$$A = 4 - \frac{13}{2}$$ $$A = \frac{8}{2} - \frac{13}{2}$$ $$A = \frac{8-13}{2}$$ $$A = -\frac{5}{2}$$

Your calculations for $A = -\frac{5}{2}$ and $B = \frac{13}{2}$ are correct.

$$\frac{4x+3}{(x+2)(x+4)} = -\frac{5}{2(x+2)} + \frac{13}{2(x+4)}$$

The partial fraction decomposition is:

$$-\frac{5}{2(x+2)} + \frac{13}{2(x+4)}$$