Add and Subtract Rational Expressions - College Algebra

Given that $\frac{A}{x-2} + \frac{B}{x+1} = \frac{4x+7}{(x-2)(x+1)}$ , what are the values of constants $A$ and $B$ ?
Answer: $A = 5, B = -1$ To find A and B, we need to combine the left side over a common denominator. $\frac{A}{x-2} + \frac{B}{x+1} = \frac{A(x+1) + B(x-2)}{(x-2)(x+1)} = \frac{Ax + A + Bx - 2B}{(x-2)(x+1)} = \frac{(A+B)x + (A-2B)}{(x-2)(x+1)}$ . For this to equal $\frac{4x+7}{(x-2)(x+1)}$ , we need the numerators to be equal: $(A+B)x + (A-2B) = 4x + 7$ . Comparing coefficients: $A + B = 4$ (coefficient of x) and $A - 2B = 7$ (constant term). From the first equation: $A = 4 - B$ . Substituting into the second: $(4-B) - 2B = 7$ , so $4 - 3B = 7$ , which gives $-3B = 3$ , so $B = -1$ . Then $A = 4 - (-1) = 5$ . Check: $A + B = 5 + (-1) = 4$ ✓ and $A - 2B = 5 - 2(-1) = 5 + 2 = 7$ ✓.