Question

Any Expert Genius Solve plz.

The fraction (5x-11)/(2x2 + x - 6) was obtained by adding the two fractions A/(x + 2) and B/(2x - 3). The values of A and B must be, respectively:

(a) 5x, -11,

(b) -11, 5x,

(c) -1, 3,

(d) 3, -1,

(e) 5, -11

Answer 1

Plz see the attachment and find ur answer __:))

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Answer 2

Answer:

The correct answer is option(D) 3 and -1

Step-by-step explanation:

Given,

$\frac{5x-11}{2x^2 + x - 6} = \frac{A}{x+2} + \frac{B}{2x-3}$

To find,

The value of A and B

Solution

We have

$\frac{5x-11}{2x^2 + x - 6} = \frac{A}{x+2} + \frac{B}{2x-3}$

Taking LCM on RHS we get

LCM = (x+2)(2x-3)

$\frac{5x-11}{2x^2 + x - 6} =\frac{A(2x-3)+B(x+2)}{(x+2)(2x-3)}$

$\frac{5x-11}{2x^2 + x - 6} =\frac{2Ax -3A + Bx + 2B}{(x+2)(2x-3)}$

$\frac{5x-11}{2x^2 + x - 6} =\frac{(2A+B)x -3A +2B}{(x+2)(2x-3)}$

$\frac{5x-11}{2x^2 + x - 6} =\frac{(2A+B)x -(3A -2B)}{(x+2)(2x-3)}$

$\frac{5x-11}{2x^2 + x - 6} =\frac{(2A+B)x -(3A -2B)}{2x^2 + x - 6}$

Since the denominators are equal, we have

5x -11 = (2A+B)x - (3A - 2B)

Comparing the coefficients of x and constant terms on both sides we get

2A + B = 5     -------------------(1)

3A - 2B = 11  ------------------(2)

Equation (1) ×2 →  4A +2B = 10 ----------(3)

Adding equations(2) and (3) we get

7A = 21

A = $\frac{21}{7}$ = 3

A = 3

Substituting the value of A in equation (1) we get

6+B = 5

B = 5 -6 = -1

B = -1

The value of A = 3 and B = -1

Hence the correct answer is option(D) 3 and -1

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