Question

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

Answer:

plzz refer to the attachment

Answer 2

Answer:

(d) a =3; b = -1

Explanation:

$\frac{a}{(x + 2)} + \frac{b}{(2x - 3)} = \frac{ (5x-11)}{(2x {}^{2} + x - 6)} \\ \frac{2ax - 3a + bx + 2b}{(x + 2)(2x - 3)} = \frac{(5x - 11)}{(2x {}^{2} + x - 6) }$

$= \frac{2ax - 3a + bx + 2b}{(2x {}^{2} + x - 6)} = \frac{(5x - 11)}{(2x {}^{2} + x - 6) }$

$x(2a + b) + 2b - 3a = 5x - 11$

When a = 3;b = -1,

$x(6 - 1) - 2 - 9 \\ = 5x - 11$

Since these values satisfy the confition, the value of a and b are 3 and -1, respectively.

---------------

Hope it helps....!

♤♤♤♤