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


SOLVE THIS. ​

Answer 1

$\huge{\underline{\underline{\mathfrak\green{Question\::}}}}$

$\bf{The\: fraction\: (5x-11)/(2x2 + x - 6) \:was \:obtained \:by \:adding,\:}$

$\bf{the \:two \:\:fractions\: A/(x + 2) \:and \:B/(2x - 3). \:The\: values \:of \:A \:and \:B \:must \:be, \:\:respectively:}$

$\bf{(a) 5x, -11}$

$\bf{ (b) -11, 5x}$

$\bf{(c) -1, 3}$

$\large\bf\blue{(d) 3, -1}$

$\bf{(e) 5, -11}$

$\huge{\underline{\underline{\mathfrak\red{Solution\::}}}}$

$\: \implies\bf{ \frac{5x - 11}{2x {}^{2 } + x - 6} } = { \frac{a}{x + 2} } + \frac{b}{2x - 3} \\ \: \implies\bf{ \frac{5x - 11}{(x + 2)(2x - 3)} } = \frac{a(2x - 3) + b(x + 2)}{(x + 2)(2x - 3)} \\ \implies\bf{5x - 11} = a(2x - 3) + b(x + 2) \\ \implies\bf{5x - 11} = (2a + b)x + ( - 3a + 2b)\\ \tiny\bf\blue{[ comparing \: the \: co - efficient \: of \: x \: and \: constant \: term \: in \: the \: both \: side]} \\ \implies \bf\:\red{ a + b \: = 5 \: ------- \: (1)} \\ \implies \bf \: \green{ - 3a + 2b \: = - 11-----(2)} \\ \implies \bf \: \orange{b = 5 - 2a------- \: (3)} \\ \bf\:\tiny \purple \:{ put \: b = 5 - 2a \: in \: (2).} \\ \implies \bf - 3a + b(5 - 2a) = - 11 \\ \implies \bf - 3a + 10 - 4a = - 11 \\ \implies \bf \cancel - 7a = \cancel - 21 \\ \implies \bf\pink{\boxed{a = 3}} \\ \tiny\bf\purple{ \: put \: \: a = 3 \: \: in \ \: \: (3).} \\ \implies\bf \: b = 5 - 2(3) \\ \implies\bf \: b = 5 - 6 \\ \implies\bf\red{\boxed{b = - 1}}$

$\large{\underline{\underline{\mathbf\orange{Therefore , A = 3\: and\: B= -1.}}}}$