Question

resolve 2x-1 /(x-1)(2x+3) into partial fractions​

Answer 1

Answer:

$\frac{2x - 1}{(x - 1)(2x + 3)} = \frac{a}{x - 1} + \frac{b}{2x + 3} \\ 2x - 1 = a(2x + 3) + b(x - 1) \\ put \: x = 1 \\ we \: get \\ 2 - 1 = a(2 + 3) \\ a = \frac{1}{5} \\ \\ put \: x \: = \frac{ - 3}{2} \\ we \: get \\ - 3 - 1 = b( \frac{ - 3}{2} - 1) \\ - 4 = \frac{ - 5b}{2} \\ b = \frac{8}{5} \\ \\ so \: \frac{2x - 1}{(x - 1)(2x + 3)} = \frac{ \frac{1}{5} }{x - 1} + \frac{ \frac{8}{5} }{2x + 3}$

Answer 2

Answer:

$\frac{2x-1}{(x-1)(2x+3)}=\frac{\frac{1}{5} }{x-1}+\frac{\frac{8}{5} }{2x+3}$

Step-by-step explanation:

To convert the given expression into partial fractions, we use the following method

$\frac{2x-1}{(x-1)(2x+3)} = \frac{a}{x-1} +\frac{b}{2x+3}$

$(2x-1)=a(2x+3)+b(x-1)$

Put $x=1$ so that we can get the value of $a$.

$2(1)-1=a(2(1)+3)+b(1-1)$

$2-1=a(2+3)+0$

$1=5a$

$a=\frac{1}{5}$.

Put $x=-\frac{3}{2}$ so that we can get the value of $b$.

$2(-\frac{3}{2} )-1=a(2(-\frac{3}{2} +3))+b(-\frac{3}{2} -1)$

$-4=0+b(-\frac{5}{2} )$

$b=\frac{8}{5}$

So the given expression becomes $\frac{2x-1}{(x-1)(2x+3)}=\frac{\frac{1}{5} }{x-1}+\frac{\frac{8}{5} }{2x+3}$.