Given that \(\frac{3x + 4}{(x – 2)(x + 3)}≡\frac{P}{x + 3}+\frac{Q}{x – 2}\),find the value…

Explanation

\(\frac{3x + 4}{(x - 2)(x + 3)}≡\frac{P}{x + 3}+\frac{Q}{x - 2}\)

\(\frac{3x + 4}{(x - 2)(x + 3)}=\frac{P(x-2)+Q(x+3)}{(x-2)(x+3)}\)

=\(3x+4=P(x-2)+Q(x+3)\)

=\(3x+4=Px-2P+Qx+3Q\)

=\(3x+4=Px+Qx-2P+3Q\)

=\(3x+4=(P+Q)x-2P+3Q\)

Equating x and the constants

P+Q=3------(i)

2P+3Q=4-------(ii)

From equation (i)

P=3-Q------(iii)

Substitute (3-Q) for P inequation (ii)

=-2(3-Q)+3Q=4

=-6+2Q+3Q=4

=-6+5Q=4

=5Q=4+6

=5Q=10

∴Q=\(\frac{10}{5}=2\)