WAEC Further Mathematics 2023 Paper

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

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

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

=$3x+4=Px-2P+Qx+3Q$

=$3x+4=Px+Qx-2P+3Q$

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

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=$\frac{10}{5}=2$