WAEC 2011 · Paper 2 · Q11

(a)
Solve for $x$ , $y$ and $z$ in the equations: $3x + 5y - 4z = -5$ ; $6x + 3y - 5z = 26$ ; $-2x + 2y + z = -11$ .
Separate values with commas, e.g. 3, −2
(b)
A function $g$ is defined by $g(x) = \dfrac{3 - 4x}{x^2 + x - 6}$ . Express $g(x)$ in partial fractions.

Worked solution (try it first)

By Cramer's rule, first find $\Delta = \begin{vmatrix} 3 & 5 & -4 \\ 6 & 3 & -5 \\ -2 & 2 & 1 \end{vmatrix}$ .
Expand along the top row: $\Delta = 3(3 + 10) - 5(6 - 10) - 4(12 + 6)$
$= 39 + 20 - 72$
$= -13$ .
Replace the $x$ column by the right-hand sides: $\Delta_x = -5(3 + 10) - 5(26 - 55) - 4(52 + 33)$
$= -65 + 145 - 340$
$= -260$ .
Replace the $y$ column: $\Delta_y = 3(26 - 55) + 5(6 - 10) - 4(-66 + 52)$
$= -87 - 20 + 56$
$= -51$ .
Replace the $z$ column: $\Delta_z = 3(-33 - 52) - 5(-66 + 52) - 5(12 + 6)$
$= -255 + 70 - 90$
$= -275$ .
So $x = \dfrac{-260}{-13} = 20$ , $y = \dfrac{-51}{-13}$
$= \dfrac{51}{13}$
$\approx 3.92$ and $z = \dfrac{-275}{-13}$
$= \dfrac{275}{13}$
$\approx 21.15$ .
Check in the third equation: $-40 + \dfrac{102}{13} + \dfrac{275}{13} = -40 + 29$
$= -11$ ✓.

Factorise the bottom: $x^2 + x - 6 = (x - 2)(x + 3)$ .
Write $\dfrac{A}{x - 2} + \dfrac{B}{x + 3}$ and multiply through: $3 - 4x = A(x + 3) + B(x - 2)$ .
Put $x = 2$ : $-5 = 5A$ , so $A = -1$ .
Put $x = -3$ : $15 = -5B$ , so $B = -3$ .
So $g(x) = -\dfrac{1}{x - 2} - \dfrac{3}{x + 3}$ .