WAEC 2017 · Paper 2 · Q12

(a)
If $\begin{pmatrix} 3x & 2 \\ 5 & 4 \end{pmatrix}\begin{pmatrix} 3 & -2 \\ 5 & 2y \end{pmatrix} = \begin{pmatrix} 28 & 4 \\ 35 & 22 \end{pmatrix}$ , find the values of $x$ and $y$ .
Separate values with commas, e.g. 3, −2
(b)
Using a scale of 2 cm to 1 unit on both axes, draw on a graph sheet the region which satisfies the following inequalities simultaneously: $y < x + 1$ ; $2y \ge -2x + 3$ ; $2x < 3$ ; $y + 1 > 0$ .
Model answer
Draw each boundary: solid for $\ge$ or $\le$ , dashed for $<$ or $>$ (the line itself is not included). $y = x + 1$ (dashed), $2y = -2x + 3$ (solid), $2x = 3$ (dashed) and $y = -1$ (dashed). The region satisfying all four is the triangle with corners $(0.25, 1.25)$ , $(1.5, 2.5)$ and $(1.5, 0)$ . Every point of it already has $y > -1$ , so that condition doesn't cut it further. Label the region.

The four inequalities; the shaded region satisfies them all.

Worked solution (try it first)

Multiply the matrices (row by column): $\begin{pmatrix} 3x & 2 \\ 5 & 4 \end{pmatrix}\begin{pmatrix} 3 & -2 \\ 5 & 2y \end{pmatrix} = \begin{pmatrix} 9x + 10 & -6x + 4y \\ 35 & -10 + 8y \end{pmatrix}$ .
Match the entries with $\begin{pmatrix} 28 & 4 \\ 35 & 22 \end{pmatrix}$ : $9x + 10 = 28$ , so $x = 2$ .
$8y - 10 = 22$ , so $y = 4$ .
(Check the other entry: $-6(2) + 4(4) = 4$ ✓.)

Draw the four boundary lines: $y = x + 1$ (dashed).
$2y = -2x + 3$ , that is $y = -x + \frac32$ (solid).
$2x = 3$ , that is $x = \frac32$ (dashed).
$y + 1 = 0$ , that is $y = -1$ (dashed).
Test the origin in each: $0 < 1$ ✓, so keep the side of $y = x + 1$ below the line.
$0 \ge 3$ ✗, so keep the side of $y = -x + \frac32$ above the line.
$0 < 3$ ✓, so keep the left of $x = \frac32$ .
$1 > 0$ ✓, so keep above $y = -1$ .
The region satisfying all four is the triangle with corners $\left(\frac14, \frac54\right)$ , $\left(\frac32, \frac52\right)$ and $\left(\frac32, 0\right)$ .
The line $y = -1$ lies below it, so it doesn't cut any of it off.
Shade and label the triangle.