Linear relationships

Fluency

$5x = 25$ , so $x = 5$ .
$2w = P - 2l$ , so $w = \dfrac{P - 2l}{2}$ .
$4x > 16$ , so $x > 4$ .
Divide by $-3$ and flip: $x \geq -4$ . Closed circle at $-4$ , shade right.
Both have gradient $2$ , so parallel.
$m_1 = 4$ , $m_2 = -\tfrac{1}{4}$ . Product $= 4 \times (-\tfrac{1}{4}) = -1$ , so perpendicular.
$m = -\tfrac{1}{5}$ .
Dashed line (strict inequality). Shade above (since $y >$ ).

Reasoning

$20 = \tfrac{5}{9}(F - 32)$ . Multiply both sides by $\tfrac{9}{5}$ : $36 = F - 32$ . $F = 68$ .
$7 - 3x \geq 2x + 22$ . $-5x \geq 15$ . Divide by $-5$ , flip: $x \leq -3$ . Closed circle at $-3$ , shade left.
Gradient $= -2$ . Through $(3, 1)$ : $y - 1 = -2(x - 3)$ , so $y = -2x + 7$ .
Given gradient $\tfrac{1}{2}$ , perpendicular gradient $= -2$ . Through $(6, 4)$ : $y - 4 = -2(x - 6)$ , so $y = -2x + 16$ .
Boundary: $y = -2x + 8$ . Solid line. Intercepts $(4, 0)$ and $(0, 8)$ . Test $(0,0)$ : $0 \leq 8$ true, shade below (toward origin). Region is the triangle with vertices $(0, 0)$ , $(4, 0)$ , $(0, 8)$ .
$2l + 2w = 40$ gives $l = 20 - w$ . Also $l \geq 2w$ : $20 - w \geq 2w$ , so $20 \geq 3w$ , $w \leq \tfrac{20}{3} \approx 6.67$ . Since $w > 0$ and $l > 0$ (so $w < 20$ ), the range is $0 < w \leq \tfrac{20}{3}$ .

Reasoning

$m_{AB} = \tfrac{2}{4} = \tfrac{1}{2}$ . $m_{AC} = \tfrac{-4}{2} = -2$ . Product: $\tfrac{1}{2} \times (-2) = -1$ . So $AB \perp AC$ , meaning the right angle is at $A$ .
Parallel means equal gradients: $k = 2k - 5$ , so $k = 5$ .
Midpoint: $\left(\tfrac{2+8}{2}, \tfrac{6+2}{2}\right) = (5, 4)$ . Gradient of $AB$ : $\tfrac{2-6}{8-2} = -\tfrac{2}{3}$ . Perpendicular gradient: $\tfrac{3}{2}$ . Equation: $y - 4 = \tfrac{3}{2}(x - 5)$ , so $y = \tfrac{3}{2}x - \tfrac{7}{2}$ .
$2x + 5y \leq 100$ , with $x \geq 0$ , $y \geq 0$ . Boundary intercepts: $(50, 0)$ and $(0, 20)$ . Three integer solutions using exactly $100$ hours: $(0, 20)$ , $(25, 10)$ , $(50, 0)$ .

Reasoning

$m_{PQ} = \tfrac{3-1}{5-1} = \tfrac{1}{2}$ . $m_{SR} = \tfrac{5-3}{4-0} = \tfrac{1}{2}$ . So $PQ \parallel SR$ . $m_{QR} = \tfrac{5-3}{4-5} = -2$ . $m_{PS} = \tfrac{3-1}{0-1} = -2$ . So $QR \parallel PS$ . Both pairs of opposite sides are parallel, confirming a parallelogram. Check rectangle: $m_{PQ} \times m_{QR} = \tfrac{1}{2} \times (-2) = -1$ . Adjacent sides are perpendicular, so it is a rectangle.
Vertices of feasible region: $(0, 0)$ , $(4, 0)$ , $(\tfrac{14}{5}, \tfrac{18}{5})$ , $(0, 5)$ . Intersection of $x + 2y = 10$ and $3x + y = 12$ : solve to get $x = \tfrac{14}{5}$ , $y = \tfrac{18}{5}$ . Evaluate $P$ : at $(0,0)$ : $0$ ; at $(4,0)$ : $16$ ; at $(\tfrac{14}{5}, \tfrac{18}{5})$ : $\tfrac{56+54}{5} = \tfrac{110}{5} = 22$ ; at $(0,5)$ : $15$ . Maximum $P = 22$ at $(\tfrac{14}{5}, \tfrac{18}{5})$ .
$m_{L_2} = \tfrac{5-(-1)}{4-1} = 2$ . Perpendicular: $m_{L_1} = -\tfrac{1}{2}$ . $m_{L_1} = \tfrac{7-2a}{3-a}$ . Set equal: $\tfrac{7-2a}{3-a} = -\tfrac{1}{2}$ . Cross-multiply: $2(7-2a) = -(3-a)$ . $14 - 4a = -3 + a$ . $17 = 5a$ . $a = \tfrac{17}{5}$ .

Year 10 core - answers