Linear equations, inequalities & graphs

Fluency

$x = 6$ . Method: subtract $9$ from both sides.
$x = 6$ . Method: divide both sides by $3$ .
$x = 6$ . Method: subtract $5$ , then divide by $2$ .
$y = 7$ . Method: add $8$ , then divide by $3$ .
$x = 24$ . Method: subtract $3$ , then multiply by $4$ .
$a = 7$ . Method: divide both sides by $4$ to get $a - 2 = 5$ , then add $2$ .
$x = 5$ . Method: subtract $2x$ from both sides, then $3x - 3 = 12$ ; $3x = 15$ .
$y = 5$ . Method: subtract $3y$ and add $16$ : $20 = 4y$ , so $y = 5$ .
$x = 5$ . Method: multiply both sides by $2$ to get $3x + 1 = 16$ ; $3x = 15$ .
$x = 24$ . Method: multiply both sides by $12$ : $4x - 3x = 24$ .

Fluency

Fluency

Reasoning

No. Dividing $12$ by $-3$ gives $-4$ , so $x = -4$ . Kai kept the sign wrong.
She forgot to flip the inequality when dividing by a negative. Correct: $-2x < 6$ gives $x > -3$ .
They both have gradient $2$ (same slope), but different $y$ -intercepts ( $0$ and $3$ ), so they rise equally but start at different heights.
The line shifts up by $c$ units on the $y$ -axis (parallel to itself). Increasing $c$ does not change the gradient.

Problem solving

After $w = 11$ weeks (since $50 + 15 \times 10 = 200$ exactly at week $10$ ; pass $200 at end of week $11$ ). Strictly $50 + 15w > 200$ gives $w > 10$ , so first at $w = 11$ .
$V = 500 - 8t$ . Empty when $V = 0$ : $t = 62.5$ hours.
$16$ km. Method: $3.80 + 2k = 19.80$ , so $2k = 16$ .
$70$ texts. Method: $30 + 0.05t = 33.50$ , $0.05t = 3.50$ .
Plan A: $C = 20 + 0.15m$ ; Plan B: $C = 35 + 0.05m$ . Equal when $20 + 0.15m = 35 + 0.05m$ , so $0.10m = 15$ , $m = 150$ minutes.

Reasoning

$(5, 9)$ . Method: $2x - 1 = x + 4$ gives $x = 5$ , then $y = 9$ .
$x = 13$ . Method: multiply by $6$ : $3(x + 1) - 2(x - 1) = 12$ ; $3x + 3 - 2x + 2 = 12$ ; $x + 5 = 12$ .
$y = 3x + 1$ . Method: $y - 7 = 3(x - 2)$ , so $y = 3x - 6 + 7$ .
$28, 30, 32$ . Method: $n + (n+2) + (n+4) = 3n + 6 = 90$ ; $n = 28$ .

Year 8 core - answers