Simultaneous equations

Fluency

$x + 3 = 2x - 1$ , $x = 4$ , $y = 7$ .
$3x + 4x = 14$ , $7x = 14$ , $x = 2$ , $y = 8$ .
Add: $2x = 14$ , $x = 7$ , $y = 3$ .
Subtract: $2x = 6$ , $x = 3$ , $y = 2$ .
Add: $3x = 9$ , $x = 3$ , $y = 1$ .
Substitute: $3x - (2x + 3) = -1$ , $x - 3 = -1$ , $x = 2$ , $y = 7$ .
Subtract: $2x = 6$ , $x = 3$ , $y = 2$ .
Subtract: $2x = 4$ , $x = 2$ , $y = 3$ .
$x + y = 20$ and $x - y = 6$ . Add: $2x = 26$ , $x = 13$ , $y = 7$ .
Let pencil $= p$ , pen $= p + 2$ . $3(p + 2) + 2p = 21$ , $5p + 6 = 21$ , $p = 3$ . Pencil $3, pen $5.

Reasoning

From eq 2: $x = y + 3$ . Substitute: $\dfrac{y+3}{2} + \dfrac{y}{3} = 4$ . Multiply by $6$ : $3(y+3) + 2y = 24$ , $5y + 9 = 24$ , $y = 3$ , $x = 6$ .
Multiply eq 1 by $2$ : $10x + 4y = 2$ . Add to eq 2: $13x = 13$ , $x = 1$ , $y = -2$ .
Multiply eq 2 by $2$ : $2x - 6y = 14$ . But eq 1 is $2x - 6y = 4$ . Since $14 \neq 4$ , there is no solution (parallel lines).
$a + b = 5$ and $3a + 2b = 12$ . From eq 1: $b = 5 - a$ . Substitute: $3a + 2(5-a) = 12$ , $a + 10 = 12$ , $a = 2$ kg apples, $b = 3$ kg bananas.
Car B position: $60t$ (where $t$ is hours after B left). Car A position: $80(t - 1)$ (A left 1 hour later). Overtake: $80(t-1) = 60t$ , $80t - 80 = 60t$ , $20t = 80$ , $t = 4$ hours after B left. Position: $60 \times 4 = 240$ km from start.
$2l + 2w = 36$ and $l = w + 4$ . Substitute: $2(w+4) + 2w = 36$ , $4w + 8 = 36$ , $w = 7$ cm, $l = 11$ cm.
Eq 2 is exactly $2 \times$ eq 1: $4x + 6y = 2(2x + 3y) = 2(1) = 2$ . The equations are identical, so there are infinitely many solutions.
Let $c$ = correct, $w$ = wrong. $c + w = 30$ and $4c - w = 75$ . Add: $5c = 105$ , $c = 21$ correct answers.

Reasoning

$2t + s = 35$ and $t + 2s = 25$ . Multiply eq 2 by $2$ : $2t + 4s = 50$ . Subtract eq 1: $3s = 15$ , $s = 5$ . $t = \dfrac{35 - 5}{2} = 15$ . Ticket $15, snack $5.
Let boat speed $= b$ , current $= c$ . Upstream: $\dfrac{30}{b - c} = 3$ , so $b - c = 10$ . Downstream: $\dfrac{30}{b + c} = 2$ , so $b + c = 15$ . Add: $2b = 25$ , $b = 12.5$ km/h, $c = 2.5$ km/h.
$500 + 8n = 20n - 0.05n^2$ . $0.05n^2 - 12n + 500 = 0$ . $n^2 - 240n + 10\,000 = 0$ . $n = \dfrac{240 \pm \sqrt{57\,600 - 40\,000}}{2} = \dfrac{240 \pm \sqrt{17\,600}}{2} = \dfrac{240 \pm 40\sqrt{11}}{2} = 120 \pm 20\sqrt{11}$ . So $n \approx 120 - 66.3 = 53.7$ or $n \approx 120 + 66.3 = 186.3$ . Break-even at approximately $54$ and $186$ units.
Through $(2,1)$ : $2a + b = 1$ . Through $(4,-1)$ : $4a - b = 1$ . Add: $6a = 2$ , $a = \dfrac{1}{3}$ , $b = 1 - \dfrac{2}{3} = \dfrac{1}{3}$ . Equation: $\dfrac{x}{3} + \dfrac{y}{3} = 1$ , so $x + y = 3$ , $y = -x + 3$ .

Reasoning

Number $= 10t + u$ . $t + u = 9$ and $10t + u - (10u + t) = 27$ , so $9t - 9u = 27$ , $t - u = 3$ . Add: $2t = 12$ , $t = 6$ , $u = 3$ . The number is $63$ .
Let $u = \dfrac{1}{x}$ , $v = \dfrac{1}{y}$ . $2u + 3v = 7$ and $5u - v = 9$ . From eq 2: $v = 5u - 9$ . Substitute: $2u + 3(5u - 9) = 7$ , $17u - 27 = 7$ , $u = 2$ , $v = 1$ . So $x = \dfrac{1}{2}$ , $y = 1$ .
Solve $2x + y = 5$ and $x - y = 1$ : add, $3x = 6$ , $x = 2$ , $y = 1$ . The intersection is $(2, 1)$ . Line through $(2, 1)$ and $(0, 4)$ : gradient $= \dfrac{4 - 1}{0 - 2} = -\dfrac{3}{2}$ . Equation: $y = -\dfrac{3}{2}x + 4$ .
$40s + 25l = 285$ and $30s + 35l = 295$ . Multiply eq 1 by $7$ and eq 2 by $5$ : $280s + 175l = 1995$ and $150s + 175l = 1475$ . Subtract: $130s = 520$ , $s = 4$ . $40(4) + 25l = 285$ , $25l = 125$ , $l = 5$ . Small $4, large $5. Revenue for 50 of each: $50(4) + 50(5) = 200 + 250 = 450$ .

Year 10 core - answers