(a)

\begin{align} y & = 2x + 1 \phantom{000} \text{ --- (1)} \\ \\ y & = x^2 + 2x - 3 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & (1) \text{ into (2),} \\ 2x + 1 & = x^2 + 2x - 3 \\ 0 & = x^2 - 4 \\ 0 & = (x - 2)(x + 2) \end{align} \begin{align} x - 2 & = 0 & \text{or } \phantom{00000} x + 2 & = 0 \\ x & = 2 & x & = - 2 \\ \\ \text{Substitute } & x = 2 \text{ into (1),} & \text{Substitute } & x = -2 \text{ into (1),} \\ y & = 2(2) + 1 & y & = 2(-2) + 1 \\ & = 5 & & = -3 \end{align}
$$ \therefore x = 2, y = 5 \text{ or } x = -2, y = -3 $$

(b)

\begin{align} y & = 2 + x \phantom{000} \text{ --- (1)} \\ \\ y & = 2x^2 - 5x - 6 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2 + x & = 2x^2 - 5x - 6 \\ 0 & = 2x^2 - 6x - 8 \\ 0 & = x^2 - 3x - 4 \\ 0 & = (x - 4)(x + 1) \end{align} \begin{align} x - 4 & = 0 &\text{or }\phantom{00000} x + 1 & = 0 \\ x & = 4 & x & = - 1 \\ \\ \text{Substitute } & x = 4 \text{ into (1),} & \text{Substitute } & x = -1 \text{ into (1),} \\ y & = 2 + (4) & y & = 2 + (-1) \\ & = 6 & & = 1 \end{align}
$$ \therefore x = 4, y = 6 \text{ or } x = -1, y = 1 $$

(c)

\begin{align} 2x + y & = 4 \\ y & = 4 - 2x \phantom{000} \text{ --- (1)} \\ \\ y^2 - 4x & = 0 \\ y^2 & = 4x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (4 - 2x)^2 & = 4x \\ (4)^2 - 2(4)(2x) + (2x)^2 & = 4x \\ 16 - 16x + 4x^2 & = 4x \\ 4x^2 - 20x + 16 & = 0 \\ x^2 - 5x + 4 & = 0 \\ (x - 4)(x - 1) & = 0 \end{align} \begin{align} x - 4 & = 0 &\text{or }\phantom{00000} x - 1 & = 0 \\ x & = 4 & x & = 1 \\ \\ \text{Substitute } & x = 4 \text{ into (1),} & \text{Substitute } & x = 1 \text{ into (1),} \\ y & = 4 - 2(4) & y & = 4 - 2(1) \\ & = -4 & & = 2 \end{align}
$$ \therefore x = 4, y = -4 \text{ or } x = 1, y = 2 $$

(a)

\begin{align} y & = 2 - x \phantom{000} \text{ --- (1)} \\ \\ 2x^2 + xy + 1 & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x^2 + x(2 - x) + 1 & = 0 \\ 2x^2 + 2x - x^2 + 1 & = 0 \\ x^2 + 2x + 1 & = 0 \\ (x)^2 + 2(x)(1) + (1)^2 & = 0 \\ (x + 1)^2 & = 0 \\ x + 1 & = 0 \\ x & = - 1 \\ \\ \text{Substitute } & \text{into (1),} \\ y & = 2 - (-1) \\ & = 3 \\ \\ \therefore \text{Point of } & \text{intersection is } (-1, 3) \end{align}

(b)

\begin{align} y & = 1 - 3x \phantom{000} \text{ --- (1)} \\ \\ x^2 + y^2 & = 5 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (1 - 3x)^2 & = 5 \\ x^2 + (1)^2 - 2(1)(3x) + (3x)^2 & = 5 \\ x^2 + 1 - 6x + 9x^2 & = 5 \\ 10x^2 - 6x - 4 & = 0 \\ 5x^2 - 3x - 2 & = 0 \\ (5x + 2)(x - 1) & = 0 \end{align} \begin{align} 5x + 2 & = 0 &\text{or } \phantom{00000} x - 1 & = 0 \\ 5x & = - 2 & x & = 1 \\ x & = -{2 \over 5} \\ \\ \text{Substitute } & x = -{2 \over 5} \text{ into (1),} & \text{Substitute } & x = 1 \text{ into (1),} \\ y & = 1 - 3\left(-{2 \over 5}\right) & y & = 1 - 3(1) \\ & = {11 \over 5} & & = -2 \end{align}
$$ \therefore \text{Points of intersection are } \left(-{2 \over 5}, {11 \over 5}\right) \text{ and } (1, - 2) $$

(c)

\begin{align} 3x + 2y & = 1 \\ 2y & = 1 - 3x \\ y & = {1 \over 2} - {3 \over 2}x \phantom{000} \text{ --- (1)} \\ \\ 3x^2 + 2y^2 & = 11 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3x^2 + 2 \left({1 \over 2} - {3 \over 2}x \right)^2 & = 11 \\ 3x^2 + 2 \left[ \left(1 \over 2\right)^2 - 2\left(1 \over 2\right)\left({3 \over 2}x\right) + \left({3 \over 2}x\right)^2 \right] & = 11 \\ 3x^2 + 2 \left( {1 \over 4} - {3 \over 2}x + {9 \over 4}x^2 \right) & = 11 \\ 3x^2 + {1 \over 2} - 3x + {9 \over 2}x^2 & = 11 \\ 6x^2 + 1 - 6x + 9x^2 & = 22 \\ 15x^2 - 6x - 21 & = 0 \\ 5x^2 - 2x - 7 & = 0 \\ (5x - 7)(x + 1) & = 0 \end{align} \begin{align} 5x - 7 & = 0 &\text{or } \phantom{00000} x + 1 & = 0 \\ 5x & = 7 & x & = -1 \\ x & = {7 \over 5} \\ \\ \text{Substitute } & x = {7 \over 5} \text{ into (1),} & \text{Substitute } & x = -1 \text{ into (1),} \\ y & = {1 \over 2} - {3 \over 2}\left(7 \over 5\right) & y & = {1 \over 2} - {3 \over 2}(-1) \\ & = -{8 \over 5} & & = 2 \end{align}
$$ \therefore \text{Points of intersection are } \left({7 \over 5}, -{8 \over 5} \right) \text{ and } (-1, 2) $$

\begin{align} {x^2 \over 4} + {y \over 3} & = 3 \\ 12\left({x^2 \over 4} + {y \over 3} \right) & = 12(3) \\ 3x^2 + 4y & = 36 \phantom{000} \text{ --- (1)} \\ \\ x + y & = 8 \\ y & = 8 - x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3x^2 + 4(8 - x) & = 36 \\ 3x^2 + 32 - 4x & = 36 \\ 3x^2 - 4x - 4 & = 0 \\ (3x + 2)(x - 2) & = 0 \end{align} \begin{align} 3x + 2 & = 0 &\text{or }\phantom{00000} x - 2 & = 0 \\ 3x & = - 2 & x & = 2 \\ x & = -{2 \over 3} \end{align}

(i)

\begin{align} \text{Perimeter of first square} & = 4x \\ \\ \text{Perimeter of second square} & = 4y \\ \\ 4x + 4y & = 32 \\ x + y & = 8 \\ y & = 8 - x \phantom{000} \text{ --- (1)} \end{align}

(ii)

\begin{align} \text{Area of first square} & = x^2 \\ \\ \text{Area of second square} & = y^2 \\ \\ x^2 + y^2 & = 34 \phantom{000} \text{ --- (2)} \end{align}

(iii)

\begin{align} \text{Substitute } & \text{(1) into (2),} \\ x^2 + (8 - x)^2 & = 34 \\ x^2 + (8)^2 - 2(8)(x) + (x)^2 & = 34 \\ x^2 + 64 - 16x + x^2 & = 34 \\ 2x^2 - 16x + 30 & = 0 \\ x^2 - 8x + 15 & = 0 \\ (x - 3)(x - 5) & = 0 \end{align} \begin{align} x - 3 & = 0 &\text{or }\phantom{00000} x - 5 & = 0 \\ x & = 3 & x & = 5 \\ \\ \\ \text{Substitute } & x = 3 \text{ into (1),} & \text{Substitute } & x = 5 \text{ into (1),} \\ y & = 8 - (3) & y & = 8 - (5) \\ & = 5 & & = 3 \end{align}
$$ \therefore 3 \text{ cm}, 5 \text{ cm} $$

(i)

\begin{align} \text{Area} & = (x)(y) \\ & = xy \\ \\ \text{Perimeter} & = 2(x) + 2(y) \\ & = 2x + 2y \end{align}

(ii)

\begin{align} xy & = 216 \phantom{000} \text{ --- (1)} \\ \\ 2x + 2y & = 60 \\ 2y & = 60 - 2x \\ y & = 30 - x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ x(30 - x) & = 216 \\ 30x - x^2 & = 216 \\ 0 & = x^2 - 30x + 216 \\ 0 & = (x - 12)(x - 18) \end{align} \begin{align} x - 12 & = 0 &\text{or }\phantom{00000} x - 18 & = 0 \\ x & = 12 & x & = 18 \\ \\ \\ \text{Substitute } & x = 12 \text{ into (2),} & \text{Substitute } & x = 18 \text{ into (2),} \\ y & = 30 - (12) & y & = 30 - (18) \\ & = 18 & & = 12 \end{align}
$$ \therefore \text{Dimensions of land is } 12 \text{ m by } 18 \text{ m} $$

(a)

\begin{align} 3x - 2y & = 1 \\ -2y & = 1 - 3x \\ 2y & = 3x - 1 \phantom{000} \text{ --- (1)} \\ \\ (x - 2)^2 + (2y + 3)^2 & = 26 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (x - 2)^2 + [(3x - 1) + 3]^2 & = 26 \\ (x - 2)^2 + (3x + 2)^2 & = 26 \\ (x)^2 - 2(x)(2) + (2)^2 + (3x)^2 + 2(3x)(2) + (2)^2 & = 26 \\ x^2 - 4x + 4 + 9x^2 + 12x + 4 & = 26 \\ 10x^2 + 8x + 8 & = 26 \\ 10x^2 + 8x - 18 & = 0 \\ 5x^2 + 4x - 9 & = 0 \\ (5x + 9)(x - 1) & = 0 \end{align} \begin{align} 5x + 9 & = 0 &\text{or } \phantom{00000} x - 1 & = 0 \\ 5x & = - 9 & x & = 1 \\ x & = -{9 \over 5} \\ \\ \\ \text{Substitute } & x = -{9 \over 5} \text{ into (1),} & \text{Substitute } & x = 1 \text{ into (1),} \\ 2y & = 3\left(-{9 \over 5}\right) - 1 & 2y & = 3(1) - 1 \\ & = -{32 \over 5} & & = 2 \\ y & = -{16 \over 5} & y & = 1 \end{align}
$$ \therefore x = -{9 \over 5}, y = -{16 \over 5} \text{ or } x = 1, y = 1 $$

(b)

\begin{align} x^2 - 2xy + y^2 & = 1 \phantom{000} \text{ --- (1)} \\ \\ x - 2y & = 2 \\ x & = 2y + 2 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (2y + 2)^2 - 2(2y + 2)y + y^2 & = 1 \\ (2y)^2 + 2(2y)(2) + (2)^2 - 2y(2y + 2) + y^2 & = 1 \\ 4y^2 + 8y + 4 - 4y^2 - 4y + y^2 & = 1 \\ y^2 + 4y + 4 & = 1 \\ y^2 + 4y + 3 & = 0 \\ (y + 3)(y + 1) & = 0 \end{align} \begin{align} y + 3 & = 0 &\text{or }\phantom{00000} y + 1 & = 0 \\ y & = -3 & y & = -1 \\ \\ \\ \text{Substitute } & y = -3 \text{ into (2),} & \text{Substitute } & y = -1 \text{ into (2),} \\ x & = 2(-3) + 2 & x & = 2(-1) + 2 \\ & = -4 & & = 0 \end{align}
$$ \therefore x = -4, y = -3 \text{ or } x = 0, y = -1 $$

(c)

\begin{align} 3y - x & = 3 \\ 3y & = x + 3 \phantom{000} \text{ --- (1)} \\ \\ {2 \over 3y} - {1 \over x} & = 2 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ {2 \over x + 3} - {1 \over x} & = 2 \\ {2(x) \over x(x + 3)} - {(x + 3) \over x(x + 3)} & = 2 \\ {2x - (x + 3) \over x(x + 3)} & = 2 \\ {2x - x - 3 \over x(x + 3)} & = 2 \\ {x - 3 \over x(x + 3)} & = 2 \\ x - 3 & = 2x(x + 3) \\ x - 3 & = 2x^2 + 6x \\ 0 & = 2x^2 + 5x + 3 \\ 0 & = (2x + 3)(x + 1) \end{align} \begin{align} 2x + 3 & = 0 &\text{or } \phantom{00000} x + 1 & = 0 \\ 2x & = - 3 & x & = - 1 \\ x & = -{3 \over 2} \\ \\ \text{Substitute } & x = -{3 \over 2} \text{ into (1),} & \text{Substitute } & x = -1 \text{ into (1),} \\ 3y & = -{3 \over 2} + 3 & 3y & = -1 + 3 \\ & = {3 \over 2} & & = 2 \\ y & = {1 \over 2} & y & = {2 \over 3} \end{align}
$$ \therefore x = -{3 \over 2}, y = {1 \over 2} \text{ or } x = -1, y = {2 \over 3} $$

(a)

\begin{align} xy + 20 & = 5x \phantom{000} \text{ --- (1)} \\ \\ x - 2y - 3 & = 0 \\ x & = 2y + 3 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (2y + 3)(y) + 20 & = 5(2y + 3) \\ 2y^2 + 3y + 20 & = 10y + 15 \\ 2y^2 - 7y + 5 & = 0 \\ (2y - 5)(y - 1) & = 0 \end{align} \begin{align} 2y - 5 & = 0 &\text{or }\phantom{00000} y - 1 & = 0 \\ 2y & = 5 & y & = 1 \\ y & = {5 \over 2} \\ \\ \text{Substitute } & y = {5 \over 2} \text{ into (2),} & \text{Substitute } & y = 1 \text{ into (2),} \\ x & = 2\left(5 \over 2\right) + 3 & x & = 2(1) + 3 \\ & = 8 & & = 5 \end{align}
$$ \therefore \text{Points of intersection are } \left(8, {5 \over 2}\right) \text{ and } (5, 1) $$

(b)

\begin{align} 2x - y & = 4 \\ -y & = 4 - 2x \\ y & = 2x - 4 \phantom{000} \text{ --- (1)} \\ \\ 2x^2 + 4xy - 3y & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x^2 + 4x(2x - 4) - 3(2x - 4) & = 0 \\ 2x^2 + 8x^2 - 16x - 6x + 12 & = 0 \\ 10x^2 - 22x + 12 & = 0 \\ 5x^2 - 11x + 6 & = 0 \\ (5x - 6)(x - 1) & = 0 \end{align} \begin{align} 5x - 6 & = 0 &\text{or }\phantom{00000} x - 1 & = 0 \\ 5x & = 6 & x & = 1 \\ x & = {6 \over 5} \\ \\ \text{Substitute } & x = {6 \over 5} \text{ into (1),} & \text{Substitute } & x = 1 \text{ into (1),} \\ y & = 2\left(6 \over 5\right) - 4 & y & = 2(1) - 4 \\ & = -{8 \over 5} & & = - 2 \end{align}
$$ \therefore \text{Points of intersection are } \left( {6 \over 5}, -{8 \over 5} \right) \text{ and } (1, -2) $$

(c)

\begin{align} 3x + y & = 1 \\ y & = 1 - 3x \phantom{000} \text{ --- (1)} \\ \\ (x + y)(x + 2y) & = 3 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ [x + (1 - 3x)][x + 2(1 - 3x)] & = 3 \\ (x + 1 - 3x)(x + 2 - 6x) & = 3 \\ (1 - 2x)(2 - 5x) & = 3 \\ 2 - 5x - 4x + 10x^2 & = 3 \\ 10x^2 - 9x + 2 & = 3 \\ 10x^2 - 9x - 1 & = 0 \\ (10x + 1)(x - 1) & = 0 \end{align} \begin{align} 10x + 1 & = 0 &\text{or }\phantom{00000} x - 1 & = 0 \\ 10x & = -1 & x & = 1 \\ x & = -{1 \over 10} \\ \\ \text{Substitute } & x = -{1 \over 10} \text{ into (1),} & \text{Substitute } & x = 1 \text{ into (1),} \\ y & = 1 - 3\left(-{1 \over 10}\right) & y & = 1 - 3(1) \\ & = {13 \over 10} & & = -2 \end{align}
$$ \therefore \text{Points of intersection are } \left( -{1 \over 10}, {13 \over 10} \right) \text{ and } (1, -2) $$

\begin{align} {2x \over y} + {y \over x} & = 3 \phantom{000} \text{ --- (1)} \\ \\ 3x - y & = 2 \\ -y & = 2 - 3x \\ y & = 3x - 2 \phantom{000} \text{ --- (2)} \\ \\ \therefore \text{Substitute } & \text{(2) into (1),} \\ {2x \over 3x - 2} + {3x - 2 \over x} & = 3 \\ {2x(x) \over x(3x - 2)} + {(3x - 2)(3x - 2) \over x(3x - 2)} & = 3 \\ {2x^2 + (3x - 2)^2 \over x(3x - 2)} & = 3 \\ {2x^2 + (3x)^2 - 2(3x)(2) + (2)^2 \over x(3x - 2)} & = 3 \\ {2x^2 + 9x^2 - 12x + 4 \over x(3x - 2)} & = 3 \\ {11x^2 - 12x + 4 \over x(3x - 2)} & = 3 \\ 11x^2 - 12x + 4 & = 3x(3x - 2) \\ 11x^2 - 12x + 4 & = 9x^2 - 6x \\ 2x^2 - 6x + 4 & = 0 \\ x^2 - 3x + 2 & = 0 \\ (x - 1)(x - 2) & = 0 \end{align} \begin{align} x - 1 & = 0 &\text{or }\phantom{00000} x - 2 & = 0 \\ x & = 1 & x & = 2 \\ \\ \text{Substitute } & x = 1 \text{ into (2),} & \text{Substitute } & x = 2 \text{ into (2),} \\ y & = 3(1) - 2 & y & = 3(2) - 2 \\ & = 1 & & = 4 \end{align}
$$ \text{Coordinates are } (1, 1) \text{ and } (2, 4) $$

\begin{align} \text{Let } x \text{ and } y \text{ denote } & \text{the width and height of the monitor respectively}. \\ \\ {x \over y} & = {4 \over 3} \\ x & = {4 \over 3}y \phantom{000} \text{ --- (1)} \\ \\ \text{By Pythagora's }& \text{theorem,} \\ x^2 + y^2 & = (\text{Diagonal})^2 \\ x^2 + y^2 & = 15^2 \\ x^2 + y^2 & = 225 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ \left({4 \over 3}y\right)^2 + y^2 & = 225 \\ {16 \over 9}y^2 + y^2 & = 225 \\ 16y^2 + 9y^2 & = 2025 \\ 25y^2 & = 2025 \\ y^2 & = {2025 \over 25} \\ & = 81 \\ y & = \pm \sqrt{81} \\ & = \pm 9 \\ & = 9 \text{ or } - 9 \text{ (Reject)} \\ \\ \text{Substitute } y & = 9 \text{ into (1),} \\ x & = {4 \over 3}(9) \\ & = 12 \end{align}
$$ \therefore \text{Dimensions are } 12 \text{ in by } 9 \text{ in} $$

(i)

\begin{align} \text{By Pythagora's} & \text{ theorem,} \\ x^2 + y^2 & = 5^2 \\ x^2 + y^2 & = 25 \phantom{000} \text{ --- (1)} \end{align}

(ii)

\begin{align} y - x & = 1 \\ y & = x + 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ x^2 + (x + 1)^2 & = 25 \\ x^2 + (x)^2 + 2(x)(1) + (1)^2 & = 25 \\ x^2 + x^2 + 2x + 1 & = 25 \\ 2x^2 + 2x - 24 & = 0 \\ x^2 + x - 12 & = 0 \\ (x + 4)(x - 3) & = 0 \end{align} \begin{align} x + 4 & = 0 &\text{or }\phantom{00000} x - 3 & = 0 \\ x & = -4 & x & = 3 \\ \\ \text{Substitute } & x = -4 \text{ into (2),} & \text{Substitute } & x = 3 \text{ into (2),} \\ y & = (-4) + 1 & y & = (3) + 1 \\ & = -3 & & = 4 \end{align}
$$ \therefore x = -4, y = -3 \text{ or } x = 3, y = 4 $$

(iii)

$$\text{Since } x \text{ and } y \text{ must be positive, } x = 3 \text{ and } y = 4$$

(i)

\begin{align} x^2 + xy + ay & = b \\ \\ \text{When } x & = 2, y = 1, \\ (2)^2 + (2)(1) + a(1) & = b \\ 4 + 2 + a & = b \\ 6 + a & = b \phantom{000} \text{ --- (1)} \\ \\ 2ax + 3y & = b \\ \\ \text{When } & x = 2, y = 1, \\ 2a(2) + 3(1) & = b \\ 4a + 3 & = b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4a + 3 & = 6 + a \\ 4a - a & = 6 - 3 \\ 3a & = 3 \\ a & = {3 \over 3} \\ & = 1 \\ \\ \text{Substitute } & a = 1 \text{ into (1),} \\ 6 + (1) & = b \\ 7 & = b \\ \\ \therefore a = 1, & \phantom{0} b = 7 \end{align}

(ii)

\begin{align} \text{When } & a = 1, \phantom{0} b = 7, \\ x^2 + xy + y & = 7 \phantom{000} \text{ --- (3)} \\ \\ 2x + 3y & = 7 \\ 3y & = 7 - 2x \\ y & = {7 \over 3} - {2 \over 3}x \phantom{000} \text{ --- (4)} \\ \\ \text{Substitute } & \text{(4) into (3),} \\ x^2 + x\left({7 \over 3} - {2 \over 3}x \right) + \left( {7 \over 3} - {2 \over 3}x \right) & = 7 \\ x^2 + {7 \over 3}x - {2 \over 3}x^2 + {7 \over 3} - {2 \over 3}x & = 7 \\ 3x^2 + 7x - 2x^2 + 7 - 2x & = 21 \\ x^2 + 5x - 14 & = 0 \\ (x + 7)(x - 2) & = 0 \end{align} \begin{align} x + 7 & = 0 &\text{or }\phantom{00000} x - 2 & = 0 \\ x & = - 7 & x & = 2 \\ \\ \\ \text{Substitute } & x = -7 \text{ into (4),} & \text{Substitute } & x = 2 \text{ into (4),} \\ y & = {7 \over 3} - {2 \over 3}(-7) & y & = {7 \over 3} - {2 \over 3}(2) \\ & = 7 & & = 1 \end{align}
$$ \therefore \text{Other solution is } x = -7, y = 7 $$

(i)

\begin{align} 2p^2x - 5y & = 7 \\ \\ \text{Using } (1, p), \text{ substitute } & x = 1 \text{ and } y = p, \\ 2p^2 (1) - 5(p) & = 7 \\ 2p^2 - 5p & = 7 \\ 2p^2 - 5p - 7 & = 0 \\ (2p - 7)(p + 1) & = 0 \end{align} \begin{align} 2p - 7 & = 0 &\text{or }\phantom{00000} p + 1 & = 0 \\ 2p & = 7 & p & = - 1 \\ p & = {7 \over 2} \end{align} \begin{align} 12x^2 - 5y^2 & = 7 \\ \\ \\ \text{Using } (1, p), \text{ substitute } & x = 1 \text{ and } y = p, \\ 12(1)^2 - 5(p)^2 & = 7 \\ 12 - 5p^2 & = 7 \\ -5p^2 & = -5 \\ p^2 & = {-5 \over -5} \\ & = 1 \\ p & = \pm \sqrt{1} \\ & = \pm 1 \\ \\ \\ \therefore p & = -1 \text{ (Common solution)} \end{align}

(ii)

\begin{align} \text{When } & p = -1, \\ 2(-1)^2x - 5y & = 7 \\ 2(1)x - 5y & = 7 \\ 2x - 5y & = 7 \\ 2x & = 7 + 5y \\ x & = {7 \over 2} + {5 \over 2}y \phantom{000} \text{ --- (1)} \\ \\ 12x^2 - 5y^2 & = 7 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 12 \left({7 \over 2} + {5 \over 2}y\right)^2 - 5y^2 & = 7 \\ 12 \left[ \left(7 \over 2\right)^2 + 2\left(7 \over 2\right)\left({5 \over 2}y\right) + \left({5 \over 2}y\right)^2 \right] - 5y^2 & = 7 \\ 12 \left( {49 \over 4} + {35 \over 2}y + {25 \over 4}y^2 \right) - 5y^2 & = 7 \\ 147 + 210y + 75y^2 - 5y^2 & = 7 \\ 70y^2 + 210y + 140 & = 0 \\ y^2 + 3y + 2 & = 0 \\ (y + 2)(y + 1) & = 0 \end{align} \begin{align} y + 2 & = 0 &\text{or } \phantom{00000} y + 1 & = 0 \\ y & = -2 & y & = - 1 \text{ (Repeated)} \end{align} \begin{align} \text{Substitute } & y = -2 \text{ into (1),} \\ x & = {7 \over 2} + {5 \over 2}(-2) \\ & = -{3 \over 2} \\ \\ \therefore \text{Other point } & \text{of intersection is } \left( -{3 \over 2}, - 2 \right). \end{align}

(i)

\begin{align} \text{Total surface area} & = 2\pi r h + 2\pi r^2 \\ 32\pi & = 2\pi rh + 2\pi r^2 \\ 32 & = 2rh + 2r^2 \\ 16 & = rh + r^2 \text{ (Shown)} \end{align}

(ii)

\begin{align} 16 & = rh + r^2 \phantom{000} \text{ --- (1)} \\ \\ h - r & = 4 \\ h & = r + 4 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 16 & = r(r + 4) + r^2 \\ 16 & = r^2 + 4r + r^2 \\ 16 & = 2r^2 + 4r \\ 0 & = 2r^2 + 4r - 16 \\ 0 & = r^2 + 2r - 8 \\ 0 & = (r + 4)(r - 2) \end{align} \begin{align} r + 4 & = 0 &\text{or }\phantom{00000} r - 2 & = 0 \\ r & = - 4 \text{ (Reject)} & r & = 2 \end{align} \begin{align} \text{Substitute } & r = 2 \text{ into (2),} \\ h & = (2) + 4 \\ & = 6 \end{align}
$$ \therefore r = 2, h = 6 $$

(i)

\begin{align} 21x + 6y & = k \\ 6y & = k - 21x \phantom{00} \text{--- (1)} \\ \\ (7x - 20)^2 + (6y + 10)^2 & = 200 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (7x - 20)^2 + (k - 21x + 10)^2 & = 200 \\ \\ \text{If } & k = 10, \\ (7x - 20)^2 + (10 - 21x + 10)^2 & = 200 \\ (7x - 20)^2 + (20 - 21x)^2 & = 200 \\ (7x)^2 - 2(7x)(20) + (20)^2 + (20)^2 - 2(20)(21x) + (21x)^2 & = 200 \\ 49x^2 - 280x + 400 + 400 - 840x + 441x^2 & = 200 \\ 490x^2 - 1120x + 600 & = 0 \\ 49x^2 - 112x + 60 & = 0 \\ (7x - 6)(7x - 10) & = 0 \end{align} \begin{align} 7x - 6 & = 0 && \text{ or } & 7x - 10 & = 0 \\ 7x & = 6 &&& 7x & = 10 \\ x & = {6 \over 7} &&& x & = {10 \over 7} \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ 6y & = 10 - 21 \left(6 \over 7\right) &&& 6y & = 10 - 21 \left(10 \over 7\right) \\ 6y & = -8 &&& 6y & = -20 \\ y & = -{4 \over 3} &&& y & = -{10 \over 3} \end{align}
$$ \therefore x = {6 \over 7}, y = -{4 \over 3} \text{ or } x = {10 \over 7}, y = -{10 \over 3}$$

(ii)

Sketch the graph representing each equation and there will be no point of intersection between the two graphs.

(iii)

\begin{align} \text{From (i), } (7x - 20)^2 + (k - 21x + 10)^2 & = 200 \\ \\ \text{There are at most 2 solutions to any qua} & \text{dratic equation} \end{align}