By substitution or by graphically. Since the equation y = x² - 4 is non-linear and the equation y = 2x - 4 is linear, it is not possible to solve by elimination.

(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 & = 2x - 5 \phantom{000} \text{---(1)} \\ x^2 + y^2 & = 10 \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ x^2 + (2x - 5)^2 & = 10 \\ x^2 + (2x)^2 - 2(2x)(5) + (5)^2 & = 10 \\ x^2 + 4x^2 - 20x + 25 & = 10 \\ 5x^2 - 20x + 15 & = 0 \\ x^2 - 4x + 3 & = 0 \\ (x - 1)(x - 3) & = 0 \\ \\ x - 1 = 0 \phantom{0} & \text{ or } \phantom{0} x - 3 = 0 \\ x = 1 \phantom{0} & \phantom{0000000} x = 3 \\ \\ \text{Substitute } & x = 1 \text{ into (1),} \\ y & = 2(1) - 5 \\ y & = -3 \\ \\ \text{Substitute } & x = 3 \text{ into (1),} \\ y & = 2(3) - 5 \\ y & = 1 \\ \\ \therefore \text{Coordinates of points of } & \text{ intersection are } (1, -3) \text{ and } (3, 1) \end{align*}

(b)

\begin{align*} y & = 11 - 2x \phantom{000} \text{---(1)} \\ 2x^2 - y^2 & = 23 \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ 2x^2 - (11 - 2x)^2 & = 23 \\ 2x^2 - [ (11)^2 - 2(11)(2x) + (2x)^2] & = 23 \\ 2x^2 - (121 - 44x + 4x^2) & = 23 \\ 2x^2 - 121 + 44x - 4x^2 & = 23 \\ -2x^2 + 44x - 144 & = 0 \\ x^2 - 22x + 72 & = 0 \\ (x - 4)(x - 18) & = 0 \\ \\ x - 4 = 0 \phantom{0} & \text{ or } \phantom{0} x - 18 = 0 \\ x = 4 \phantom{0} & \phantom{00000000} x = 18 \\ \\ \text{Substitute } & x = 4 \text{ into (1),} \\ y & = 11 - 2(4) \\ y & = 3 \\ \\ \text{Substitute } & x = 18 \text{ into (1),} \\ y & = 11 - 2(18) \\ y & = -25 \\ \\ \therefore \text{Coordinates of points of } & \text{ intersection are } (4, 3) \text{ and } (18, -25) \end{align*}

(c)

\begin{align*} x & = 5 - y \phantom{000} \text{---(1)} \\ \\ x^2 + x + y^2 & = 51 \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ (5 - y)^2 + (5 - y) + y^2 & = 51 \\ (5)^2 - 2(5)(y) + (y)^2 + 5 - y + y^2 & = 51 \\ 25 - 10y + y^2 + 5 - y + y^2 & = 51 \\ 2y^2 - 11y + 30 & = 51 \\ 2y^2 - 11y - 21 & = 0 \\ (y - 7)(2y + 3) & = 0 \\ \\ y - 7 = 0 \phantom{0} & \text{ or } \phantom{0} 2y + 3 = 0 \\ y = 7 \phantom{0} & \phantom{0000000} 2y = -3 \\ & \phantom{00000000} y = -{3 \over 2} \\ \\ \text{Substitute } & y = 7 \text{ into (1),} \\ x & = 5 - (7) \\ x & = -2 \\ \\ \text{Substitute } & x = -{3 \over 2} \text{ into (1),} \\ x & = 5 - \left(-{3 \over 2}\right) \\ x & = {13 \over 2} \\ \\ \therefore \text{Coordinates of points of } & \text{ intersection are } (-2, 7) \text{ and } \left({13 \over 2}, -{3 \over 2}\right) \end{align*}

\begin{align*} 2x^2 + xy - y^2 & = 26 \phantom{000} \text{---(1)} \\ \\ 4x - 2y & = 26 \\ -2y & = 26 - 4x \\ y & = 2x - 13 \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ 2x^2 + x(2x - 13) - (2x - 13)^2 & = 26 \\ 2x^2 + 2x^2 - 13x - [ (2x)^2 - 2(2x)(13) + (13)^2] & = 26 \\ 4x^2 - 13x - (4x^2 - 52x + 169) & = 26 \\ 4x^2 - 13x - 4x^2 + 52x - 169 & = 26 \\ 39x - 169 & = 26 \\ 39x & = 195 \\ x & = {195 \over 39} \\ x & = 5 \\ \\ \text{Substitute } & x = 5 \text{ into (2),} \\ y & = 2(5) -13 \\ y & = -3 \end{align*}

(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}
$$ \therefore \text{Points of intersection are } \left( -{1 \over 10}, {13 \over 10} \right) \text{ and } (1, -2) $$

(i)

There are no points of intersection between the curve y = x² + 1 and the line y = x.

(ii)

Draw a line parallel to y = x and has y-intercept 2, i.e. y = x + 2

(iii)

$$ x^2 + 1 = x + 2 $$

(i) Since the equation y = 2x² + bx + c has a minimum value, the minimum value of y = 2x² + bx + c is - 4

2 solutions

(ii)

0 solution

\begin{align*} x + x + y & = 18 \\ 2x + y & = 18 \\ y & = 18 - 2x \phantom{000} \text{---(1)} \\ \\ xy & = 40 \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ x(18 - 2x )& = 40 \\ 18x - 2x^2 & = 40 \\ 0 & = 2x^2 - 18x + 40 \\ 0 & = x^2 - 9x + 20 \\ 0 & = (x - 4)(x - 5) \\ \\ x - 4 = 0 \phantom{0} & \text{ or } \phantom{0} x - 5 = 0 \\ x = 4 \phantom{0} & \phantom{0000000} x = 5 \\ \\ \text{Substitute } & x = 4 \text{ into (1),} \\ y & = 18 - 2(4) \\ y & = 10 \\ \\ \text{Substitute } & x = 5 \text{ into (1),} \\ y & = 18 - 2(5) \\ y & = 8 \\ \\ \therefore \text{Dimensions of rectangle} & \text{ is } 4 \text{ m by 10 m or 5 m by 8 m} \end{align*}

(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}
$$ \therefore \text{Dimensions of land is } 12 \text{ m by } 18 \text{ m} $$

(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)

Since the length of the base of the ladder from the wall ($x$) and the length of the top of the ladder from the floor ($y$) must be positive, only $x = 3$ and $y = 4$ can be accepted.

(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*} 12x^2 - 5y^2 & = 7 \\ \\ \text{Using } & (1, p), \\ 12(1)^2 - 5(p)^2 & = 7 \\ 12 - 5p^2 & = 7 \\ -5p^2 & = 7 - 12 \\ -5p^2 & = -5 \\ p^2 & = {-5 \over -5} \\ p^2 & = 1 \\ p & = \pm \sqrt{1} \\ p & = \pm 1 \\ \\ 2p^2 x - 5y & = 7 \\ \\ \text{Using } & (1, p), \\ 2p^2 (1) - 5(p) & = 7 \\ 2p^2 - 5p & = 7 \\ 2p^2 - 5p - 7 & = 0 \\ (p + 1)(2p - 7) & = 0 \\ \\ p + 1 = 0 \phantom{00} & \text{ or } \phantom{0} 2p - 7 = 0 \\ p = - 1 \phantom{.} & \phantom{00000000} p = {7 \over 2} \\ \\ \therefore p & = -1 \end{align*}

(ii)

\begin{align*} 12x^2 - 5y^2 & = 7 \phantom{000} \text{---(1)} \\ \\ 2p^2 x - 5y & = 7 \\ 2(-1)^2 x - 5y & = 7 \\ 2x - 5y & = 7 \\ 2x & = 5y + 7 \\ x & = 2.5y + 3.5 \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & (2) \text{ into } (1), \\ 12(2.5y + 3.5)^2 - 5y^2 & = 7 \\ 12[ (2.5y)^2 + 2(2.5y)(3.5) + (3.5)^2] - 5y^2 & = 7 \\ 12(6.25y^2 + 17.5y + 12.25) - 5y^2 & = 7 \\ 75y^2 + 210y + 147 - 5y^2 & = 7 \\ 70y^2 + 210y + 140 & = 0 \\ y^2 + 3y + 2 & = 0 \\ (y + 1)(y + 2) & = 0 \\ \\ y + 1 = 0 \phantom{00.} & \text{ or } \phantom{0} y + 2 = 0 \\ y = -1 \phantom{0} & \phantom{0000000} y = -2 \\ \\ \text{Substitute } & y = -2 \text{ into (2),} \\ x & = 2.5(-2) + 3.5 \\ x & = -1.5 \\ \\ \therefore \text{Other coordinates} & \text{ is } (-1.5, -2) \end{align*}

\begin{align*} {x \over 2} + {y \over 5} & = 5 \\ 10\left({x \over 2} + {y \over 5} \right) & = 10(5) \\ 5x + 2y & = 50 \\ 2y & = 50 - 5x \\ y & = 25 - 2.5x \phantom{000} \text{---(1)} \\ \\ {2 \over x} + {5 \over y} & = {5 \over 6} \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ {2 \over x} + {5 \over 25 - 2.5x} & = {5 \over 6} \\ {2(25 - 2.5x) \over x(25 - 2.5x)} + {5(x) \over x(25- 2.5x)} & = {5 \over 6} \\ {2(25 - 2.5x) + 5x \over x(25 - 2.5x)} & = {5 \over 6} \\ {50 - 5x + 5x \over x(25 - 2.5x)} & = {5 \over 6} \\ {50 \over x(25 - 2.5x)} & = {5 \over 6} \\ 6(50) & = 5x(25 - 2.5x) \\ 300 & = 125x - 12.5x^2 \\ 0 & = -12.5x^2 + 125x - 300 \\ 0 & = 12.5x^2 - 125x + 300 \\ 0 & = x^2 - 10x + 24 \\ 0 & = (x - 4)(x - 6) \\ \\ x - 4 = 0 \phantom{0} & \text{ or } \phantom{0} x - 6 = 0 \\ x = 4 \phantom{0} & \phantom{0000000} x = 6 \\ \\ \text{Substitute } & x = 4 \text{ into (1),} \\ y & = 25 - 2.5(4) \\ y & = 15 \\ \\ \text{Substitute } & x = 6 \text{ into (1),} \\ y & = 25 - 2.5(6) \\ y & = 10 \\ \\ \therefore x = 4, y = 15 & \text{ or } \phantom{.} x = 6, y = 10 \end{align*}

\begin{align*} s & = x^2 + y^2 \phantom{000} \text{--- (1)} \\ \\ t & = xy \phantom{000} \text{---(2)} \\ \\ (1) & + 2 \times (2), \\ s + 2t & = x^2 + y^2 + 2xy \\ s + 2t & = x^2 + 2xy + y^2 \\ s + 2t & = (x + y)^2 \\ \pm \sqrt{s + 2t} & = x + y \\ \sqrt{s + 2t} & = x + y \phantom{000} \text{ --- (3)} [\text{Reject negative value since } x > y > 0] \\ \\ (1) & - 2 \times (2), \\ s - 2t & = x^2 + y^2 - 2xy \\ s - 2t & = x^2 - 2xy + y^2 \\ s - 2t & = (x - y)^2 \\ \pm \sqrt{s - 2t} & = x - y \\ \sqrt{s-2t} & = x - y \phantom{000} \text{ --- (4)} \phantom{0} [\text{Reject negative value since } x > y > 0] \\ \\ (3) & + (4), \\ \sqrt{s + 2t} + \sqrt{s - 2t} & = x + y + (x - y) \\ \sqrt{s + 2t} + \sqrt{s - 2t} & = 2x \\ {\sqrt{s + 2t} + \sqrt{s - 2t} \over 2} & = x \\ \\ (3) & - (4), \\ \sqrt{s + 2t} - \sqrt{s - 2t} & = x + y - (x - y) \\ \sqrt{s + 2t} - \sqrt{s - 2t} & = x + y - x + y \\ \sqrt{s + 2t} - \sqrt{s - 2t} & = 2y \\ {\sqrt{s + 2t} - \sqrt{s - 2t} \over 2} & = y \end{align*}

(i)

\begin{align*} 21x + 6y & = 10 \\ 6y & = 10 - 21x \phantom{000} \text{--- (1)} \\ \\ (7x - 20)^2 + (6y + 10)^2 & = 200 \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & \text{(1) into (2)}, \\ (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 \\ \\ 7x - 6 = 0 \phantom{0} & \text{ or } \phantom{0} 7x - 10 = 0 \\ x = {6 \over 7} \phantom{.} & \phantom{000000000} x = {10 \over 7} \\ \\ \text{Substitute } & x = {6 \over 7} \text{ into (1),} \\ 6y & = 10 - 21 \left(6 \over 7\right) \\ 6y & = -8 \\ y & = {-8 \over 6} \\ y & = -{4 \over 3} \\ \\ \text{Substitute } & x = {10 \over 7} \text{ into (1),} \\ 6y & = 10 - 21 \left(10 \over 7\right) \\ 6y & = -20 \\ y & = {-20 \over 6} \\ y & = -{10 \over 3} \\ \\ \therefore x = {6 \over 7}, y = -{4 \over 3} \text{ or } \phantom{.} x = {10 \over 7}, y = -{10 \over 3} \end{align*}

(ii)

There are no points of intersection between the line and the ellipse (graph on Desmos)

(iii)

\begin{align*} 21x + 6y & = k \\ 6y & = k - 21x \phantom{000} \text{--- (1)} \\ \\ (7x - 20)^2 + (6y + 10)^2 & = 200 \phantom{000} \text{---(2)} \\ \\ \text{Substitute } & \text{(1) into (2)}, \\ (7x - 20)^2 + (k - 21x + 10)^2 & = 200 \\ (7x - 20)^2 + [(k + 10) - 21x]^2 & = 200 \\ (7x)^2 - 2(7x)(20) + (20)^2 + (k + 10)^2 - 2(k + 10)(21x) + (21x)^2 & = 200 \\ 49x^2 - 280x + 400 + k^2 + 2(k)(10) + 10^2 - 42kx - 420x + 441x^2 & = 200 \\ 490x^2 - 700x - 42kx + 400 + k^2 + 20k + 100 & = 200 \\ 490x^2 + (-700 - 42k)x + (k^2 + 20k + 300) & = 0 \end{align*}
\begin{align*} x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ x & = {-(-700-42k) \pm \sqrt{ (-700 - 42k)^2 - 4(490)(k^2 + 20k + 300)} \over 2(490)} \\ x & = { (700 + 42k) \pm \sqrt{ (-700)^2 - 2(-700)(42k) + (42k)^2 - 1960k^2 - 39200k - 588000} \over 980} \\ x & = { (700 + 42k) \pm \sqrt{ 490000 + 58800k + 1764k^2 - 1960k^2 - 39200k - 588000} \over 980} \\ x & = { (700 + 42k) \pm \sqrt{ -196k^2 + 19600k - 98000} \over 980} \end{align*}
\begin{align*} \text{For there to }& \text{be only 1 solution, } \\ -196k^2 + 19600k - 98000 & = 0 \\ 196k^2 - 19600k + 98000 & = 0 \\ k^2 - 100k + 500 & = 0 \\ \left(k - {100 \over 2}\right)^2 - \left(100 \over 2\right)^2 + 500 & = 0 \\ (k - 50)^2 - 2500 + 500 & = 0 \\ (k - 50)^2 - 2000 & = 0 \\ (k - 50)^2 & = 2000 \\ k - 50 & = \pm \sqrt{2000} \\ k & = \pm \sqrt{2000} + 50 \\ k & = 94.721 \text{ or } 5.2786 \\ k & \approx 94.7 \text{ or } 5.28 \end{align*}

(iv)

The line can meet the ellipse twice or once or never.