Question paper, Set 9

Take all the terms to the left side: $2x + 7y + 9 = 0$

Compare with $ax + by + c = 0$: $a = 2$, $b = 7$, $c = 9$

Answer: $2x + 7y + 9 = 0$; $a = 2$, $b = 7$, $c = 9$

Put $x = 1$, $y = -5$: $2(1) + k(-5) = 32$

$\Rightarrow 2 - 5k = 32$

$\Rightarrow -5k = 32 - 2 = 30$

$\Rightarrow k = -6$

Answer: $k = -6$

$\displaystyle \frac{a_1}{a_2} = \frac{4}{9}$, $\displaystyle \frac{b_1}{b_2} = \frac{-3}{8}$

$\displaystyle \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$, so the lines intersect: a unique solution.

Answer: $\displaystyle \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$: a unique solution; intersecting lines

For no solution: $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$

$\displaystyle \frac{k}{3} = \frac{-4}{-2}$

$\Rightarrow -2k = -12$

$\Rightarrow k = 6$

Check: $\displaystyle \frac{c_1}{c_2} = \frac{9}{-4} = -\frac{9}{4} \ne 2$, so the pair has no solution.

Answer: $k = 6$

$\displaystyle \frac{a_1}{a_2} = \frac{5}{5} = 1$, $\displaystyle \frac{b_1}{b_2} = \frac{2}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{7}{13}$

$\displaystyle \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$, so the lines intersect: a unique solution.

Answer: One (a unique solution)

$x + 5y = -7$

$\Rightarrow x = -7 - 5y$

Answer: $x = -7 - 5y$

$x - 6y = 11 \quad (1)$

$9x + 8y = -13 \quad (2)$

$(1) \Rightarrow x = 11 + 6y$

Substitute in $(2)$: $9(11 + 6y) + 8y = -13$

$\Rightarrow 99 + 54y + 8y = -13$

$\Rightarrow 99 + 62y = -13$

$\Rightarrow 62y = -13 - 99$

$\Rightarrow 62y = -112$

$\displaystyle \Rightarrow y = -\frac{56}{31}$

$\displaystyle (1) \Rightarrow x = 11 + 6(-\frac{56}{31}) = \frac{5}{31}$

$\displaystyle \therefore (x, y) = (\frac{5}{31}, -\frac{56}{31})$

Answer: $\displaystyle (x, y) = (\frac{5}{31}, -\frac{56}{31})$

$4x + 3y = 13 \quad (1)$

$4x - y = 1 \quad (2)$

Subtracting, $4y = 12$

$\Rightarrow y = 3$

$(1) \Rightarrow 4x + 3(3) = 13$

$\Rightarrow 4x + 9 = 13$

$\Rightarrow 4x = 13 - 9$

$\Rightarrow 4x = 4$

$\Rightarrow x = 1$

$\therefore (x, y) = (1, 3)$

Answer: $(x, y) = (1, 3)$

$x + 3y = 6 \quad (1)$

$2x + 3y = 9 \quad (2)$

$(1) \times 2 \Rightarrow 2x + 6y = 12$

$(2) \Rightarrow 2x + 3y = 9$

Subtracting, $3y = 3$

$\Rightarrow y = 1$

$(1) \Rightarrow x + 3(1) = 6$

$\Rightarrow x + 3 = 6$

$\Rightarrow x = 6 - 3$

$\Rightarrow x = 3$

$\therefore (x, y) = (3, 1)$

Answer: $(x, y) = (3, 1)$

$a_1 = 3$, $b_1 = -2$, $a_2 = 4$, $b_2 = -1$

$a_1b_2 - a_2b_1 = (3)(-1) - (4)(-2) = -3 + 8 = 5$

Answer: $5$

$4x + y + 4 = 0 \quad (1)$

$3x + 6y + 7 = 0 \quad (2)$

$\displaystyle \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

$\displaystyle \frac{x}{7 - 24} = \frac{y}{12 - 28} = \frac{1}{24 - 3}$

$\displaystyle \frac{x}{-17} = \frac{y}{-16} = \frac{1}{21}$

$\displaystyle \Rightarrow x = \frac{-17}{21}, \; y = \frac{-16}{21}$

$\displaystyle \therefore (x, y) = (-\frac{17}{21}, -\frac{16}{21})$

Answer: $\displaystyle (x, y) = (-\frac{17}{21}, -\frac{16}{21})$

$\displaystyle \frac{2x}{3} - 3y = 7 \Rightarrow \frac{2x - 9y}{3} = 7 \Rightarrow 2x - 9y = 21$

$\displaystyle \frac{x}{2} - y = -2 \Rightarrow \frac{x - 2y}{2} = -2 \Rightarrow x - 2y = -4$

Answer: $2x - 9y = 21$, $x - 2y = -4$

Put $\displaystyle \frac{1}{x} = p$, $\displaystyle \frac{1}{y} = q$ in $\displaystyle \frac{2}{x} + \frac{4}{y} = -1$:

$2p + 4q = -1$

Answer: $2p + 4q = -1$

Let $\displaystyle \frac{1}{x} = p$ and $\displaystyle \frac{1}{y} = q$.

$9p + 2q = 4 \quad (1)$

$3p + 2q = 2 \quad (2)$

$(1) \Rightarrow 9p + 2q = 4$

$(2) \times 3 \Rightarrow 9p + 6q = 6$

Subtracting, $-4q = -2$

$\displaystyle \Rightarrow q = \frac{1}{2}$

$\displaystyle (2) \Rightarrow 3p + 2(\frac{1}{2}) = 2$

$\Rightarrow 3p + 1 = 2$

$\Rightarrow 3p = 2 - 1$

$\Rightarrow 3p = 1$

$\displaystyle \Rightarrow p = \frac{1}{3}$

$\displaystyle \therefore (p, q) = (\frac{1}{3}, \frac{1}{2})$

$\displaystyle \frac{1}{x} = \frac{1}{3} \Rightarrow x = 3$

$\displaystyle \frac{1}{y} = \frac{1}{2} \Rightarrow y = 2$

$\therefore (x, y) = (3, 2)$

Answer: $(x, y) = (3, 2)$

Let $\displaystyle \frac{1}{x} = p$.

$p + 4y + 1 = 0 \quad (1)$

$p - 2y - 5 = 0 \quad (2)$

$\displaystyle \frac{p}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

$\displaystyle \frac{p}{-20 + 2} = \frac{y}{1 + 5} = \frac{1}{-2 - 4}$

$\displaystyle \frac{p}{-18} = \frac{y}{6} = \frac{1}{-6}$

$\displaystyle \Rightarrow p = \frac{-18}{-6}, \; y = \frac{6}{-6}$

$\therefore (p, y) = (3, -1)$

$\displaystyle \frac{1}{x} = 3 \Rightarrow x = \frac{1}{3}$

$\displaystyle \therefore (x, y) = (\frac{1}{3}, -1)$

Answer: $\displaystyle (x, y) = (\frac{1}{3}, -1)$

Let $\displaystyle \frac{1}{x - 2} = p$ and $\displaystyle \frac{1}{y - 5} = q$.

$6p + q = 3 \quad (1)$

$3p + 3q = 4 \quad (2)$

Answer: $6p + q = 3$, $3p + 3q = 4$

Let $\displaystyle \frac{1}{x - 2} = p$ and $\displaystyle \frac{1}{y + 2} = q$.

$4p + q = 1 \quad (1)$

$2p + 3q = 1 \quad (2)$

$(1) \Rightarrow q = 1 - 4p$

Substitute in $(2)$: $2p + 3(1 - 4p) = 1$

$\Rightarrow 2p + 3 - 12p = 1$

$\Rightarrow -10p + 3 = 1$

$\Rightarrow -10p = 1 - 3$

$\Rightarrow -10p = -2$

$\displaystyle \Rightarrow p = \frac{1}{5}$

$\displaystyle (1) \Rightarrow q = 1 - 4(\frac{1}{5}) = \frac{1}{5}$

$\displaystyle \therefore (p, q) = (\frac{1}{5}, \frac{1}{5})$

$\displaystyle \frac{1}{x - 2} = \frac{1}{5} \Rightarrow x - 2 = 5 \Rightarrow x = 7$

$\displaystyle \frac{1}{y + 2} = \frac{1}{5} \Rightarrow y + 2 = 5 \Rightarrow y = 3$

$\therefore (x, y) = (7, 3)$

Answer: $(x, y) = (7, 3)$

$\displaystyle \frac{x + 8y}{xy} = \frac{x}{xy} + \frac{8y}{xy}$

$\displaystyle = \frac{1}{y} + \frac{8}{x}$

Answer: $\displaystyle \frac{1}{y} + \frac{8}{x}$

$\displaystyle \frac{x + 4y}{xy} = -1 \Rightarrow \frac{1}{y} + \frac{4}{x} = -1$

$\displaystyle \frac{3x - 2y}{xy} = 4 \Rightarrow \frac{3}{y} - \frac{2}{x} = 4$

Let $\displaystyle \frac{1}{x} = p$ and $\displaystyle \frac{1}{y} = q$.

$4p + q = -1 \quad (1)$

$-2p + 3q = 4 \quad (2)$

Answer: $4p + q = -1$, $-2p + 3q = 4$

Let $\displaystyle \frac{1}{\sqrt{x}} = p$ and $\displaystyle \frac{1}{\sqrt{y}} = q$.

$4p + 2q = 1 \quad (1)$

$p + 5q = 1 \quad (2)$

$(1) \Rightarrow 4p + 2q = 1$

$(2) \times 4 \Rightarrow 4p + 20q = 4$

Subtracting, $-18q = -3$

$\displaystyle \Rightarrow q = \frac{1}{6}$

$\displaystyle (2) \Rightarrow p + 5(\frac{1}{6}) = 1$

$\displaystyle \Rightarrow p + \frac{5}{6} = 1$

$\displaystyle \Rightarrow p = 1 - \frac{5}{6}$

$\displaystyle \Rightarrow p = \frac{1}{6}$

$\displaystyle \therefore (p, q) = (\frac{1}{6}, \frac{1}{6})$

$\displaystyle \frac{1}{\sqrt{x}} = \frac{1}{6} \Rightarrow \sqrt{x} = 6 \Rightarrow x = 36$

$\displaystyle \frac{1}{\sqrt{y}} = \frac{1}{6} \Rightarrow \sqrt{y} = 6 \Rightarrow y = 36$

$\therefore (x, y) = (36, 36)$

Answer: $(x, y) = (36, 36)$