Question paper, Set 5

Take all the terms to the left side: $5x + 4y - 5 = 0$

Compare with $ax + by + c = 0$: $a = 5$, $b = 4$, $c = -5$

Answer: $5x + 4y - 5 = 0$; $a = 5$, $b = 4$, $c = -5$

Put $x = 2$, $y = -2$: $k(2) + 4(-2) = -4$

$\Rightarrow 2k - 8 = -4$

$\Rightarrow 2k = -4 + 8 = 4$

$\Rightarrow k = 2$

Answer: $k = 2$

$\displaystyle \frac{a_1}{a_2} = \frac{4}{6} = \frac{2}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{-9}{-1} = 9$

$\displaystyle \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$, 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: $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$

$\displaystyle \frac{k}{20} = \frac{4}{16}$

$\Rightarrow 16k = 80$

$\Rightarrow k = 5$

Check: $\displaystyle \frac{c_1}{c_2} = \frac{-4}{4} = -1 \ne \frac{1}{4}$, so the pair has no solution.

Answer: $k = 5$

$\displaystyle \frac{a_1}{a_2} = \frac{1}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{-9}{-7} = \frac{9}{7}$, $\displaystyle \frac{c_1}{c_2} = \frac{-31}{-37} = \frac{31}{37}$

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

Answer: One (a unique solution)

$3x + y = 5$

$\Rightarrow y = 5 - 3x$

Answer: $y = 5 - 3x$

$5x + y = -9 \quad (1)$

$9x - 9y = 5 \quad (2)$

$(1) \Rightarrow y = -9 - 5x$

Substitute in $(2)$: $9x - 9(-9 - 5x) = 5$

$\Rightarrow 9x + 81 + 45x = 5$

$\Rightarrow 54x + 81 = 5$

$\Rightarrow 54x = 5 - 81$

$\Rightarrow 54x = -76$

$\displaystyle \Rightarrow x = -\frac{38}{27}$

$\displaystyle (1) \Rightarrow y = -9 - 5(-\frac{38}{27}) = -\frac{53}{27}$

$\displaystyle \therefore (x, y) = (-\frac{38}{27}, -\frac{53}{27})$

Answer: $\displaystyle (x, y) = (-\frac{38}{27}, -\frac{53}{27})$

$x + 5y = -6 \quad (1)$

$x - y = 6 \quad (2)$

Subtracting, $6y = -12$

$\Rightarrow y = -2$

$(1) \Rightarrow x + 5(-2) = -6$

$\Rightarrow x - 10 = -6$

$\Rightarrow x = -6 + 10$

$\Rightarrow x = 4$

$\therefore (x, y) = (4, -2)$

Answer: $(x, y) = (4, -2)$

$2x + 5y = 23 \quad (1)$

$x - 4y = -21 \quad (2)$

$(1) \Rightarrow 2x + 5y = 23$

$(2) \times 2 \Rightarrow 2x - 8y = -42$

Subtracting, $13y = 65$

$\Rightarrow y = 5$

$(2) \Rightarrow x - 4(5) = -21$

$\Rightarrow x - 20 = -21$

$\Rightarrow x = -21 + 20$

$\Rightarrow x = -1$

$\therefore (x, y) = (-1, 5)$

Answer: $(x, y) = (-1, 5)$

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

$a_1b_2 - a_2b_1 = (1)(-1) - (4)(1) = -1 - 4 = -5$

Answer: $-5$

$5x - 4y + 15 = 0 \quad (1)$

$3x - 8y - 4 = 0 \quad (2)$

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

$\displaystyle \frac{x}{16 + 120} = \frac{y}{45 + 20} = \frac{1}{-40 + 12}$

$\displaystyle \frac{x}{136} = \frac{y}{65} = \frac{1}{-28}$

$\displaystyle \Rightarrow x = \frac{136}{-28}, \; y = \frac{65}{-28}$

$\displaystyle \therefore (x, y) = (-\frac{34}{7}, -\frac{65}{28})$

Answer: $\displaystyle (x, y) = (-\frac{34}{7}, -\frac{65}{28})$

$\displaystyle x + \frac{7y}{3} = 4 \Rightarrow \frac{3x + 7y}{3} = 4 \Rightarrow 3x + 7y = 12$

$\displaystyle \frac{3x}{4} - \frac{7y}{2} = 2 \Rightarrow \frac{3x - 14y}{4} = 2 \Rightarrow 3x - 14y = 8$

Answer: $3x + 7y = 12$, $3x - 14y = 8$

Put $\displaystyle \frac{1}{x} = p$, $\displaystyle \frac{1}{y} = q$ in $\displaystyle \frac{6}{x} - \frac{2}{y} = 1$:

$6p - 2q = 1$

Answer: $6p - 2q = 1$

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

$8p - 2q = 15 \quad (1)$

$3p + 8q = 10 \quad (2)$

$(1) \times 3 \Rightarrow 24p - 6q = 45$

$(2) \times 8 \Rightarrow 24p + 64q = 80$

Subtracting, $-70q = -35$

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

$\displaystyle (2) \Rightarrow 3p + 8(\frac{1}{2}) = 10$

$\Rightarrow 3p + 4 = 10$

$\Rightarrow 3p = 10 - 4$

$\Rightarrow 3p = 6$

$\Rightarrow p = 2$

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

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

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

$\displaystyle \therefore (x, y) = (\frac{1}{2}, 2)$

Answer: $\displaystyle (x, y) = (\frac{1}{2}, 2)$

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

$4p + y - 14 = 0 \quad (1)$

$p - 5y - 14 = 0 \quad (2)$

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

$\displaystyle \frac{p}{-14 - 70} = \frac{y}{-14 + 56} = \frac{1}{-20 - 1}$

$\displaystyle \frac{p}{-84} = \frac{y}{42} = \frac{1}{-21}$

$\displaystyle \Rightarrow p = \frac{-84}{-21}, \; y = \frac{42}{-21}$

$\therefore (p, y) = (4, -2)$

$\displaystyle \frac{1}{x} = 4 \Rightarrow x = \frac{1}{4}$

$\displaystyle \therefore (x, y) = (\frac{1}{4}, -2)$

Answer: $\displaystyle (x, y) = (\frac{1}{4}, -2)$

$\displaystyle \frac{1}{x + 1} = \frac{1}{4} \Rightarrow x + 1 = 4 \Rightarrow x = 3$

$\displaystyle \frac{1}{y - 5} = \frac{1}{3} \Rightarrow y - 5 = 3 \Rightarrow y = 8$

Answer: $x = 3$, $y = 8$

Let $\displaystyle \frac{1}{x + 3} = p$ and $\displaystyle \frac{1}{y + 2} = q$.

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

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

$(1) \Rightarrow q = -1 - 2p$

Substitute in $(2)$: $5p - 2(-1 - 2p) = -1$

$\Rightarrow 5p + 2 + 4p = -1$

$\Rightarrow 9p + 2 = -1$

$\Rightarrow 9p = -1 - 2$

$\Rightarrow 9p = -3$

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

$\displaystyle (1) \Rightarrow q = -1 - 2(-\frac{1}{3}) = -\frac{1}{3}$

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

$\displaystyle \frac{1}{x + 3} = -\frac{1}{3} \Rightarrow x + 3 = -3 \Rightarrow x = -6$

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

$\therefore (x, y) = (-6, -5)$

Answer: $(x, y) = (-6, -5)$

$\displaystyle \frac{x - 7y}{xy} = \frac{x}{xy} - \frac{7y}{xy}$

$\displaystyle = \frac{1}{y} - \frac{7}{x}$

Answer: $\displaystyle \frac{1}{y} - \frac{7}{x}$

$\displaystyle \frac{2x - 4y}{xy} = -9 \Rightarrow \frac{2}{y} - \frac{4}{x} = -9$

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

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

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

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

Answer: $-4p + 2q = -9$, $2p + 2q = 3$

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

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

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

$(1) \times 3 \Rightarrow 15p - 6q = 6$

$(2) \times 5 \Rightarrow 15p + 10q = 10$

Subtracting, $-16q = -4$

$\displaystyle \Rightarrow q = \frac{1}{4}$

$\displaystyle (2) \Rightarrow 3p + 2(\frac{1}{4}) = 2$

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

$\displaystyle \Rightarrow 3p = 2 - \frac{1}{2}$

$\displaystyle \Rightarrow 3p = \frac{3}{2}$

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

$\displaystyle \therefore (p, q) = (\frac{1}{2}, \frac{1}{4})$

$\displaystyle \frac{1}{\sqrt{x}} = \frac{1}{2} \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4$

$\displaystyle \frac{1}{\sqrt{y}} = \frac{1}{4} \Rightarrow \sqrt{y} = 4 \Rightarrow y = 16$

$\therefore (x, y) = (4, 16)$

Answer: $(x, y) = (4, 16)$