Question paper, Set 8

$2x - (5) = 3$

$\Rightarrow 2x - 5 = 3$

$\Rightarrow 2x = 3 + 5 = 8$

$\Rightarrow x = 4$

Answer: $x = 4$

$4x + y + 6 = 0 \Rightarrow y = -6 - 4x$

$x = -2$: $y = -6 - 4(-2) = 2$, point $(-2, 2)$

$x = 0$: $y = -6 - 4(0) = -6$, point $(0, -6)$

$x = 2$: $y = -6 - 4(2) = -14$, point $(2, -14)$

All its solutions lie on one straight line, so the graph is a straight line.

Answer: $y = -6 - 4x$; $(-2, 2)$, $(0, -6)$, $(2, -14)$; a straight line

Here $a_1 = 8$, $b_1 = -3$, $a_2 = 4$, $b_2 = 9$.

$\displaystyle \frac{a_1}{a_2} = \frac{8}{4} = 2$, $\displaystyle \frac{b_1}{b_2} = \frac{-3}{9} = -\frac{1}{3}$

Answer: $\displaystyle \frac{a_1}{a_2} = 2$, $\displaystyle \frac{b_1}{b_2} = -\frac{1}{3}$

$\displaystyle \frac{a_1}{a_2} = \frac{9}{1} = 9$, $\displaystyle \frac{b_1}{b_2} = \frac{8}{-5} = -\frac{8}{5}$

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

$\displaystyle \frac{1}{k} = \frac{5}{20}$

$\Rightarrow 5k = 20$

$\Rightarrow k = 4$

Check: $\displaystyle \frac{c_1}{c_2} = \frac{9}{-12} = -\frac{3}{4} \ne \frac{1}{4}$, so the pair has no solution.

Answer: $k = 4$

$\displaystyle \frac{a_1}{a_2} = \frac{4}{8} = \frac{1}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{6}{12} = \frac{1}{2}$, $\displaystyle \frac{c_1}{c_2} = \frac{-4}{-8} = \frac{1}{2}$

$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, so the lines coincide: infinitely many solutions.

Answer: Infinitely many solutions (coincident lines)

$x - 6y = 3$

$\Rightarrow x = 3 + 6y$

Answer: $x = 3 + 6y$

$2(6 - 5y) + 3y = 33$

$\Rightarrow 12 - 10y + 3y = 33$

$\Rightarrow 12 - 7y = 33$

$\Rightarrow -7y = 33 - 12$

$\Rightarrow -7y = 21$

$\Rightarrow y = -3$

Answer: $y = -3$

$7x + 5y = 32 \quad (1)$

$9x - 2y = -1 \quad (2)$

$(1) \times 9 \Rightarrow 63x + 45y = 288$

$(2) \times 7 \Rightarrow 63x - 14y = -7$

Subtracting, $59y = 295$

$\Rightarrow y = 5$

$(1) \Rightarrow 7x + 5(5) = 32$

$\Rightarrow 7x + 25 = 32$

$\Rightarrow 7x = 32 - 25$

$\Rightarrow 7x = 7$

$\Rightarrow x = 1$

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

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

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

$a_1b_2 - a_2b_1 = (2)(-2) - (3)(1) = -4 - 3 = -7$

Answer: $-7$

$5x - y - 7 = 0 \quad (1)$

$2x - 5y + 11 = 0 \quad (2)$

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

$\displaystyle \frac{x}{-11 - 35} = \frac{y}{-14 - 55} = \frac{1}{-25 + 2}$

$\displaystyle \frac{x}{-46} = \frac{y}{-69} = \frac{1}{-23}$

$\displaystyle \Rightarrow x = \frac{-46}{-23}, \; y = \frac{-69}{-23}$

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

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

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

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

Answer: $3x + 2y = -8$, $2x - y = 2$

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

$5p - 2q = 12 \quad (1)$

$2p - 5q = -12 \quad (2)$

$(1) \times 2 \Rightarrow 10p - 4q = 24$

$(2) \times 5 \Rightarrow 10p - 25q = -60$

Subtracting, $21q = 84$

$\Rightarrow q = 4$

$(2) \Rightarrow 2p - 5(4) = -12$

$\Rightarrow 2p - 20 = -12$

$\Rightarrow 2p = -12 + 20$

$\Rightarrow 2p = 8$

$\Rightarrow p = 4$

$\therefore (p, q) = (4, 4)$

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

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

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

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

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

$6p - 3q = 19 \quad (1)$

$8p - 3q = 25 \quad (2)$

$(1) \times 4 \Rightarrow 24p - 12q = 76$

$(2) \times 3 \Rightarrow 24p - 9q = 75$

Subtracting, $-3q = 1$

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

$\displaystyle (1) \Rightarrow 6p - 3(-\frac{1}{3}) = 19$

$\Rightarrow 6p + 1 = 19$

$\Rightarrow 6p = 19 - 1$

$\Rightarrow 6p = 18$

$\Rightarrow p = 3$

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

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

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

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

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

Put $\displaystyle \frac{1}{x} = p$: $2p + 5y = 5$

$\Rightarrow 2p + 5y - 5 = 0$

Answer: $2p + 5y - 5 = 0$

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

$5p - 4y - 7 = 0 \quad (1)$

$p + 4y - 11 = 0 \quad (2)$

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

$\displaystyle \frac{p}{44 + 28} = \frac{y}{-7 + 55} = \frac{1}{20 + 4}$

$\displaystyle \frac{p}{72} = \frac{y}{48} = \frac{1}{24}$

Answer: $\displaystyle \frac{p}{72} = \frac{y}{48} = \frac{1}{24}$

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

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

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

$(1) \Rightarrow q = 1 - 6p$

Substitute in $(2)$: $2p - 3(1 - 6p) = 2$

$\Rightarrow 2p - 3 + 18p = 2$

$\Rightarrow 20p - 3 = 2$

$\Rightarrow 20p = 2 + 3$

$\Rightarrow 20p = 5$

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

$\displaystyle (1) \Rightarrow q = 1 - 6(\frac{1}{4}) = -\frac{1}{2}$

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

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

$\displaystyle \frac{1}{y + 2} = -\frac{1}{2} \Rightarrow y + 2 = -2 \Rightarrow y = -4$

$\therefore (x, y) = (3, -4)$

Answer: $(x, y) = (3, -4)$

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

$\displaystyle \frac{6x + 9y}{xy} = 20 \Rightarrow \frac{6}{y} + \frac{9}{x} = 20$

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

$2p + 3q = 5 \quad (1)$

$9p + 6q = 20 \quad (2)$

$(1) \times 9 \Rightarrow 18p + 27q = 45$

$(2) \times 2 \Rightarrow 18p + 12q = 40$

Subtracting, $15q = 5$

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

$\displaystyle (1) \Rightarrow 2p + 3(\frac{1}{3}) = 5$

$\Rightarrow 2p + 1 = 5$

$\Rightarrow 2p = 5 - 1$

$\Rightarrow 2p = 4$

$\Rightarrow p = 2$

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

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

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

$\displaystyle \therefore (x, y) = (\frac{1}{2}, 3)$

Answer: $\displaystyle (x, y) = (\frac{1}{2}, 3)$

$\displaystyle \frac{3}{\sqrt{x}} = \frac{3}{5} \Rightarrow \sqrt{x} = 3 \times \frac{5}{3} = 5$

$\Rightarrow x = 5^2 = 25$

Answer: $x = 25$

$\displaystyle \frac{1}{\sqrt{x}} = \frac{1}{3} \Rightarrow \sqrt{x} = 3 \Rightarrow x = 9$

$\displaystyle \frac{1}{\sqrt{y}} = \frac{1}{5} \Rightarrow \sqrt{y} = 5 \Rightarrow y = 25$

Answer: $x = 9$, $y = 25$