Question paper, Set 2

$5(-4) + 3y = -29$

$\Rightarrow -20 + 3y = -29$

$\Rightarrow 3y = -29 + 20 = -9$

$\Rightarrow y = -3$

Answer: $y = -3$

$4x - y + 8 = 0 \Rightarrow y = 8 + 4x$

$x = -1$: $y = 8 + 4(-1) = 4$, point $(-1, 4)$

$x = 1$: $y = 8 + 4(1) = 12$, point $(1, 12)$

$x = 2$: $y = 8 + 4(2) = 16$, point $(2, 16)$

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

Answer: $y = 8 + 4x$; $(-1, 4)$, $(1, 12)$, $(2, 16)$; a straight line

Here $a_1 = 7$, $b_1 = 1$, $a_2 = 5$, $b_2 = 3$.

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

Answer: $\displaystyle \frac{a_1}{a_2} = \frac{7}{5}$, $\displaystyle \frac{b_1}{b_2} = \frac{1}{3}$

$\displaystyle \frac{a_1}{a_2} = \frac{15}{10} = \frac{3}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{6}{4} = \frac{3}{2}$, $\displaystyle \frac{c_1}{c_2} = \frac{-5}{5} = -1$

$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$, so the lines are parallel: no solution.

Answer: No solution (parallel lines)

$\displaystyle \frac{a_1}{a_2} = \frac{8}{16} = \frac{1}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{6}{12} = \frac{1}{2}$, $\displaystyle \frac{c_1}{c_2} = \frac{-14}{-28} = \frac{1}{2}$

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

Answer: Infinitely many solutions (coincident lines)

$4(-2 + 3y) - 5y = 27$

$\Rightarrow -8 + 12y - 5y = 27$

$\Rightarrow -8 + 7y = 27$

$\Rightarrow 7y = 27 + 8$

$\Rightarrow 7y = 35$

$\Rightarrow y = 5$

Answer: $y = 5$

$3x + y = -5 \quad (1)$

$3x - 5y = -29 \quad (2)$

$(1) \Rightarrow y = -5 - 3x$

Substitute in $(2)$: $3x - 5(-5 - 3x) = -29$

$\Rightarrow 3x + 25 + 15x = -29$

$\Rightarrow 18x + 25 = -29$

$\Rightarrow 18x = -29 - 25$

$\Rightarrow 18x = -54$

$\Rightarrow x = -3$

$(1) \Rightarrow y = -5 - 3(-3) = 4$

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

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

The LCM of $9$ and $2$ is $18$.

$(1) \times 2 \Rightarrow 18x + 12y = 28$

$(2) \times 9 \Rightarrow 18x + 18y = -99$

Now the $x$ terms are equal, and subtracting removes $x$.

Answer: (1) $\times 2$ and (2) $\times 9$

$7x - 8y = -12 \quad (1)$

$3x - 7y = -23 \quad (2)$

$(1) \times 3 \Rightarrow 21x - 24y = -36$

$(2) \times 7 \Rightarrow 21x - 49y = -161$

Subtracting, $25y = 125$

$\Rightarrow y = 5$

$(2) \Rightarrow 3x - 7(5) = -23$

$\Rightarrow 3x - 35 = -23$

$\Rightarrow 3x = -23 + 35$

$\Rightarrow 3x = 12$

$\Rightarrow x = 4$

$\therefore (x, y) = (4, 5)$

Answer: $(x, y) = (4, 5)$

$2x + y + 7 = 0 \quad (1)$

$2x + 3y + 13 = 0 \quad (2)$

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

$\displaystyle \frac{x}{13 - 21} = \frac{y}{14 - 26} = \frac{1}{6 - 2}$

$\displaystyle \frac{x}{-8} = \frac{y}{-12} = \frac{1}{4}$

Answer: $\displaystyle \frac{x}{-8} = \frac{y}{-12} = \frac{1}{4}$

$x - 3y + 4 = 0 \quad (1)$

$2x + y - 13 = 0 \quad (2)$

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

$\displaystyle \frac{x}{39 - 4} = \frac{y}{8 + 13} = \frac{1}{1 + 6}$

$\displaystyle \frac{x}{35} = \frac{y}{21} = \frac{1}{7}$

$\displaystyle \Rightarrow x = \frac{35}{7}, \; y = \frac{21}{7}$

$\therefore (x, y) = (5, 3)$

Answer: $(x, y) = (5, 3)$

Multiply by the LCM $3$:

$\displaystyle \frac{x}{3} + 5y = 6 \Rightarrow \frac{x + 15y}{3} = 6 \Rightarrow x + 15y = 18$

Answer: $x + 15y = 18$

$\displaystyle \frac{x}{4} - \frac{7y}{6} = 4 \Rightarrow \frac{3x - 14y}{12} = 4 \Rightarrow 3x - 14y = 48 \quad (1)$

$\displaystyle \frac{x}{2} - \frac{5y}{3} = 6 \Rightarrow \frac{3x - 10y}{6} = 6 \Rightarrow 3x - 10y = 36 \quad (2)$

Subtracting, $-4y = 12$

$\Rightarrow y = -3$

$(1) \Rightarrow 3x - 14(-3) = 48$

$\Rightarrow 3x + 42 = 48$

$\Rightarrow 3x = 48 - 42$

$\Rightarrow 3x = 6$

$\Rightarrow x = 2$

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

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

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

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

$4p + 5q = 6 \quad (2)$

Answer: $4p + q = 2$, $4p + 5q = 6$

Put $\displaystyle \frac{1}{x} = p$: $5p - y = 9$

$\Rightarrow 5p - y - 9 = 0$

Answer: $5p - y - 9 = 0$

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

$3p - 8y - 22 = 0 \quad (1)$

$5p - y - 12 = 0 \quad (2)$

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

$\displaystyle \frac{p}{96 - 22} = \frac{y}{-110 + 36} = \frac{1}{-3 + 40}$

$\displaystyle \frac{p}{74} = \frac{y}{-74} = \frac{1}{37}$

$\displaystyle \Rightarrow p = \frac{74}{37}, \; y = \frac{-74}{37}$

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

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

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

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

$x + 2 = 7$

$\Rightarrow x = 5$

Answer: $x = 5$

$\displaystyle \frac{4x - 9y}{xy} = -13 \Rightarrow \frac{4}{y} - \frac{9}{x} = -13$

$\displaystyle \frac{9x + 2y}{xy} = -7 \Rightarrow \frac{9}{y} + \frac{2}{x} = -7$

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

$-9p + 4q = -13 \quad (1)$

$2p + 9q = -7 \quad (2)$

$(1) \times 9 \Rightarrow -81p + 36q = -117$

$(2) \times 4 \Rightarrow 8p + 36q = -28$

Subtracting, $-89p = -89$

$\Rightarrow p = 1$

$(1) \Rightarrow 4q - 9(1) = -13$

$\Rightarrow 4q - 9 = -13$

$\Rightarrow 4q = -13 + 9$

$\Rightarrow 4q = -4$

$\Rightarrow q = -1$

$\therefore (p, q) = (1, -1)$

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

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

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

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

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

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

Answer: $x = 36$, $y = 9$

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

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

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

$(1) \times 4 \Rightarrow 12p - 12q = 8$

$(2) \times 3 \Rightarrow 12p - 9q = 9$

Subtracting, $-3q = -1$

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

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

$\Rightarrow 3p - 1 = 2$

$\Rightarrow 3p = 2 + 1$

$\Rightarrow 3p = 3$

$\Rightarrow p = 1$

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

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

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

$\therefore (x, y) = (1, 9)$

Answer: $(x, y) = (1, 9)$