Question paper, Set 1

Put $x = 4$, $y = 1$ in $4x - y$:

LHS $= 4(4) - (1) = 16 - 1 = 15$

RHS $= 15$. LHS = RHS, so $(4, 1)$ is a solution.

Answer: Yes, $(4, 1)$ is a solution

Put $x = 4$, $y = 5$: $(4) + k(5) = -31$

$\Rightarrow 4 + 5k = -31$

$\Rightarrow 5k = -31 - 4 = -35$

$\Rightarrow k = -7$

Answer: $k = -7$

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

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

$\displaystyle \frac{k}{5} = \frac{12}{6}$

$\Rightarrow 6k = 60$

$\Rightarrow k = 10$

Check: $\displaystyle \frac{c_1}{c_2} = \frac{9}{9} = 1 \ne 2$, so the pair has no solution.

Answer: $k = 10$

$\displaystyle \frac{a_1}{a_2} = \frac{15}{5} = 3$, $\displaystyle \frac{b_1}{b_2} = \frac{3}{1} = 3$, $\displaystyle \frac{c_1}{c_2} = \frac{-11}{1} = -11$

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

Answer: None (no solution)

$x - 7y = 12$

$\Rightarrow x = 12 + 7y$

Answer: $x = 12 + 7y$

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

$x + 9y = -16 \quad (2)$

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

Substitute in $(2)$: $(11 + y) + 9y = -16$

$\Rightarrow 11 + y + 9y = -16$

$\Rightarrow 11 + 10y = -16$

$\Rightarrow 10y = -16 - 11$

$\Rightarrow 10y = -27$

$\displaystyle \Rightarrow y = -\frac{27}{10}$

$\displaystyle (1) \Rightarrow x = 11 + (-\frac{27}{10}) = \frac{83}{10}$

$\displaystyle \therefore (x, y) = (\frac{83}{10}, -\frac{27}{10})$

Answer: $\displaystyle (x, y) = (\frac{83}{10}, -\frac{27}{10})$

$5x - y = 27 \quad (1)$

$5x + 3y = 19 \quad (2)$

Subtracting, $-4y = 8$

$\Rightarrow y = -2$

$(1) \Rightarrow 5x - (-2) = 27$

$\Rightarrow 5x + 2 = 27$

$\Rightarrow 5x = 27 - 2$

$\Rightarrow 5x = 25$

$\Rightarrow x = 5$

$\therefore (x, y) = (5, -2)$

Answer: $(x, y) = (5, -2)$

$9x + 4y = -2 \quad (1)$

$3x + 4y = 10 \quad (2)$

$(1) \Rightarrow 9x + 4y = -2$

$(2) \times 3 \Rightarrow 9x + 12y = 30$

Subtracting, $-8y = -32$

$\Rightarrow y = 4$

$(2) \Rightarrow 3x + 4(4) = 10$

$\Rightarrow 3x + 16 = 10$

$\Rightarrow 3x = 10 - 16$

$\Rightarrow 3x = -6$

$\Rightarrow x = -2$

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

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

$a_1 = 4$, $b_1 = 5$, $a_2 = 4$, $b_2 = -3$

$a_1b_2 - a_2b_1 = (4)(-3) - (4)(5) = -12 - 20 = -32$

Answer: $-32$

$5x + y = 24 \quad (1)$

$2x - 3y = 30 \quad (2)$

(i) Substitution method:

$(1) \Rightarrow y = 24 - 5x$

Substitute in $(2)$: $2x - 3(24 - 5x) = 30$

$\Rightarrow 2x - 72 + 15x = 30$

$\Rightarrow 17x - 72 = 30$

$\Rightarrow 17x = 30 + 72$

$\Rightarrow 17x = 102$

$\Rightarrow x = 6$

$(1) \Rightarrow y = 24 - 5(6) = -6$

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

(ii) Elimination method:

$(1) \times 2 \Rightarrow 10x + 2y = 48$

$(2) \times 5 \Rightarrow 10x - 15y = 150$

Subtracting, $17y = -102$

$\Rightarrow y = -6$

$(2) \Rightarrow 2x - 3(-6) = 30$

$\Rightarrow 2x + 18 = 30$

$\Rightarrow 2x = 30 - 18$

$\Rightarrow 2x = 12$

$\Rightarrow x = 6$

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

(iii) Cross multiplication method:

$5x + y - 24 = 0 \quad (1)$

$2x - 3y - 30 = 0 \quad (2)$

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

$\displaystyle \frac{x}{-30 - 72} = \frac{y}{-48 + 150} = \frac{1}{-15 - 2}$

$\displaystyle \frac{x}{-102} = \frac{y}{102} = \frac{1}{-17}$

$\displaystyle \Rightarrow x = \frac{-102}{-17}, \; y = \frac{102}{-17}$

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

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

$\displaystyle x + \frac{7y}{2} = 5 \Rightarrow \frac{2x + 7y}{2} = 5 \Rightarrow 2x + 7y = 10$

$\displaystyle 3x + \frac{7y}{3} = 4 \Rightarrow \frac{9x + 7y}{3} = 4 \Rightarrow 9x + 7y = 12$

Answer: $2x + 7y = 10$, $9x + 7y = 12$

Put $\displaystyle \frac{1}{x} = p$, $\displaystyle \frac{1}{y} = q$ in $\displaystyle \frac{5}{x} - \frac{7}{y} = 11$:

$5p - 7q = 11$

Answer: $5p - 7q = 11$

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

$5p - 3q = 11 \quad (1)$

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

$(1) \times 3 \Rightarrow 15p - 9q = 33$

$(2) \times 5 \Rightarrow 15p + 15q = 25$

Subtracting, $-24q = 8$

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

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

$\Rightarrow 3p - 1 = 5$

$\Rightarrow 3p = 5 + 1$

$\Rightarrow 3p = 6$

$\Rightarrow p = 2$

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

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

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

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

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

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

$5p + 3y - 1 = 0 \quad (1)$

$p - y + 3 = 0 \quad (2)$

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

$\displaystyle \frac{p}{9 - 1} = \frac{y}{-1 - 15} = \frac{1}{-5 - 3}$

$\displaystyle \frac{p}{8} = \frac{y}{-16} = \frac{1}{-8}$

$\displaystyle \Rightarrow p = \frac{8}{-8}, \; y = \frac{-16}{-8}$

$\therefore (p, y) = (-1, 2)$

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

$\therefore (x, y) = (-1, 2)$

Answer: $(x, y) = (-1, 2)$

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

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

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

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

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

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

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

$(1) \Rightarrow q = 1 - 3p$

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

$\Rightarrow p + 3 - 9p = 1$

$\Rightarrow -8p + 3 = 1$

$\Rightarrow -8p = 1 - 3$

$\Rightarrow -8p = -2$

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

$\displaystyle (1) \Rightarrow q = 1 - 3(\frac{1}{4}) = \frac{1}{4}$

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

$\displaystyle \frac{1}{x - 5} = \frac{1}{4} \Rightarrow x - 5 = 4 \Rightarrow x = 9$

$\displaystyle \frac{1}{y - 5} = \frac{1}{4} \Rightarrow y - 5 = 4 \Rightarrow y = 9$

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

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

$\displaystyle \frac{5x - y}{xy} = \frac{5x}{xy} - \frac{y}{xy}$

$\displaystyle = \frac{5}{y} - \frac{1}{x}$

Answer: $\displaystyle \frac{5}{y} - \frac{1}{x}$

$\displaystyle \frac{5x - 3y}{xy} = -4 \Rightarrow \frac{5}{y} - \frac{3}{x} = -4$

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

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

$-3p + 5q = -4 \quad (1)$

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

Answer: $-3p + 5q = -4$, $2p + 4q = -1$

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

$5p - 4q = -1 \quad (1)$

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

Subtracting, $-6q = -3$

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

$\displaystyle (1) \Rightarrow 5p - 4(\frac{1}{2}) = -1$

$\Rightarrow 5p - 2 = -1$

$\Rightarrow 5p = -1 + 2$

$\Rightarrow 5p = 1$

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

$\displaystyle \therefore (p, q) = (\frac{1}{5}, \frac{1}{2})$

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

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

$\therefore (x, y) = (25, 4)$

Answer: $(x, y) = (25, 4)$