Question paper, Set 7

Put $x = -2$, $y = -1$ in $x - y$:

LHS $= (-2) - (-1) = -2 + 1 = -1$

RHS $= -4$. LHS $\ne$ RHS, so $(-2, -1)$ is not a solution.

Answer: No, $(-2, -1)$ is not a solution

$2x + y - 4 = 0$

$\Rightarrow y = 4 - 2x$

When $x = 4$: $y = 4 - 2(4) = -4$

Answer: $y = 4 - 2x$; $y = -4$

Here $a_1 = 5$, $b_1 = 6$, $a_2 = 2$, $b_2 = 5$.

$\displaystyle \frac{a_1}{a_2} = \frac{5}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{6}{5}$

Answer: $\displaystyle \frac{a_1}{a_2} = \frac{5}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{6}{5}$

$\displaystyle \frac{a_1}{a_2} = \frac{24}{12} = 2$, $\displaystyle \frac{b_1}{b_2} = \frac{20}{10} = 2$, $\displaystyle \frac{c_1}{c_2} = \frac{-7}{12}$

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

Answer: No solution (parallel lines)

(i) $\displaystyle \frac{a_1}{a_2} = \frac{12}{16} = \frac{3}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{-3}{-4} = \frac{3}{4}$, $\displaystyle \frac{c_1}{c_2} = \frac{-11}{-5} = \frac{11}{5}$: no solution

(ii) $\displaystyle \frac{a_1}{a_2} = \frac{7}{8}$, $\displaystyle \frac{b_1}{b_2} = \frac{-6}{9} = -\frac{2}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{1}{17}$: a unique solution

(iii) $\displaystyle \frac{a_1}{a_2} = \frac{1}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{-3}{-12} = \frac{1}{4}$, $\displaystyle \frac{c_1}{c_2} = \frac{5}{20} = \frac{1}{4}$: infinitely many solutions

Answer: (i) no solution; (ii) a unique solution; (iii) infinitely many solutions

$5x - 3y = -13 \quad (1)$

$x + 3y = -17 \quad (2)$

$(2) \Rightarrow x = -17 - 3y$

Substitute in $(1)$: $5(-17 - 3y) - 3y = -13$

$\Rightarrow -85 - 15y - 3y = -13$

$\Rightarrow -85 - 18y = -13$

$\Rightarrow -18y = -13 + 85$

$\Rightarrow -18y = 72$

$\Rightarrow y = -4$

$(2) \Rightarrow x = -17 - 3(-4) = -5$

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

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

The LCM of $5$ and $7$ is $35$.

$(1) \times 7 \Rightarrow 35x - 56y = 91$

$(2) \times 5 \Rightarrow 35x - 25y = 60$

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

Answer: (1) $\times 7$ and (2) $\times 5$

$3x - 5y = -18 \quad (1)$

$6x + 5y = 9 \quad (2)$

$(1) \times 2 \Rightarrow 6x - 10y = -36$

$(2) \Rightarrow 6x + 5y = 9$

Subtracting, $-15y = -45$

$\Rightarrow y = 3$

$(1) \Rightarrow 3x - 5(3) = -18$

$\Rightarrow 3x - 15 = -18$

$\Rightarrow 3x = -18 + 15$

$\Rightarrow 3x = -3$

$\Rightarrow x = -1$

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

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

$x - 4y + 9 = 0 \quad (1)$

$3x + 2y - 1 = 0 \quad (2)$

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

$\displaystyle \frac{x}{4 - 18} = \frac{y}{27 + 1} = \frac{1}{2 + 12}$

$\displaystyle \frac{x}{-14} = \frac{y}{28} = \frac{1}{14}$

Answer: $\displaystyle \frac{x}{-14} = \frac{y}{28} = \frac{1}{14}$

$3x + y = 25 \quad (1)$

$2x - y = 15 \quad (2)$

(i) Substitution method:

$(1) \Rightarrow y = 25 - 3x$

Substitute in $(2)$: $2x - (25 - 3x) = 15$

$\Rightarrow 2x - 25 + 3x = 15$

$\Rightarrow 5x - 25 = 15$

$\Rightarrow 5x = 15 + 25$

$\Rightarrow 5x = 40$

$\Rightarrow x = 8$

$(1) \Rightarrow y = 25 - 3(8) = 1$

$\therefore (x, y) = (8, 1)$

(ii) Elimination method:

$(1) \times 2 \Rightarrow 6x + 2y = 50$

$(2) \times 3 \Rightarrow 6x - 3y = 45$

Subtracting, $5y = 5$

$\Rightarrow y = 1$

$(2) \Rightarrow 2x - (1) = 15$

$\Rightarrow 2x - 1 = 15$

$\Rightarrow 2x = 15 + 1$

$\Rightarrow 2x = 16$

$\Rightarrow x = 8$

$\therefore (x, y) = (8, 1)$

(iii) Cross multiplication method:

$3x + y - 25 = 0 \quad (1)$

$2x - y - 15 = 0 \quad (2)$

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

$\displaystyle \frac{x}{-15 - 25} = \frac{y}{-50 + 45} = \frac{1}{-3 - 2}$

$\displaystyle \frac{x}{-40} = \frac{y}{-5} = \frac{1}{-5}$

$\displaystyle \Rightarrow x = \frac{-40}{-5}, \; y = \frac{-5}{-5}$

$\therefore (x, y) = (8, 1)$

Answer: $(x, y) = (8, 1)$

Multiply by the LCM $6$:

$\displaystyle \frac{5x}{6} - \frac{3y}{2} = -1 \Rightarrow \frac{5x - 9y}{6} = -1 \Rightarrow 5x - 9y = -6$

Answer: $5x - 9y = -6$

$\displaystyle 5x + \frac{5y}{2} = -10 \Rightarrow \frac{10x + 5y}{2} = -10 \Rightarrow 10x + 5y = -20 \quad (1)$

$4x - y = -14 \quad (2)$

$(1) \times 2 \Rightarrow 20x + 10y = -40$

$(2) \times 5 \Rightarrow 20x - 5y = -70$

Subtracting, $15y = 30$

$\Rightarrow y = 2$

$(2) \Rightarrow 4x - (2) = -14$

$\Rightarrow 4x - 2 = -14$

$\Rightarrow 4x = -14 + 2$

$\Rightarrow 4x = -12$

$\Rightarrow x = -3$

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

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

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

$4p - 5q = 18 \quad (1)$

$3p - 2q = 10 \quad (2)$

Answer: $4p - 5q = 18$, $3p - 2q = 10$

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

$8p + 7y - 11 = 0 \quad (1)$

$4p + y + 7 = 0 \quad (2)$

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

$\displaystyle \frac{p}{49 + 11} = \frac{y}{-44 - 56} = \frac{1}{8 - 28}$

$\displaystyle \frac{p}{60} = \frac{y}{-100} = \frac{1}{-20}$

$\displaystyle \Rightarrow p = \frac{60}{-20}, \; y = \frac{-100}{-20}$

$\therefore (p, y) = (-3, 5)$

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

$\displaystyle \therefore (x, y) = (-\frac{1}{3}, 5)$

Answer: $\displaystyle (x, y) = (-\frac{1}{3}, 5)$

$x + 4 = -3$

$\Rightarrow x = -7$

Answer: $x = -7$

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

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

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

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

$\displaystyle \frac{2x - 6y}{xy} = \frac{2x}{xy} - \frac{6y}{xy}$

$\displaystyle = \frac{2}{y} - \frac{6}{x}$

Answer: $\displaystyle \frac{2}{y} - \frac{6}{x}$

$\displaystyle \frac{5x - 4y}{xy} = 13 \Rightarrow \frac{5}{y} - \frac{4}{x} = 13$

$\displaystyle \frac{3x + 4y}{xy} = 11 \Rightarrow \frac{3}{y} + \frac{4}{x} = 11$

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

$-4p + 5q = 13 \quad (1)$

$4p + 3q = 11 \quad (2)$

$(1) \times 3 \Rightarrow -12p + 15q = 39$

$(2) \times 5 \Rightarrow 20p + 15q = 55$

Subtracting, $-32p = -16$

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

$\displaystyle (2) \Rightarrow 3q + 4(\frac{1}{2}) = 11$

$\Rightarrow 3q + 2 = 11$

$\Rightarrow 3q = 11 - 2$

$\Rightarrow 3q = 9$

$\Rightarrow q = 3$

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

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

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

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

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

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

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

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

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

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

$4p + 8q = 5 \quad (1)$

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

$(1) \Rightarrow 4p + 8q = 5$

$(2) \times 2 \Rightarrow 4p - 18q = -8$

Subtracting, $26q = 13$

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

$\displaystyle (2) \Rightarrow 2p - 9(\frac{1}{2}) = -4$

$\displaystyle \Rightarrow 2p - \frac{9}{2} = -4$

$\displaystyle \Rightarrow 2p = -4 + \frac{9}{2}$

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

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

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

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

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

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

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