Question paper, Set 10

Put $x = -2$, $y = 3$ in $5x - 4y$:

LHS $= 5(-2) - 4(3) = -10 - 12 = -22$

RHS $= -23$. LHS $\ne$ RHS, so $(-2, 3)$ is not a solution.

Answer: No, $(-2, 3)$ is not a solution

$3x - y + 5 = 0 \Rightarrow y = 5 + 3x$

$x = -2$: $y = 5 + 3(-2) = -1$, point $(-2, -1)$

$x = 2$: $y = 5 + 3(2) = 11$, point $(2, 11)$

$x = 3$: $y = 5 + 3(3) = 14$, point $(3, 14)$

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

Answer: $y = 5 + 3x$; $(-2, -1)$, $(2, 11)$, $(3, 14)$; a straight line

Here $a_1 = 1$, $b_1 = -5$, $a_2 = 3$, $b_2 = -2$.

$\displaystyle \frac{a_1}{a_2} = \frac{1}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{-5}{-2} = \frac{5}{2}$

Answer: $\displaystyle \frac{a_1}{a_2} = \frac{1}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{5}{2}$

$\displaystyle \frac{a_1}{a_2} = \frac{3}{9} = \frac{1}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{12}{-7} = -\frac{12}{7}$

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

Answer: No solution (parallel lines)

$\displaystyle \frac{a_1}{a_2} = \frac{2}{1} = 2$, $\displaystyle \frac{b_1}{b_2} = \frac{-10}{-5} = 2$, $\displaystyle \frac{c_1}{c_2} = \frac{-16}{-8} = 2$

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

Answer: Infinitely many solutions (coincident lines)

$2x + 3(5 - 5x) = -37$

$\Rightarrow 2x + 15 - 15x = -37$

$\Rightarrow -13x + 15 = -37$

$\Rightarrow -13x = -37 - 15$

$\Rightarrow -13x = -52$

$\Rightarrow x = 4$

Answer: $x = 4$

$x - 5y = 12 \quad (1)$

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

$(2) \Rightarrow x = 3 + 2y$

Substitute in $(1)$: $(3 + 2y) - 5y = 12$

$\Rightarrow 3 + 2y - 5y = 12$

$\Rightarrow 3 - 3y = 12$

$\Rightarrow -3y = 12 - 3$

$\Rightarrow -3y = 9$

$\Rightarrow y = -3$

$(2) \Rightarrow x = 3 + 2(-3) = -3$

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

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

The LCM of $9$ and $5$ is $45$.

$(1) \times 5 \Rightarrow 45x + 40y = -10$

$(2) \times 9 \Rightarrow 45x + 45y = -117$

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

Answer: (1) $\times 5$ and (2) $\times 9$

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

$7x - 4y = -12 \quad (2)$

$(1) \times 7 \Rightarrow 35x - 49y = 7$

$(2) \times 5 \Rightarrow 35x - 20y = -60$

Subtracting, $-29y = 67$

$\displaystyle \Rightarrow y = -\frac{67}{29}$

$\displaystyle (1) \Rightarrow 5x - 7(-\frac{67}{29}) = 1$

$\displaystyle \Rightarrow 5x + \frac{469}{29} = 1$

$\displaystyle \Rightarrow 5x = 1 - \frac{469}{29}$

$\displaystyle \Rightarrow 5x = -\frac{440}{29}$

$\displaystyle \Rightarrow x = -\frac{88}{29}$

$\displaystyle \therefore (x, y) = (-\frac{88}{29}, -\frac{67}{29})$

Answer: $\displaystyle (x, y) = (-\frac{88}{29}, -\frac{67}{29})$

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

$2x - y - 7 = 0 \quad (2)$

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

$\displaystyle \frac{x}{28 - 14} = \frac{y}{-28 + 7} = \frac{1}{-1 + 8}$

$\displaystyle \frac{x}{14} = \frac{y}{-21} = \frac{1}{7}$

Answer: $\displaystyle \frac{x}{14} = \frac{y}{-21} = \frac{1}{7}$

$3x - 5y + 6 = 0 \quad (1)$

$x + y - 6 = 0 \quad (2)$

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

$\displaystyle \frac{x}{30 - 6} = \frac{y}{6 + 18} = \frac{1}{3 + 5}$

$\displaystyle \frac{x}{24} = \frac{y}{24} = \frac{1}{8}$

$\displaystyle \Rightarrow x = \frac{24}{8}, \; y = \frac{24}{8}$

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

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

Multiply by the LCM $2$:

$\displaystyle \frac{x}{2} - 5y = 4 \Rightarrow \frac{x - 10y}{2} = 4 \Rightarrow x - 10y = 8$

Answer: $x - 10y = 8$

$\displaystyle \frac{x}{3} + 3y = 5 \Rightarrow \frac{x + 9y}{3} = 5 \Rightarrow x + 9y = 15 \quad (1)$

$\displaystyle \frac{2x}{3} - 3y = -8 \Rightarrow \frac{2x - 9y}{3} = -8 \Rightarrow 2x - 9y = -24 \quad (2)$

$(1) \times 2 \Rightarrow 2x + 18y = 30$

$(2) \Rightarrow 2x - 9y = -24$

Subtracting, $27y = 54$

$\Rightarrow y = 2$

$(1) \Rightarrow x + 9(2) = 15$

$\Rightarrow x + 18 = 15$

$\Rightarrow x = 15 - 18$

$\Rightarrow x = -3$

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

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

$\displaystyle \frac{1}{x} = \frac{3}{2} \Rightarrow x = \frac{2}{3}$

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

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

Put $\displaystyle \frac{1}{x} = p$: $3p - 5y = -2$

$\Rightarrow 3p - 5y + 2 = 0$

Answer: $3p - 5y + 2 = 0$

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

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

$4p - 5y + 4 = 0 \quad (2)$

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

$\displaystyle \frac{p}{12 - 20} = \frac{y}{-16 - 16} = \frac{1}{-20 - 12}$

$\displaystyle \frac{p}{-8} = \frac{y}{-32} = \frac{1}{-32}$

$\displaystyle \Rightarrow p = \frac{-8}{-32}, \; y = \frac{-32}{-32}$

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

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

$\therefore (x, y) = (4, 1)$

Answer: $(x, y) = (4, 1)$

$x - 2 = 6$

$\Rightarrow x = 8$

Answer: $x = 8$

$\displaystyle \frac{7x + 2y}{xy} = -12 \Rightarrow \frac{7}{y} + \frac{2}{x} = -12$

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

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

$2p + 7q = -12 \quad (1)$

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

$(1) \times 2 \Rightarrow 4p + 14q = -24$

$(2) \Rightarrow 4p + 3q = -2$

Subtracting, $11q = -22$

$\Rightarrow q = -2$

$(1) \Rightarrow 2p + 7(-2) = -12$

$\Rightarrow 2p - 14 = -12$

$\Rightarrow 2p = -12 + 14$

$\Rightarrow 2p = 2$

$\Rightarrow p = 1$

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

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

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

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

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

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

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

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

Answer: $4p - q = 1$, $p - 4q = -1$

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

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

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

Subtracting, $-6q = -1$

$\displaystyle \Rightarrow q = \frac{1}{6}$

$\displaystyle (1) \Rightarrow 5p - 3(\frac{1}{6}) = 2$

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

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

$\displaystyle \Rightarrow 5p = \frac{5}{2}$

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

$\displaystyle \therefore (p, q) = (\frac{1}{2}, \frac{1}{6})$

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

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

$\therefore (x, y) = (4, 36)$

Answer: $(x, y) = (4, 36)$