MCQ test, Set 7

Answer: (C) $(5, -3)$

Put each point in $x - y$; only $(5, -3)$ gives $8$:

$(5) - (-3) = 5 + 3 = 8$

Answer: (C) $-4$

Put $x = 5$, $y = 2$: $k(5) + 6(2) = -8$

$\Rightarrow 5k + 12 = -8$

$\Rightarrow 5k = -8 - 12 = -20$

$\Rightarrow k = -4$

Answer: (D) $12x - 20y = -36$

With $12x - 20y = -36$: $\displaystyle \frac{a_1}{a_2} = \frac{3}{12} = \frac{1}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{5}{-20} = -\frac{1}{4}$, $\displaystyle \frac{c_1}{c_2} = \frac{-9}{-36} = \frac{1}{4}$

$\displaystyle \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$, so the lines intersect: a unique solution.

Answer: (B) $\displaystyle \frac{5}{3}$

Here $b_1 = 5$, $b_2 = 3$.

$\displaystyle \frac{b_1}{b_2} = \frac{5}{3}$

Answer: (D) no solution

$\displaystyle \frac{a_1}{a_2} = \frac{8}{6} = \frac{4}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{12}{9} = \frac{4}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{-1}{5}$

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

Answer: (A) infinitely many solutions

$\displaystyle \frac{a_1}{a_2} = \frac{1}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{4}{16} = \frac{1}{4}$, $\displaystyle \frac{c_1}{c_2} = \frac{-3}{-12} = \frac{1}{4}$

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

Answer: (C) $-2$

For infinitely many solutions: $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

$\displaystyle \frac{-4}{k} = \frac{16}{8}$

$\Rightarrow 16k = -32$

$\Rightarrow k = -2$

Also $\displaystyle \frac{c_1}{c_2} = \frac{-16}{-8} = 2$, equal to the other ratios.

Answer: (A) $-11 - 3x$

$3x + y = -11$

Move the $x$ term to the right: $y = -11 - 3x$

Answer: (A) $-18 + 20y = 42$

$3(-6 + 5y) + 5y = 42$

$\Rightarrow -18 + 15y + 5y = 42$

$\Rightarrow -18 + 20y = 42$

Answer: (B) $(-2, 5)$

$2x - 5y = -29 \quad (1)$

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

$(2) \Rightarrow x = -12 + 2y$

Substitute in $(1)$: $2(-12 + 2y) - 5y = -29$

$\Rightarrow -24 + 4y - 5y = -29$

$\Rightarrow -24 - y = -29$

$\Rightarrow -y = -29 + 24$

$\Rightarrow -y = -5$

$\Rightarrow y = 5$

$(2) \Rightarrow x = -12 + 2(5) = -2$

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

Answer: (B) (1) by $7$ and (2) by $9$

The LCM of $9$ and $7$ is $63$.

$(1) \times 7 \Rightarrow 63x + 56y = 63$

$(2) \times 9 \Rightarrow 63x + 9y = -9$

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

Answer: (A) $(1, -2)$

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

$6x + 5y = -4 \quad (2)$

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

$(2) \Rightarrow 6x + 5y = -4$

Subtracting, $-3y = 6$

$\Rightarrow y = -2$

$(1) \Rightarrow 3x + (-2) = 1$

$\Rightarrow 3x - 2 = 1$

$\Rightarrow 3x = 1 + 2$

$\Rightarrow 3x = 3$

$\Rightarrow x = 1$

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

Answer: (C) $(3, 5)$

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

$3x - y - 4 = 0 \quad (2)$

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

$\displaystyle \frac{x}{8 + 1} = \frac{y}{3 + 12} = \frac{1}{-3 + 6}$

$\displaystyle \frac{x}{9} = \frac{y}{15} = \frac{1}{3}$

$\displaystyle \Rightarrow x = \frac{9}{3}, \; y = \frac{15}{3}$

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

Answer: (D) $(2, 3)$

$\displaystyle x = \frac{-12}{-6} = 2$

$\displaystyle y = \frac{-18}{-6} = 3$

Answer: (B) $(4, 2)$

$\displaystyle 3x + \frac{y}{2} = 13 \Rightarrow \frac{6x + y}{2} = 13 \Rightarrow 6x + y = 26 \quad (1)$

$6x - 6y = 12 \quad (2)$

Subtracting, $7y = 14$

$\Rightarrow y = 2$

$(1) \Rightarrow 6x + (2) = 26$

$\Rightarrow 6x + 2 = 26$

$\Rightarrow 6x = 26 - 2$

$\Rightarrow 6x = 24$

$\Rightarrow x = 4$

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

Answer: (C) $3x - 2y = -10$

Multiply every term by the LCM $2$:

$\displaystyle \frac{3x}{2} - y = -5 \Rightarrow \frac{3x - 2y}{2} = -5 \Rightarrow 3x - 2y = -10$

Answer: (A) $\displaystyle (\frac{1}{4}, \frac{1}{4})$

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

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

$2p + 3q = 20 \quad (2)$

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

$(2) \times 2 \Rightarrow 4p + 6q = 40$

Subtracting, $-11q = -44$

$\Rightarrow q = 4$

$(2) \Rightarrow 2p + 3(4) = 20$

$\Rightarrow 2p + 12 = 20$

$\Rightarrow 2p = 20 - 12$

$\Rightarrow 2p = 8$

$\Rightarrow p = 4$

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

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

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

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

Answer: (C) $6p + 5q = -7$

$\displaystyle \frac{6}{x} = 6 \times \frac{1}{x} = 6p$ and $\displaystyle \frac{5}{y} = 5q$

So $6p + 5q = -7$

Answer: (D) $2p - 3y = 12$

$\displaystyle \frac{2}{x} = 2 \times \frac{1}{x} = 2p$

So $2p - 3y = 12$

Answer: (D) $\displaystyle (-\frac{1}{3}, -1)$

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

$5p - 3y + 12 = 0 \quad (1)$

$p + 5y + 8 = 0 \quad (2)$

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

$\displaystyle \frac{p}{-24 - 60} = \frac{y}{12 - 40} = \frac{1}{25 + 3}$

$\displaystyle \frac{p}{-84} = \frac{y}{-28} = \frac{1}{28}$

$\displaystyle \Rightarrow p = \frac{-84}{28}, \; y = \frac{-28}{28}$

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

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

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

Answer: (B) $3$

$\displaystyle \frac{1}{x + 4} = \frac{1}{7} \Rightarrow x + 4 = 7$

$\Rightarrow x = 3$

Answer: (A) $(1, 0)$

Let $\displaystyle \frac{1}{x - 4} = p$ and $\displaystyle \frac{1}{y + 1} = q$.

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

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

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

Substitute in $(2)$: $3p - 3(-1 - 6p) = -4$

$\Rightarrow 3p + 3 + 18p = -4$

$\Rightarrow 21p + 3 = -4$

$\Rightarrow 21p = -4 - 3$

$\Rightarrow 21p = -7$

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

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

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

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

$\displaystyle \frac{1}{y + 1} = 1 \Rightarrow y + 1 = 1 \Rightarrow y = 0$

$\therefore (x, y) = (1, 0)$

Answer: (D) $(3, 1)$

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

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

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

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

$-3p + 5q = 4 \quad (2)$

$(1) \times 5 \Rightarrow 15p + 10q = 15$

$(2) \times 2 \Rightarrow -6p + 10q = 8$

Subtracting, $21p = 7$

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

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

$\Rightarrow 2q + 1 = 3$

$\Rightarrow 2q = 3 - 1$

$\Rightarrow 2q = 2$

$\Rightarrow q = 1$

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

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

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

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

Answer: (A) $(36, 4)$

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

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

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

Subtracting, $-4q = -2$

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

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

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

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

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

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

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

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

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

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

Answer: (D) $64$

$\sqrt{x} = 8$

$\Rightarrow x = 8^2 = 64$