MCQ test, Set 9

Answer: (A) $-6$

$5(-4) + 3y = -38$

$\Rightarrow -20 + 3y = -38$

$\Rightarrow 3y = -38 + 20 = -18$

$\Rightarrow y = -6$

Answer: (C) $-2$

Put $x = -1$, $y = -4$: $6(-1) + k(-4) = 2$

$\Rightarrow -6 - 4k = 2$

$\Rightarrow -4k = 2 + 6 = 8$

$\Rightarrow k = -2$

Answer: (C) $20x + 24y = 20$

With $20x + 24y = 20$: $\displaystyle \frac{a_1}{a_2} = \frac{5}{20} = \frac{1}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{-6}{24} = -\frac{1}{4}$, $\displaystyle \frac{c_1}{c_2} = \frac{5}{20} = \frac{1}{4}$

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

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

Here $a_1 = 1$, $a_2 = 3$.

$\displaystyle \frac{a_1}{a_2} = \frac{1}{3}$

Answer: (D) $12$

For no solution: $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$

$\displaystyle \frac{k}{6} = \frac{4}{2}$

$\Rightarrow 2k = 24$

$\Rightarrow k = 12$

Check: $\displaystyle \frac{c_1}{c_2} = \frac{1}{-6} = -\frac{1}{6} \ne 2$, so the pair has no solution.

Answer: (B) $4x - 6y = -14$

With $4x - 6y = -14$: $\displaystyle \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$, $\displaystyle \frac{c_1}{c_2} = \frac{-8}{-14} = \frac{4}{7}$

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

Answer: (C) infinitely many solutions

$\displaystyle \frac{a_1}{a_2} = \frac{6}{9} = \frac{2}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{10}{15} = \frac{2}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{16}{24} = \frac{2}{3}$

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

Answer: (C) $18x - 12y = -15$

With $18x - 12y = -15$: $\displaystyle \frac{a_1}{a_2} = \frac{6}{18} = \frac{1}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{-4}{-12} = \frac{1}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{-5}{-15} = \frac{1}{3}$

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

Answer: (A) $(4, -2)$

$5x - 2y = 24 \quad (1)$

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

$(2) \Rightarrow y = 6 - 2x$

Substitute in $(1)$: $5x - 2(6 - 2x) = 24$

$\Rightarrow 5x - 12 + 4x = 24$

$\Rightarrow 9x - 12 = 24$

$\Rightarrow 9x = 24 + 12$

$\Rightarrow 9x = 36$

$\Rightarrow x = 4$

$(2) \Rightarrow y = 6 - 2(4) = -2$

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

Answer: (D) $-9 + 3y$

$x - 3y = -9$

Move the $y$ term to the right: $x = -9 + 3y$

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

$3x + y = 14 \quad (1)$

$9x - 4y = 7 \quad (2)$

$(1) \times 3 \Rightarrow 9x + 3y = 42$

$(2) \Rightarrow 9x - 4y = 7$

Subtracting, $7y = 35$

$\Rightarrow y = 5$

$(1) \Rightarrow 3x + (5) = 14$

$\Rightarrow 3x + 5 = 14$

$\Rightarrow 3x = 14 - 5$

$\Rightarrow 3x = 9$

$\Rightarrow x = 3$

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

Answer: (A) (1) by $1$ and (2) by $3$

The LCM of $9$ and $3$ is $9$.

$(1) \times 1 \Rightarrow 9x - 2y = 3$

$(2) \times 3 \Rightarrow 9x - 24y = 3$

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

Answer: (B) $(-3, 1)$

$\displaystyle x = \frac{-6}{2} = -3$

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

Answer: (D) $9$

$a_1b_2 - a_2b_1 = (1)(1) - (2)(-4) = 1 + 8 = 9$

Answer: (B) $2x - 5y = 15$

Multiply every term by the LCM $5$:

$\displaystyle \frac{2x}{5} - y = 3 \Rightarrow \frac{2x - 5y}{5} = 3 \Rightarrow 2x - 5y = 15$

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

$x - 3y = -10 \quad (1)$

$\displaystyle \frac{2x}{3} + \frac{y}{6} = 2 \Rightarrow \frac{4x + y}{6} = 2 \Rightarrow 4x + y = 12 \quad (2)$

$(1) \times 4 \Rightarrow 4x - 12y = -40$

$(2) \Rightarrow 4x + y = 12$

Subtracting, $-13y = -52$

$\Rightarrow y = 4$

$(1) \Rightarrow x - 3(4) = -10$

$\Rightarrow x - 12 = -10$

$\Rightarrow x = -10 + 12$

$\Rightarrow x = 2$

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

Answer: (B) $\displaystyle (-\frac{1}{3}, 4)$

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

$5p + 4q = -14 \quad (1)$

$3p + 4q = -8 \quad (2)$

$(1) \times 3 \Rightarrow 15p + 12q = -42$

$(2) \times 5 \Rightarrow 15p + 20q = -40$

Subtracting, $-8q = -2$

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

$\displaystyle (2) \Rightarrow 3p + 4(\frac{1}{4}) = -8$

$\Rightarrow 3p + 1 = -8$

$\Rightarrow 3p = -8 - 1$

$\Rightarrow 3p = -9$

$\Rightarrow p = -3$

$\displaystyle \therefore (p, q) = (-3, \frac{1}{4})$

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

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

$\displaystyle \therefore (x, y) = (-\frac{1}{3}, 4)$

Answer: (C) $7p + 6q = -9$

$\displaystyle \frac{7}{x} = 7 \times \frac{1}{x} = 7p$ and $\displaystyle \frac{6}{y} = 6q$

So $7p + 6q = -9$

Answer: (A) $(-4, 1)$

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

$4p - 5y + 6 = 0 \quad (1)$

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

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

$\displaystyle \frac{p}{15 - 24} = \frac{y}{24 + 12} = \frac{1}{16 + 20}$

$\displaystyle \frac{p}{-9} = \frac{y}{36} = \frac{1}{36}$

$\displaystyle \Rightarrow p = \frac{-9}{36}, \; y = \frac{36}{36}$

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

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

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

Answer: (A) $7p - 4y = -4$

$\displaystyle \frac{7}{x} = 7 \times \frac{1}{x} = 7p$

So $7p - 4y = -4$

Answer: (D) $9$

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

$\Rightarrow x = 9$

Answer: (B) $\displaystyle \frac{9}{y} + \frac{1}{x}$

Divide each term by $xy$: $\displaystyle \frac{9x + y}{xy} = \frac{9x}{xy} + \frac{y}{xy}$

$\displaystyle = \frac{9}{y} + \frac{1}{x}$

Answer: (C) $(-1, 4)$

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

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

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

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

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

Subtracting, $p = -1$

$(1) \Rightarrow 4q - (-1) = 2$

$\Rightarrow 4q + 1 = 2$

$\Rightarrow 4q = 2 - 1$

$\Rightarrow 4q = 1$

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

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

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

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

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

Answer: (C) $(1, 9)$

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

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

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

$(1) \times 4 \Rightarrow 20p - 12q = 16$

$(2) \times 5 \Rightarrow 20p + 15q = 25$

Subtracting, $-27q = -9$

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

$\displaystyle (2) \Rightarrow 4p + 3(\frac{1}{3}) = 5$

$\Rightarrow 4p + 1 = 5$

$\Rightarrow 4p = 5 - 1$

$\Rightarrow 4p = 4$

$\Rightarrow p = 1$

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

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

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

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

Answer: (D) $(1, 25)$

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

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

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

$(1) \times 4 \Rightarrow 20p - 20q = 16$

$(2) \times 5 \Rightarrow 20p + 25q = 25$

Subtracting, $-45q = -9$

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

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

$\Rightarrow 4p + 1 = 5$

$\Rightarrow 4p = 5 - 1$

$\Rightarrow 4p = 4$

$\Rightarrow p = 1$

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

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

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

$\therefore (x, y) = (1, 25)$