MCQ test, Set 2

Answer: (A) straight line

Each solution $(x, y)$ is a point, and all these points lie on one straight line.

So $ax + by + c = 0$ represents a straight line.

Answer: (B) $-1$

$6x + (1) = -5$

$\Rightarrow 6x + 1 = -5$

$\Rightarrow 6x = -5 - 1 = -6$

$\Rightarrow x = -1$

Answer: (D) $-3$

Put $x = 4$, $y = 2$: $5(4) + k(2) = 14$

$\Rightarrow 20 + 2k = 14$

$\Rightarrow 2k = 14 - 20 = -6$

$\Rightarrow k = -3$

Answer: (B) $\displaystyle \frac{4}{5}$

Here $a_1 = 4$, $a_2 = 5$.

$\displaystyle \frac{a_1}{a_2} = \frac{4}{5}$

Answer: (D) no solution

$\displaystyle \frac{a_1}{a_2} = \frac{12}{3} = 4$, $\displaystyle \frac{b_1}{b_2} = \frac{16}{4} = 4$, $\displaystyle \frac{c_1}{c_2} = \frac{-3}{2}$

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

Answer: (C) $20$

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

$\displaystyle \frac{5}{k} = \frac{-3}{-12}$

$\Rightarrow -3k = -60$

$\Rightarrow k = 20$

Check: $\displaystyle \frac{c_1}{c_2} = \frac{-2}{3} \ne \frac{1}{4}$, so the pair has no solution.

Answer: (A) $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

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

Answer: (B) $16x - 4y = -20$

With $16x - 4y = -20$: $\displaystyle \frac{a_1}{a_2} = \frac{4}{16} = \frac{1}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{-1}{-4} = \frac{1}{4}$, $\displaystyle \frac{c_1}{c_2} = \frac{-5}{-20} = \frac{1}{4}$

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

Answer: (C) $6 + 5y = 1$

$3(2 + 3y) - 4y = 1$

$\Rightarrow 6 + 9y - 4y = 1$

$\Rightarrow 6 + 5y = 1$

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

$x - 4y = -7 \quad (1)$

$5x + 2y = -13 \quad (2)$

$(1) \Rightarrow x = -7 + 4y$

Substitute in $(2)$: $5(-7 + 4y) + 2y = -13$

$\Rightarrow -35 + 20y + 2y = -13$

$\Rightarrow -35 + 22y = -13$

$\Rightarrow 22y = -13 + 35$

$\Rightarrow 22y = 22$

$\Rightarrow y = 1$

$(1) \Rightarrow x = -7 + 4(1) = -3$

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

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

$3x - 4y = -14 \quad (1)$

$9x - 5y = -28 \quad (2)$

$(1) \times 3 \Rightarrow 9x - 12y = -42$

$(2) \Rightarrow 9x - 5y = -28$

Subtracting, $-7y = -14$

$\Rightarrow y = 2$

$(1) \Rightarrow 3x - 4(2) = -14$

$\Rightarrow 3x - 8 = -14$

$\Rightarrow 3x = -14 + 8$

$\Rightarrow 3x = -6$

$\Rightarrow x = -2$

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

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

The LCM of $4$ and $8$ is $8$.

$(1) \times 2 \Rightarrow 8x - 18y = -14$

$(2) \times 1 \Rightarrow 8x - 2y = 7$

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

Answer: (A) $5$

$a_1b_2 - a_2b_1 = (5)(-1) - (5)(-2) = -5 + 10 = 5$

Answer: (B) $(1, 5)$

$\displaystyle x = \frac{-13}{-13} = 1$

$\displaystyle y = \frac{-65}{-13} = 5$

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

$2x + y = 5 \quad (1)$

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

$(1) \Rightarrow 2x + y = 5$

$(2) \times 2 \Rightarrow 2x - 6y = 12$

Subtracting, $7y = -7$

$\Rightarrow y = -1$

$(2) \Rightarrow x - 3(-1) = 6$

$\Rightarrow x + 3 = 6$

$\Rightarrow x = 6 - 3$

$\Rightarrow x = 3$

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

Answer: (B) $5x + 2y = 18$

Multiply every term by the LCM $6$:

$\displaystyle \frac{5x}{6} + \frac{y}{3} = 3 \Rightarrow \frac{5x + 2y}{6} = 3 \Rightarrow 5x + 2y = 18$

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

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

$4p - 5q = 21 \quad (1)$

$3p - 2q = 14 \quad (2)$

$(1) \times 3 \Rightarrow 12p - 15q = 63$

$(2) \times 4 \Rightarrow 12p - 8q = 56$

Subtracting, $-7q = 7$

$\Rightarrow q = -1$

$(2) \Rightarrow 3p - 2(-1) = 14$

$\Rightarrow 3p + 2 = 14$

$\Rightarrow 3p = 14 - 2$

$\Rightarrow 3p = 12$

$\Rightarrow p = 4$

$\therefore (p, q) = (4, -1)$

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

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

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

Answer: (D) $4p - 2q = 11$

$\displaystyle \frac{4}{x} = 4 \times \frac{1}{x} = 4p$ and $\displaystyle \frac{2}{y} = 2q$

So $4p - 2q = 11$

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

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

$2p + 4y + 5 = 0 \quad (1)$

$4p + 3y + 5 = 0 \quad (2)$

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

$\displaystyle \frac{p}{20 - 15} = \frac{y}{20 - 10} = \frac{1}{6 - 16}$

$\displaystyle \frac{p}{5} = \frac{y}{10} = \frac{1}{-10}$

$\displaystyle \Rightarrow p = \frac{5}{-10}, \; y = \frac{10}{-10}$

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

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

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

Answer: (D) $5p + y = 6$

$\displaystyle \frac{5}{x} = 5 \times \frac{1}{x} = 5p$

So $5p + y = 6$

Answer: (D) $(6, 3)$

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

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

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

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

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

$\Rightarrow 4p - 8 + 16p = -3$

$\Rightarrow 20p - 8 = -3$

$\Rightarrow 20p = -3 + 8$

$\Rightarrow 20p = 5$

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

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

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

$\displaystyle \frac{1}{x - 2} = \frac{1}{4} \Rightarrow x - 2 = 4 \Rightarrow x = 6$

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

$\therefore (x, y) = (6, 3)$

Answer: (C) $4$

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

$\Rightarrow x = 4$

Answer: (C) $\displaystyle \frac{7}{y} - \frac{9}{x}$

Divide each term by $xy$: $\displaystyle \frac{7x - 9y}{xy} = \frac{7x}{xy} - \frac{9y}{xy}$

$\displaystyle = \frac{7}{y} - \frac{9}{x}$

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

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

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

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

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

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

$(1) \times 3 \Rightarrow 12p + 15q = -9$

$(2) \times 5 \Rightarrow -10p + 15q = -20$

Subtracting, $22p = 11$

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

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

$\Rightarrow 3q - 1 = -4$

$\Rightarrow 3q = -4 + 1$

$\Rightarrow 3q = -3$

$\Rightarrow q = -1$

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

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

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

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

Answer: (C) $81$

$\sqrt{x} = 9$

$\Rightarrow x = 9^2 = 81$