MCQ test, Set 8

Answer: (D) $-4$

$5x + (4) = -16$

$\Rightarrow 5x + 4 = -16$

$\Rightarrow 5x = -16 - 4 = -20$

$\Rightarrow x = -4$

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

Put each point in $3x + 5y$; only $(5, -2)$ gives $5$:

$3(5) + 5(-2) = 15 - 10 = 5$

Answer: (C) a unique solution

$\displaystyle \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{-3}{3} = -1$, $\displaystyle \frac{c_1}{c_2} = \frac{22}{8} = \frac{11}{4}$

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

Answer: (D) $6x - 2y = 11$

With $6x - 2y = 11$: $\displaystyle \frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}$, $\displaystyle \frac{c_1}{c_2} = \frac{5}{11}$

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

Answer: (B) no solution

$\displaystyle \frac{a_1}{a_2} = \frac{5}{20} = \frac{1}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{2}{8} = \frac{1}{4}$, $\displaystyle \frac{c_1}{c_2} = \frac{5}{-12} = -\frac{5}{12}$

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

Answer: (B) infinitely many solutions

$\displaystyle \frac{a_1}{a_2} = \frac{20}{10} = 2$, $\displaystyle \frac{b_1}{b_2} = \frac{-12}{-6} = 2$, $\displaystyle \frac{c_1}{c_2} = \frac{-28}{-14} = 2$

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

Answer: (A) $18x + 9y = 15$

With $18x + 9y = 15$: $\displaystyle \frac{a_1}{a_2} = \frac{6}{18} = \frac{1}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{5}{15} = \frac{1}{3}$

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

Answer: (A) $-8x + 24 = 16$

$7x + 3(8 - 5x) = 16$

$\Rightarrow 7x + 24 - 15x = 16$

$\Rightarrow -8x + 24 = 16$

Answer: (D) $-7 + 7y$

$x - 7y = -7$

Move the $y$ term to the right: $x = -7 + 7y$

Answer: (D) $4$

$3x - 4y = -13 \quad (1)$

$9x + 4y = 25 \quad (2)$

$(1) \times 3 \Rightarrow 9x - 12y = -39$

$(2) \Rightarrow 9x + 4y = 25$

Subtracting, $-16y = -64$

$\Rightarrow y = 4$

$(1) \Rightarrow 3x - 4(4) = -13$

$\Rightarrow 3x - 16 = -13$

$\Rightarrow 3x = -13 + 16$

$\Rightarrow 3x = 3$

$\Rightarrow x = 1$

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

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

$3x - 5y = 7 \quad (1)$

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

$(1) \Rightarrow 3x - 5y = 7$

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

Subtracting, $y = 1$

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

$\Rightarrow x - 2 = 2$

$\Rightarrow x = 2 + 2$

$\Rightarrow x = 4$

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

Answer: (B) (1) by $4$ and (2) by $7$

The LCM of $7$ and $4$ is $28$.

$(1) \times 4 \Rightarrow 28x - 8y = 48$

$(2) \times 7 \Rightarrow 28x + 35y = -14$

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

Answer: (B) $-13$

$a_1b_2 - a_2b_1 = (5)(-5) - (3)(-4) = -25 + 12 = -13$

Answer: (A) $(5, 5)$

$5x + 2y - 35 = 0 \quad (1)$

$5x - y - 20 = 0 \quad (2)$

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

$\displaystyle \frac{x}{-40 - 35} = \frac{y}{-175 + 100} = \frac{1}{-5 - 10}$

$\displaystyle \frac{x}{-75} = \frac{y}{-75} = \frac{1}{-15}$

$\displaystyle \Rightarrow x = \frac{-75}{-15}, \; y = \frac{-75}{-15}$

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

Answer: (B) $3x - 10y = -30$

Multiply every term by the LCM $5$:

$\displaystyle \frac{3x}{5} - 2y = -6 \Rightarrow \frac{3x - 10y}{5} = -6 \Rightarrow 3x - 10y = -30$

Answer: (A) $(5, 5)$

$6x - 4y = 10 \quad (1)$

$\displaystyle 2x + \frac{3y}{5} = 13 \Rightarrow \frac{10x + 3y}{5} = 13 \Rightarrow 10x + 3y = 65 \quad (2)$

$(1) \times 5 \Rightarrow 30x - 20y = 50$

$(2) \times 3 \Rightarrow 30x + 9y = 195$

Subtracting, $-29y = -145$

$\Rightarrow y = 5$

$(1) \Rightarrow 6x - 4(5) = 10$

$\Rightarrow 6x - 20 = 10$

$\Rightarrow 6x = 10 + 20$

$\Rightarrow 6x = 30$

$\Rightarrow x = 5$

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

Answer: (B) $5p - q = 1$

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

So $5p - q = 1$

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

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

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

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

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

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

Subtracting, $6q = -2$

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

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

$\Rightarrow 3p + 1 = -2$

$\Rightarrow 3p = -2 - 1$

$\Rightarrow 3p = -3$

$\Rightarrow p = -1$

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

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

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

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

Answer: (C) $6p - 7y = -4$

$\displaystyle \frac{6}{x} = 6 \times \frac{1}{x} = 6p$

So $6p - 7y = -4$

Answer: (A) $-2$

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

$\Rightarrow x = -2$

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

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

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

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

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

Substitute in $(2)$: $3p - 2(5 - 4p) = 1$

$\Rightarrow 3p - 10 + 8p = 1$

$\Rightarrow 11p - 10 = 1$

$\Rightarrow 11p = 1 + 10$

$\Rightarrow 11p = 11$

$\Rightarrow p = 1$

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

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

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

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

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

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

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

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

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

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

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

Subtracting, $-9p = -9$

$\Rightarrow p = 1$

$(1) \Rightarrow 3q - 5(1) = -4$

$\Rightarrow 3q - 5 = -4$

$\Rightarrow 3q = -4 + 5$

$\Rightarrow 3q = 1$

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

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

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

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

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

Answer: (D) $\displaystyle \frac{1}{y} + \frac{8}{x}$

Divide each term by $xy$: $\displaystyle \frac{x + 8y}{xy} = \frac{x}{xy} + \frac{8y}{xy}$

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

Answer: (C) $(16, 4)$

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

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

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

$(1) \times 2 \Rightarrow 4p - 6q = -2$

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

Subtracting, $-10q = -5$

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

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

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

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

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

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

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

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

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

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

Answer: (B) $4$

$\sqrt{x} = 2$

$\Rightarrow x = 2^2 = 4$