2022 CEMC Cayley Problems/Problem 9

Problem

The operation $a \nabla b$ is defined by $a \nabla b = \frac{a + b}{a - b}$ for all integers $a$ and $b$ with $a \neq b$ . For example, $2 \nabla 3 = \frac{2 + 3}{2 - 3} = -5$

$\text{ (A) }\ 5 \qquad\text{ (B) }\ -7 \qquad\text{ (C) }\ 7 \qquad\text{ (D) }\ -5 \qquad\text{ (E) }\ 3$

Solution 1

Using the definition of the operation, we can see that $3 \nabla b = \frac{3 + b}{3 - b}$ . Plugging this into the equation, we have:

$\frac{3 + b}{3 - b} = -4$

Multiplying both sides by $3 - b$ , we have:

$3 + b = -4(3 - b)$

$3 + b = -4 \times 3 - 4 \times -b$

$3 + b = -12 + 4b$

$b - 4b = -12 - 3$

$-3b = -15$

$b = \frac{-15}{-3} = \boxed {\textbf {(A) } 5}$

~anabel.disher

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2022 CEMC Cayley Problems/Problem 9

Problem

Solution 1