Question

\(G\) is a set of all rational numbers except \(-1\) and \(*\) is defined by \(a*b = a + b + ab\) for all \(a,b \in G\). In the group \((G,*)\) the solution of \(2^{-1} * x * 3^{-1} = 5\) is

• A. 71
• B. 68
• C. 63/5
• D. 72/5

Hint:

Find inverses using \(a * a^{-1} = 0\) (identity is 0).

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \text{Inverse: } a^{-1} = \frac{-a}{1+a} \]
Step 2: Calculation / Simplification}
\(2^{-1} = \frac{-2}{1+2} = -\frac{2}{3}\)
\(3^{-1} = \frac{-3}{1+3} = -\frac{3}{4}\)
\(2^{-1} * x * 3^{-1} = (-\frac{2}{3}) * x * (-\frac{3}{4})\)
First: \((-\frac{2}{3}) * x = -\frac{2}{3} + x - \frac{2}{3}x = \frac{-2 + 3x - 2x}{3} = \frac{x - 2}{3}\)
Then: \(\frac{x-2}{3} * (-\frac{3}{4}) = \frac{x-2}{3} - \frac{3}{4} + \frac{x-2}{3} \cdot (-\frac{3}{4})\)
\(= \frac{x-2}{3} - \frac{3}{4} - \frac{x-2}{4} = \frac{4(x-2) - 9 - 3(x-2)}{12} = \frac{x - 11}{12} = 5\)
\(x - 11 = 60 \Rightarrow x = 71\)
Step 3: Final Answer
\[ 71 \]