Step 1: Understanding the Question:
We are given a pair of linear equations in two variables: \(\frac{3x}{2} + \frac{5y}{3} = 7\) and \(9x + 10y = 14\). We need to determine the nature of the system (consistent, inconsistent, unique solution, or infinite solutions).
Step 2: Key Formula or Approach:
For a pair of linear equations in standard form:
\[ a_1x + b_1y + c_1 = 0 \]
\[ a_2x + b_2y + c_2 = 0 \]
We compare the ratios of their coefficients:
- If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\): The lines intersect at one point, and the system is consistent with a unique solution.
- If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\): The lines coincide, and the system is consistent with infinitely many solutions.
- If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\): The lines are parallel, and the system has no solution, making it inconsistent.
Step 3: Detailed Explanation:
1. Convert the first equation \(\frac{3x}{2} + \frac{5y}{3} = 7\) into standard integer form:
Multiply the entire equation by the L.C.M. of the denominators, which is 6:
\[ 6 \left( \frac{3x}{2} + \frac{5y}{3} \right) = 6 \times 7 \]
\[ 9x + 10y = 42 \]
So, the two linear equations are:
\[ 9x + 10y - 42 = 0 \implies a_1 = 9, \quad b_1 = 10, \quad c_1 = -42 \]
\[ 9x + 10y - 14 = 0 \implies a_2 = 9, \quad b_2 = 10, \quad c_2 = -14 \]
2. Compare the coefficient ratios:
- Ratio of \(x\)-coefficients:
\[ \frac{a_1}{a_2} = \frac{9}{9} = 1 \]
- Ratio of \(y\)-coefficients:
\[ \frac{b_1}{b_2} = \frac{10}{10} = 1 \]
- Ratio of constant terms:
\[ \frac{c_1}{c_2} = \frac{-42}{-14} = 3 \]
3. Since we have:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \quad (1 = 1 \neq 3) \]
The lines are parallel to each other and do not intersect at any point.
Therefore, the system has no solution, which is classified as inconsistent.
Step 4: Final Answer:
The pair of linear equations is inconsistent, which corresponds to option (B).