Question

The value of k for which the system of equations \(3x - 7y = 1\) and \(kx + 14y = 6\) is inconsistent, is

• A. --6
• B. \(\frac{2}{3}\)
• C. 6
• D. \(-\frac{3}{2}\)

Answer 1

The Correct Option is A

Given:
Two linear equations:
1) \(3x - 7y = 1\) with \(a_1 = 3, b_1 = -7, c_1 = 1\)
2) \(kx + 14y = 6\) with \(a_2 = k, b_2 = 14, c_2 = 6\)

Step 1: Inconsistency Condition
A system of two equations is inconsistent (has no solution) if:
\[\frac{a_1}{a_2} = \frac{b_1}{b_2} eq \frac{c_1}{c_2}\]

Step 2: Solve for \(k\) using \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\)
\[\frac{3}{k} = \frac{-7}{14}\] Simplify:
\[\frac{3}{k} = -\frac{1}{2}\] Cross-multiply:
\[3 \times 2 = -1 \times k \implies 6 = -k \implies k = -6\]

Step 3: Verify \(\frac{b_1}{b_2} eq \frac{c_1}{c_2}\)
Calculate:
\[\frac{b_1}{b_2} = \frac{-7}{14} = -\frac{1}{2}\] \[\frac{c_1}{c_2} = \frac{1}{6}\] Since \(-\frac{1}{2} eq \frac{1}{6}\), the inconsistency condition is met.

Final Answer:
\[\boxed{k = -6}\]