Question

The value of k for which the system of linear equations \(\frac{x}{2} + \frac{y}{3} = 5\) and \(2x + ky = 7\) is inconsistent, is

• A. \(\frac{3}{4}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{1}{3}\)
• D. 3

Hint:

To easily work with the first equation, multiply it by 6 to clear the fractions:
\[ 6\left(\frac{x}{2} + \frac{y}{3}\right) = 6(5) \implies 3x + 2y = 30 \] Now compare coefficients with \(2x + ky = 7\):
\[ \frac{3}{2} = \frac{2}{k} \implies 3k = 4 \implies k = \frac{4}{3} \] This reduces fraction handling and saves time!

Answer 1

The Correct Option is B

Step 1: Clear fractions and try eliminating x.
Multiplying the first equation by 4 gives $2x + \frac{4}{3}y = 20$.
Step 2: Subtract this from the second equation.
\[ (2x + ky) - \left(2x + \frac{4}{3}y\right) = 7 - 20 \]
\[ \left(k - \frac{4}{3}\right)y = -13 \]
Step 3: Force the y-term to vanish for no solution to exist.
If $k - \frac{4}{3} \neq 0$, we could solve uniquely for $y$, so the system would be consistent. For the system to be inconsistent (no solution), this coefficient must be zero:
\[ k = \frac{4}{3} \]
\[ \boxed{k = \frac{4}{3}} \]