Value of k for 2x + 3y = 6 and 4x + 6y = 3k Having at Least One Solution

Answer: 2

The linear equations
2x + 3y = 6 and
4x + 6y = 3k
have at least one solution when the system is consistent.
A consistent system has either a unique solution or infinitely many solutions.

Condition Analysis for Consistency

For two linear equations:

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

The system is consistent if either:

• 
  a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  
  (unique solution)
• 
  a₁/a₂ = b₁/b₂ = c₁/c₂
  
  (infinitely many solutions)

Rewrite the Given Equations

Convert the equations to standard form:

2x + 3y − 6 = 0
4x + 6y − 3k = 0

Identify coefficients:

• a₁/a₂ = 2/4 = 1/2
• b₁/b₂ = 3/6 = 1/2
• c₁/c₂ = (−6)/(−3k) = 2/k

The coefficients of x and y are proportional
(1/2 = 1/2), so the system is either consistent or dependent.

Solution Requirement

For at least one solution:

• If 1/2 ≠ 2/k, the system has a unique solution
  (valid for all k ≠ 2).
• If 1/2 = 2/k, then k = 2 and the system has
  infinitely many solutions.

When k = 2, the second equation becomes:

4x + 6y = 6

This is exactly twice the first equation, confirming infinitely many solutions
along the same line.

Final Answer

The integer value of k that satisfies the condition is:

k = 2