Question

$4(81+x)=\left(3k\right)x$
In the above equation, $k$ is a constant. For what value of $k$, is there a unique solution to the equation?

• A. k = 4/3
• B. Any real value
• C. Any real value except 4/3
• D. Any real value between -4/3 and 4/3

Answer 1

The correct option is C Any real value except 4/3

The equation is $4(81+x)=\left(3k\right)x$
There will be no solution if the $x$ terms cancelled, leaving a false statement.
$4\times 81+4x=3kx$
This can only happen if $4x=3kx$ that is $4=3k$
Solving for $k$, we get $k=\frac{4}{3}$.
There are no solution if $k=\frac{4}{3}$.
Therefore, for any real value except k = 4/3 will give a unique solution to the given equation.
Hence, the answer option C is correct.