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4(80+ n) = (3k)n In the equation above, k is a constant. For what value of k are there no solutions to the equation?

User ROrlig
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1 Answer

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Final answer:

The equation 4(80+ n) = (3k)n has no solutions when the value of k is such that the coefficient of n is zero. Solving for k, we find that when k equals 4/3, there are no solutions to the equation.

Step-by-step explanation:

You are looking to solve the equation 4(80+ n) = (3k)n and want to find the value of k for which there are no solutions. As per the given equation, we can distribute the 4 and move all terms to one side, resulting in a quadratic equation:

320 + 4n - 3kn = 0

This simplifies to:

(4 - 3k)n = -320

If k is such that the coefficient of n becomes zero, then we would be left with an equation that states a non-zero constant is equal to zero, which is impossible. Therefore, for there to be no solution, the coefficient of n must be zero:

4 - 3k = 0 \Rightarrow k = \frac{4}{3}

If k is \frac{4}{3}, there are no solutions to the equation since we would be saying -320 = 0, which is never true. Thus, the equation 4(80+ n) = (3k)n has no solutions when k equals \frac{4}{3}.

User Moise
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