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

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

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Distribute the 4: First, let's distribute the 4 on the left side of the equation. 4(80+n) = 4*80 + 4*n This gives us: 320 + 4n = (3k)nCompare coefficients of n: Now, we want to find the value of k for which there are no solutions to the equation. This would occur if the equation is inconsistent, meaning that the variable terms cancel each other out and we are left with a false statement. Let's compare the coefficients of n on both sides of the equation. On the left side, the coefficient of n is 4. On the right side, the coefficient of n is 3k. For the equation to have no solution, these coefficients must be equal, and the constant terms must not be equal. So we set 4 equal to 3k: 4 = 3kSolve for k: Now, we solve for k by dividing both sides of the equation by 3. k = 4/3Check constant terms: However, we must check if the constant terms are equal or not. If they are equal, then the equation would have infinitely many solutions instead of no solution. The constant term on the left side is 320, and since there is no constant term on the right side (it is multiplied by n), they are not equal. Therefore, the value of k = 4/3 does not make the equation have no solution. Instead, it makes the coefficients of n equal on both sides.Find value of k for no solution: To have no solution, the coefficients of n must be equal, and the constant terms must be different. Since the constant terms are already different (320 vs 0), we need to find a value of k that does not make the coefficients of n equal. However, for any value of k other than 4/3, the coefficients of n will not be equal, and the equation will have a solution. Therefore, there is no value of k that will make the equation have no solution, as long as k is not equal to 4/3.

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

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4(80+n)=(3k)n4(80+n)=(3k)n4(80+n)=(3k)n\newlineIn the given equation, \newlinekkk is a constant. For what value of \newlinekkk are there no solutions to the equation?

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