Question

4(80+n)=(3k)nspace, 4, left parenthesis, 80, plus, n, right parenthesis, equals, left parenthesis, 3, k, right parenthesis, n in the equation above, kkk is a constant. for what value of kkk are there no solutions to the equation?

Answer 1

We need to find the value of k at which the equation 4(80+n)=(3k)n has no solution.
Let's solve the above equation for n. So,
$4(80+n)=(3k)n$
$320+4n=3kn$ By using the distributing property.

$320+4n-4n=3kn-4n$ Subtract 4n from each sides.
$320=3kn-4n$
$320=(3k-4)n$ Take out n as a common factor
$\frac{320}{3k-4} =\frac{(3k-4)*n}{3k-n}$ Divide each sides by 3k-4.
$\frac{320}{3k-4} =n$
So, $n=\frac{320}{3k-4}$
Denominator of a rational equation cannot be 0 because if it will be 0 then the equation will be undefined and there will be no solution.
Here the denominator is 3k-4. Let's set up 3k-4=0 to get the answer. So,


$3k-4=0$
$3k-4+4=0+4$
$3k=4$
$\frac{3k}{3} =\frac{4}{3}$
So, $k=\frac{4}{3}$
So, for $k=\frac{4}{3}$ there will be no solution.