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

`(320)/(3k-4) =((3k-4)*n)/(3k-n)` Divide each sides by 3k-4.

`(320)/(3k-4) =n`
So,
`n=(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`

`(3k)/(3) =(4)/(3)`
So,
`k=(4)/(3)`
So, for
`k=(4)/(3)` there will be no solution.