# Can someone solve X in this

#### floydy12

Does a formula exist for this or am I out in left field? Here goes:

Where X + 13% of X = 200. Solve X

Note - The percentage and the number 200 are arbitrary. Either of these could be infinite.

#### romsek

Math Team
$X + p\% \text{ of } X = C$

$\left(1+\dfrac{p}{100}\right)X = C$

$X = \dfrac{C}{1+\frac{p}{100}}$

letting

$p=13,~C = 200$

$X = \dfrac{200}{1+\frac{13}{100}} \approx 176.99$

3 people

#### floydy12

Alright then - I figured that there must be a formula as opposed to putting in random numbers to eventually obtain an outcome.

Thank You!

#### Otis

â€¦ The percentage and the number 200 are arbitrary. Either of these could be infinite.
Well, the number 200 can't be infinite, but I know what you mean. In cases where either the percent or the given constant is infinite, there is no solution for X.