User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 August 15th, 2019, 02:07 PM #1 Newbie   Joined: Aug 2019 From: Terrace Posts: 2 Thanks: 0 Can someone solve X in this 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. August 15th, 2019, 02:18 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,571 Thanks: 1415 $X + p\% \text{ of } X = C$ $\left(1+\dfrac{p}{100}\right)X = C$ $X = \dfrac{C}{1+\frac{p}{100}}$ letting $p=13,~C = 200$ $X = \dfrac{200}{1+\frac{13}{100}} \approx 176.99$ Thanks from greg1313, topsquark and idontknow August 15th, 2019, 02:26 PM #3 Newbie   Joined: Aug 2019 From: Terrace Posts: 2 Thanks: 0 Alright then - I figured that there must be a formula as opposed to putting in random numbers to eventually obtain an outcome. Thank You!  August 16th, 2019, 12:46 PM   #4
Member

Joined: Jun 2019
From: AZ, Seattle, San Diego

Posts: 31
Thanks: 24

Quote:
 Originally Posted by floydy12 … 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. Tags solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gen_shao Algebra 12 November 2nd, 2014 06:11 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      