Solution: x + 3{x} = 11 let x = k + {x}, where k is an integer, so we have k + {x} + 3{x} = 11 k + 4{x} = 11 since both k and 11 are integers, 4{x} also has to be an integer. also, the boundaries of {x} are 0 ≤ {x} < 1. therefore {x} can only be 0, 1/4, 2/4 or 3/4, as no other fraction in that range, multiplied by 4, results in an integer. {x} = 0 → x + 3 * 0 = 11 → x = 11 {x} = 1/4 → x + 3/4 = 11 → x = 41/4 {x} = 2/4 → x + 6/4 = 11 → x = 28/2 → x = 19/2 {x} = 3/4 → x + 9/4 = 11 → x = 35/4 therefore x ∈ {35/4, 19/2, 41/4, 11}
@@PrimeNewtons Sorry, I don't make videos. I don't have time for it. I am a software developer with focus on writing and optimizing algorithms to analyze data (mostly statistical, but often also logical). So this kind of math is literally my bread and butter...
From experience, if other viewers find your solution easier to understand, they would give it a thumbs up. Use that as a yardstick. I'd pin your comment.
Notice that the coefficient of k divided by the denominator of x is the difference between two consecutive solutions of an equation of this type(whatever you want to call or name this equation). Furthermore, the number of solutions is the sum of the coefficients of x and {x} in the original equation. I always enjoy your videos.
i thought about writing x as [x] + {x} so the equation becomes [x] + 4{x} =11 which means 4{x} is an integer but {x} is also lower than 1 this can happen only if {x} ={0/4,1/4,2/4,3/4} and then you just solve for each case
Please keep the high quality of your videos (i.e. interesting math problems presented in your great unique way)! Even if it means to lower the quantity ;-)
Nice job! I have never learned anywhere (other than here!!) that {x} = the fractional part of x. Thanks. Here's a topic for a new video. Show how you can use the the floor of x to round off numbers to any decimal place of your choice.
You're the best. I thought that knew everything. Say you have 7.8. Add .5, then just take the integer part and you'll round off 7.8 to 8. If you have 7.3, add .5, get 7.8, take integer part and get 7. If you want to round off 7.3892 to two decimals, then multiply 100, do the trick like above and then divide by 100.It's very cool. I bet that you knew this(??). let me know.
The shortcut for this problem is to notice that 4{x} must be an integer, which means {x} can only take some modulo multiple of 1/4 - however, your video is better IMO, as it allows u to solve any general problem of this kind e.g. if we had a mix of fractional and negative terms, or even if we had a simultaneous system of such equations, your approach is fool-proof
I have taken many advanced math classes back in the early 80s and can't remember LambertW, or equations with floors or fractional parts. Did I sleep through those classes or am I suffering from senility?Your videos are awesome, great handwriting, and teaching style.
My biggest hang up was not knowing the definition of {x}. I spent a little too much time trying to think what times of rational numbers added to 3 times the fractional part gives us an integer. So I knew that the answer would be in fourths. After that, I got stuck. Great video though
I solved it by spliting x into floor(x) and fractional(x) giving floor(x)+4*fractional(x)=11 and since floor(x) is an integer it must then imply that 4*fractional(x) is also an integer and since the fractional part of a number is allways less than 1 and greater than or equal to 0 fractional(x) must be 0/4 or 1/4 or 2/4 or 3/4 plug those into fractional(x) for the original equation and solve for each x giving x = 11 or 10,25 or 9,5 or 8,75
Using simple symbols the whole thing even becomes trivial to formally write and solve: x + p + 3p = 11 4p = 11 - x p = (11 - x) / 4 = any_int / 4 yields directly 0, 0.25, 0.5, 0.75 (0
Here is another way : x + 3.{x} = 11 (k = floor(x)) so that means k + 4.{x} = 11 so we know that {x} is between 0 and 1 so 4.{x} must be between 0 and 4 That means k must be between 7 and 11 but we know that 4.{x} can not be 4 so we got options that k = 8,9,10,11 and when you get these you can find {x} and mix them and get the andwer SOLUTION 2 : ------------------ x + 3{x} = 11 floor(x) + 4.{x} = 11 so 4.{x} € Z {x} € Z/4 0 ≤ {x} < 1 so 4 values : 0/4 1/4 2/4 3/4 If you got 0/4 it means x = 11 If you got 1/4 it means x = 10.25 If you got 2/4 it means x = 9.5 If you got 3/4 it means x = 8.75 Thanks!
Can't we just put the value of x = 11 as in fractional part function when we will put {11}= it will give answer as 0 , so the equation will directly be x=11 and value of x =11 Edit: i am commenting on this by seeing the thumbnail 👍
9.5 is my guess. Cuz .5 times 3 is 1.5 and 9.5 plus 1.5 is 11. But this isn’t systematic. I just know .5 times 3 is 1.5, and 1.5 plus .5 is 2, and 11 minus 2 is 9, so 9.5 is good
Judging from the results, it looks like any equation of the form x + k{x} = n (assuming k, n ∈ ℤ) is going to have k+1 solutions, with one of them being n and the other k solutions each descending in value from n by k/(k+1). So the solutions will all be of the value n - mk/(k+1), where m are the integers from 0 to k.
