- #1

solakis1

- 422

- 0

An easy one:

x+[2x]+[3x]=7

x+[2x]+[3x]=7

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- Thread starter solakis1
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In summary, the conversation discusses a problem involving the equation x+[2x]+[3x]=7 and different sets of real numbers that can be divided into based on their properties. The conclusion is that the correct solution is x=4/3 and not 7/6 as previously suggested.

- #1

solakis1

- 422

- 0

An easy one:

x+[2x]+[3x]=7

x+[2x]+[3x]=7

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- #2

HOI

- 921

- 2

I would start by dividing the real numbers into classes:

1) The set of integers.

2) The set of non-integer, x, such that 2x is an integer

3) The set of non-integers, x, such that 2x is not an integer but 3x is.

4) The set of non-integers, x, such that neither 2x nor 3x is an integer but 6x is.

5) The set all other real numbers.

- #3

kaliprasad

Gold Member

MHB

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As $\lfloor 2x \rfloor$ and $\lfloor 3x \rfloor$ are integers so x is integer so $\lfloor 2x \rfloor = 2x $ and $\lfloor 3x \rfloor = 3x$

so x + 2x + 3x = 6x = 7 so $x = \frac{7}{6}$ which is not integer so NO solution

- #4

jonah1

- 107

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Desmos somehow gives the impression that $x + \lfloor 2x \rfloor + \lfloor 3x \rfloor = 7$ is basically the line $x = \frac{4}{3}$.solakis said:An easy one:

x+[2x]+[3x]=7

https://www.desmos.com/calculator/stz3o2wn2h

- #5

solakis1

- 422

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The answer given by Kaliprasad is the right one 7\6 is not an integer

4/3 is not a solution of the above equation LHS is 8 RHS Is 7

4/3 is not a solution of the above equation LHS is 8 RHS Is 7

Last edited:

The value of x in this equation is not a specific number, but rather a variable that can take on different values depending on the context of the problem. In order to solve for x, we need more information or additional equations.

To solve this equation, you can use algebraic methods such as combining like terms, distributing, and isolating the variable on one side of the equation. It may also be helpful to use a calculator or graphing software to visualize the equation and find possible solutions.

Yes, this equation can have multiple solutions. Since there is only one variable, there can be an infinite number of values that satisfy this equation. However, in order to find a specific solution, we need more information or additional equations.

Yes, there are specific methods for solving equations with multiple variables. One common method is substitution, where one variable is isolated and substituted into another equation to solve for the other variable. Another method is elimination, where equations are combined and variables are eliminated to solve for the remaining variable.

It is possible to solve this equation without using algebra, but it may be more difficult and time-consuming. Graphing or using a calculator can help find approximate solutions, but algebraic methods are typically more accurate and efficient in solving equations.

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