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Solving equations with greatest integer function

  • Thread starter Bea77
  • Start date
  • #1
3
0

Homework Statement


I can't find a step by step explanation for solving these types of equations

eg.
99 = [2x+1]/3


Homework Equations



eg.
99 = [2x+1]/3

or

48 = 4[2x/3]

How do you handle the multipliers iand constants inside the brackets?
thx
3. The Attempt at a Solution
 

Answers and Replies

  • #2
statdad
Homework Helper
1,495
35
Think about how the greatest integer function works. For example,

[tex]
\lfloor 3.2 \rfloor = \lfloor 3.582 \rfloor = 3
[/tex]

and in fact, if [tex] 3 \le x < 4 [/tex] it is true that

[tex]
\lfloor x \rfloor = 3
[/tex]

So, if you know that

[tex]
\frac{\lfloor 2x+1\rfloor}{3} = 99
[/tex]

you also know that

[tex]
\lfloor 2x+1 \rfloor = 297
[/tex]

(the [tex] 3 [/tex] in the denominator is not in the function). What does the final
statement above tell you about how large [tex] 2x + 1 [/tex] must be?
 
  • #3
3
0
Think about how the greatest integer function works. For example,

[tex]
\lfloor 3.2 \rfloor = \lfloor 3.582 \rfloor = 3
[/tex]

and in fact, if [tex] 3 \le x < 4 [/tex] it is true that

[tex]
\lfloor x \rfloor = 3
[/tex]

So, if you know that

[tex]
\frac{\lfloor 2x+1\rfloor}{3} = 99
[/tex]

you also know that

[tex]
\lfloor 2x+1 \rfloor = 297
[/tex]

(the [tex] 3 [/tex] in the denominator is not in the function). What does the final
statement above tell you about how large [tex] 2x + 1 [/tex] must be?
--------------------
so 297 <= 2x+1 < 298

296 <=2x and 2x < 297
148 <=x and x < 148.5

Did I get it?
 
  • #4
statdad
Homework Helper
1,495
35
Yup.
 
  • #5
3
0
Yup.

Thanks!
 

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