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avatar+321 

Given \(3x+\dfrac{2}{x}-4=2(x+9)-\left(7-\dfrac{1}{x}\right)\)  what is \(\dfrac{x^2+1}{x}?\)

 Apr 9, 2021
 #1
avatar+527 
+3

First, simplify the equation:
\(3x + \frac{2}{x} - 4 = 2(x + 9) - (7 - \frac{1}{x})\)

\(3x + \frac{2}{x} - 4 = 2x + 18 - 7 + \frac{1}{x}\)

\(x - 15 = -\frac{1}{x}\)

\(x^2 - 15x = -1\)

\(x^2 + 1 = 15x\)

 

Now plug it into \(\frac{x^2 + 1}{x}\):

\(= \frac{(15x)}{x}\)

\(= 15\)

 Apr 9, 2021
 #3
avatar
+2

Finalllllllly my answer is the same as everyone else...

Guest Apr 9, 2021
 #5
avatar+527 
+1

:D 

Logarhythm  Apr 9, 2021
 #2
avatar
+3

3x + 2/x - 4 = 2(x + 9) - (7 - 1/x)  ==>   3x + 2/x - 4 = 2x + 18 - 7 + 1/x   ==>   x + 1/x = 15   ==>    x^2 + 1 = 15x. Therefore, (x^2 + 1) / x = 15x / x = 15 smiley

 Apr 9, 2021
 #4
avatar+118669 
+3

3x  + 2/x   - 4  =  2(x + 9)  - (7  - 1/x)       simplify

 

3x  + 2/x  - 4  =  2x + 18  - 7  + 1/x

 

3x  + 2/x  -  4  = 2x + 11  +  1/x

 

3x - 2x  - 4 - 11 + 2/x - 1/x  =  0

 

x  -  15  +  1/x   =   0            

 

x + 1/x   =  15

 

x^2 + 1

______  =    15

    x

 

 

cool cool cool

 Apr 9, 2021
 #6
avatar
+1

So we all agree it's 15?

 Apr 9, 2021

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