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# Application of Linear Equations

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

Guest Apr 9, 2021
#5
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:D

Logarhythm  Apr 9, 2021
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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

Apr 9, 2021
#4
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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

Apr 9, 2021
#6
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So we all agree it's 15?

Apr 9, 2021