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Solving an equation and checking solution

  • Thread starter mistalopez
  • Start date
  • #1
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Homework Statement



(1/x-3)+(1/x+3)=(10/x²-9)


2. The attempt at a solution

(1/x-3)+(1/x+3)=(10/x²-9)

(1/x-3)+(1/x+3)=(10/(x+3)(x-3)) - I factored the x^2-9 to make it easier to multiply by LCD.
x+3+x-3=10 - I multipled both sides by the LCD [(x+3)(x-3)]
2x=10 - Combined like terms
x=5 - Divided both sides by 2.

Am I wrong or is the book wrong for having the answer as x=2 ?
 

Answers and Replies

  • #2
66
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Looks good. x=2 is clearly not a solution.
 
  • #3
HallsofIvy
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Homework Helper
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Homework Statement



(1/x-3)+(1/x+3)=(10/x²-9)


2. The attempt at a solution

(1/x-3)+(1/x+3)=(10/x²-9)

(1/x-3)+(1/x+3)=(10/(x+3)(x-3)) - I factored the x^2-9 to make it easier to multiply by LCD.
x+3+x-3=10 - I multipled both sides by the LCD [(x+3)(x-3)]
2x=10 - Combined like terms
x=5 - Divided both sides by 2.

Am I wrong or is the book wrong for having the answer as x=2 ?
Is the problem 1/x- 3+ 1/x+ 3= 10/x²- 9 or 1/(x-3)+ 1/(x+3)= 10/(x²- 9)? Those are quite different equations! You appear to be doing the second. If you multiply both sides by x²- 9 you get x+ 3+ x- 3= 2x= 10 so x= 5, as you say, not 2.

And, of course, it is easy to put x= 2 into the original equation and check: 1/(2-3)+ 1/(2+3)= -1+ 1/5= (-5+ 1)/5= -4/5 which is NOT 10/(4- 9)= 10/(-5)= -2.
 

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