Finding x with fractions

  • Thread starter wrx
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  • #1
wrx
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Homework Statement


1)
____1___ __ ___2___ __ ____3____ =0
3x-7 5x-5 3x+1

here's the other way of presenting the problem.

1/(3x-7) - 2/(5x-5) - 3/(3x+1) =0

ans: x=2/3, 3

2) _____2_____ + _____1_____ __ _____1_____ =0
(x^2-36)^1/2 (x+6)^1/2 (x-6)^1/2




2/sqrt(x^2-36) + 1/sqrt(x+6) - 1/sqrt(x-6) =0

ans:x=10


3) (x+7)^1/2 - (x+2)^1/2 = (x-1)^1/2 - (x-2)^1/2

ans:x=2

Homework Equations





The Attempt at a Solution



on problem #3, i tried squaring both side of the equations to cancel out those sqrt root but im thinking im doing it the wrong way because the solution is way too long.
 
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Answers and Replies

  • #2
Simon Bridge
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Homework Statement


1)
____1___ __ ___2___ __ ____3____ =0
3x-7 5x-5 3x+1

here's the other way of presenting the problem.

1/(3x-7) - 2/(5x-5) - 3/(3x+1) =0
Or, possibly:[tex]\frac{1}{3x-7}-\frac{2}{5x-5}-\frac{3}{3x+1}=0[/tex]... have you tried #1 and #2 at all?

The Attempt at a Solution


on problem #3, i tried squaring both side of the equations to cancel out those sqrt root but im thinking im doing it the wrong way because the solution is way too long.
#3 was:[tex] (x+7)^{1/2} - (x+2)^{1/2} = (x-1)^{1/2}-(x-2)^{1/2}[/tex]... notice that x ≥ 2 for the RHS to be real?
 
  • #3
ehild
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It becomes easier if you rearrange the equation before squaring:

[tex](x+7)^{1/2} - (x-1)^{1/2}= (x+2)^{1/2} -(x-2)^{1/2}[/tex]

ehild
 
  • #4
ehild
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It becomes easier if you rearraqnge the equation before squaring:

[tex](x+7)^{1/2} - (x-1)^{1/2}= (x+2)^{1/2} -(x-2)^{1/2}[/tex]

Taking the square of both sides and expand

[tex](x+7) + (x-1)-2\sqrt{(x+7)(x-1)}= (x+2) +(x-2)-2\sqrt{(x-2)(x+2)}[/tex]
the x terms cancel, simplify and divide by 2:
[tex]3-\sqrt{(x+7)(x-1)}= -\sqrt{x^2-4}[/tex]
or [tex]3+\sqrt{x^2-4}= \sqrt{x^2+6x-7}[/tex]
Square again. The x2 terms cancel and you get a simple equation for x.

ehild
 
  • #5
SammyS
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[tex]3+\sqrt{x^2-4}= \sqrt{x^2+6x-7}[/tex]
Square again. The x2 terms cancel and you get a simple equation for x.

ehild
An additional comment:

Once you do that last step, you will still have a square root. Fortunately at that point, the equation you are left with is fairly simple (uncomplicated).
 

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