Finding x with fractions

wrx · Aug 30, 2012

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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.

Simon Bridge · Aug 30, 2012

wrx said: ↑

1. The problem statement, all variables and given/known data
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?

3. 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?

ehild · Aug 30, 2012

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

ehild · Aug 31, 2012

ehild said: ↑

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

SammyS · Aug 31, 2012

ehild said: ↑

...
[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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Finding x with fractions

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