How Do You Solve Complex Fraction and Radical Equations?

[wrx]

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=103) (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 I am thinking I am doing it the wrong way because the solution is way too long.

 

[Simon Bridge]

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:\frac{1}{3x-7}-\frac{2}{5x-5}-\frac{3}{3x+1}=0... 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 I am thinking I am doing it the wrong way because the solution is way too long.

#3 was:(x+7)^{1/2} - (x+2)^{1/2} = (x-1)^{1/2}-(x-2)^{1/2}... notice that x ≥ 2 for the RHS to be real?

 

[ehild]

It becomes easier if you rearrange the equation before squaring:

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

ehild

 

[ehild]

It becomes easier if you rearraqnge the equation before squaring:

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

Taking the square of both sides and expand

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

ehild

 

[SammyS]

...
3+\sqrt{x^2-4}= \sqrt{x^2+6x-7}
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).