# Finding x with fractions

## 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

## 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.

Last edited:

Simon Bridge
Science Advisor
Homework Helper

## 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 im thinking im 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
Homework Helper
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
Homework Helper
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
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
...
$$3+\sqrt{x^2-4}= \sqrt{x^2+6x-7}$$
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).