# 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.

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Simon Bridge
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
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