I've been trying this problem for like a hour, and I've always gotten impossible solutions. Can you help?

Question

1/x+1 + 2/x-1 = 3/x+1

whats true? Is it impossible?

Answer 1

1(x-1) + 2(x+1) = 3(x-1)

3x + 1 = 3x - 1

No solution.

Answer 2

If you mean the fractions to be in the type of 1/(x+1), where (x+1) as a whole is the denominator, then there's no solution. It eventually boils down to x = x + 2, which can't be solved.

However, the way you have it written would mean (1/x) + 1 + (2/x) - 1 = (3/x) + 1, which collapses to (3/x) = (3/x) + 1 and further to 3=3+x or x=0.

Answer 3

Let us see whether this is a trick question or if it just tricky to solve.

Our goal is to get 'x' by itself on one side of the equals sign.

Having 'x' as a denominator complicates things unnecessarily, so we need to take a step to eliminate that. We can do that by multipying both sides by 'x'.

That would give us:

1 + 1x + 2 - 1x = 3 + 1x

We gather like terms together:

1 + 2 + 1x - 1x = 3 + 1x

Simplify:

3 + x - x = 3 + x

Simplify:

3 = 3 + x

Subtract 3 from both sides to get 'x' by itself on one side.

0 = x

And this must be where you get your impossible solution because when you plug in the value of 'x' into the original equation, you find that you cannot divide by zero; thus you cannot solve the equation.

I must say that this was interesting. The answer appears to be:

x = 0

however, you cannot prove it by substitution.

What is the solution?

You could use an old trick by stating:

1/0 = y

and restate the original equation as:

y + 1 + 2y - 1 = 3y + 1.

The solution to this equation works out to be:

y = all numbers.

Thus,

1 / all numbers = 0.

That's not much help, now is it?

Moral of the story: Some teachers like to mess with your mind; good for them! It makes you think!

Answer 4

To see that 1/(x+1) + 2/(x-1) = 3/(x+1) has no solutions, subtract 1/(x+1) from both sides to get:

2/(x-1) = 2/(x+1)

which is only true when x-1 = x+1. Subtracting x from both sides gives you -1 = 1, which is impossible!

Answer 5

You need to put parentheses for clarity
I will put them where I think they belong

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

The LCD = (x + 1)(x - 1)
By multiplying both sides by the LCD, the fractions
will be eliminated

(x + 1)(x - 1)[1/(x + 1) + 2/(x - 1)] = (x + 1)(x - 1)[3/(x + 1)]

1(x - 1) + 2(x + 1) = 3(x - 1)

x - 1 + 2x + 2 = 3x - 3

x + 2x - 3x = -3 + 1 - 2

0x = -4

No solution

Answer 6

The answer to this problem is not possible. You can not find any number for x that would make this problem true. If you reduce it down you are left with 3/x = 3/x +1. Therefore the answer is impossible because no number can ever be equal to itself plus one. The answer will always be one number higher no matter what variable is used for x. I hope this helps.

Answer 7

This has no solution. Here is why

1/(x+1) can be multiplied by (x-1)/(x-1) to equal (x-1`)/(x^2-1)

2/(x-1) can be multiplie by (x+1)/(x+1) to give (2x+2)/(x^2-1)

3/(x+1) can be multiplied by (x-1)/(x-1) to give (3x-3)/(x^2-1)

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

Since the denominators are the same we can ignore them and state:

x-1+2x+2=3x-3
3x+1=3x-3
(Subtracting 3x from both sides tells us)

1=-3

This is a false statement so there is no solution.

Answer 8

False

common denominator would be (x+1) (x-1)
left numerator would be 1(x-1) + 2(x+1) = x-1 +2x +2 = 3x+1

3x+1/(x+1)(x-1) does not = 3/(x+1)

Answer 9

1 / (x + 1) + 2 / (x - 1) = 3 / (x + 1)
Multiply both sides by (x - 1)(x + 1):-
(x - 1) + 2(x + 1) = 3 (x - 1)
3x + 1 = 3x - 3
Error in question

Answer 10

Its Impossible