solve 4/(2x-1) + 1/(x+1) = 1

[girl_in_black]

Well, the question is to solve the equation

4/2x-1 + 1/x+1 = 1

This is what I got so far, but I don't think it is right:

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

4(x+1) + 2x -1 -1/(2x -1)(x+1) = 0

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

6x +2 = 2x^2 +x - 1

2x^2 - 5x - 3 = 0

This is where the problem is because after that I get (2x+1)(x-3) which would mean that x = 3 or x = -1/2, but the answer is x=2 or x= - (3/2).

Can someone please tell me where I've gone wrong?

 

[stapel]

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

4(x+1) + 2x -1 -1/(2x -1)(x+1) = 0

The first line does not yield the second line.

Are you trying to convert to a common denominator? If so, you should end up with fractions. Are you trying to multiply through to clear the denominators? Then you have to multiply each term (including the "1" on the right-hand side) by the same value.

Eliz.

 

[Mrspi]

Well, the question is to solve the equation

4/2x-1 + 1/x+1 = 1

This is what I got so far, but I don't think it is right:

4(x+1)/(2x-1)(x+1) + 2x-1/(2x-1)(x+1)=1OK to here....

4(x+1) + 2x -1 -1/(2x -1)(x+1) = 0

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

6x +2 = 2x^2 +x - 1

2x^2 - 5x - 3 = 0

This is where the problem is because after that I get (2x+1)(x-3) which would mean that x = 3 or x = -1/2, but the answer is x=2 or x= - (3/2).

Can someone please tell me where I've gone wrong?

You are correct up to this point:
4(x+1)/(2x-1)(x+1) + 1(2x-1)/(x+1)(2x-1) = 1

Now, to eliminate the denominators, multiply BOTH SIDES by (2x-1)(x+1)

You should end up with

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

4(x + 1) + 2x - 1 = 2x<SUP>2</SUP> + x - 1
Somehow, you got this....but by a lucky combination of errors, I think

You made an error in combining like terms....
4x + 4 + 2x - 1 = 2x<SUP>2</SUP> + x - 1
6x + 3 = 2x<SUP>2</SUP> + x - 1

Get 0 on one side:
0 = 2x<SUP>2</SUP> - 5x - 4

Now, this does NOT factor....so you will need to use the quadratic formula. And who told you that 2 and -3/2 were the correct answers? Did you check them? I did....and neither works in the original equation!

 

[girl_in_black]

Get 0 on one side:
0 = 2x<SUP>2</SUP> - 5x - 4

Now, this does NOT factor....so you will need to use the quadratic formula. And who told you that 2 and -3/2 were the correct answers? Did you check them? I did....and neither works in the original equation!

I didn't check them but the answer sheet did say that it was 2 and -3/2...
The first time I tried to solve the equation I also got 0 = 2x<SUP>2</SUP> - 5x - 4 but when I used the formula I didn't get 2 and -3/2 so I thought I was wrong which is why I ended up doing it the way I did. I guesse the answer sheet must be wrong. I feel so clever for once lol

Thanks Mrspi

 

[stapel]

...the answer sheet did say that [the solutions were x=] 2 and -3/2.

You can always check solutions by plugging them back into the original question. If the exercise is as posted (with the parentheses around the "2x - 1" and the "x + 1"), then check the proposed solutions. Plug the x-values into the left-hand side (LHS) of the equation, and see if it simplifies to the required value of "1":

. . .LHS with x = 2:

. . . . .\(\displaystyle \L \frac{4}{2(2)\, -1}\, +\, \frac{1}{(2)\, +\, 1}\, =\, \frac{4}{3}\, +\, \frac{1}{3}\, =\, \frac{5}{3}\)

...which does not equal "1".

. . .LSH with x = -3/2:

. . . . .\(\displaystyle \L \frac{4}{2\left(-\frac{3}{2}\right)\, -\, 1}\, +\, \frac{1}{\left(-\frac{3}{2}\right)\, +\, 1}\, =\, \frac{4}{-4}\, +\, \frac{1}{\left(-\frac{1}{2}\right)}\, =\, -1\, +\, -2\, =\, -3\)

...which again does not equal "1".

Do the exercise yourself, and check your own solutions in the same manner. The solution that works in the original equation is the correct one, whether or not it is listed on the answer key.

Eliz.

 

[Denis]

For x = 2 or x = -3/2, then your equation:
4/(2x-1) + 1/(x+1) = 1 should be: 4/(2x-1) - 1/(x+1) = 1

Who typoed the +: you or the book? :shock: