# Equation solving/solving for x

1. Oct 13, 2015

### Weightofananvil

1. The problem statement, all variables and given/known data
1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3

solve for x. check solution in original equation

3. The attempt at a solution

I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.

I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.

Last edited by a moderator: Oct 13, 2015
2. Oct 13, 2015

### symbolipoint

Maybe it is not written correctly.

3. Oct 13, 2015

### ehild

I guess you omitted a few couples of parentheses .
What you wrote is equivalent to 0.5 x -2 -3 =x+0.5 x^2-5x-3 - are you sure you wanted to write that?

Last edited by a moderator: Oct 13, 2015
4. Oct 13, 2015

### Staff: Mentor

As ehild already said, you need quite a few more parentheses here.

What you wrote is properly interpreted to mean this:
$\frac 1 2 x + 1 - \frac{2x}{x} - 3 = x + \frac 1 2 x^2 - 5x - 3$
Again, you need some parentheses, plus show your intervening steps in getting to the above.

Regarding parentheses, when you write fractions on a single line, if the numerator or denominator has more than a single term, put parentheses around it.
For example: $\frac{x^2 + 3x + 2}{x + 1}$ should be written as (x^2 + 3x + 2)/(x + 1), if written on a single line.

5. Oct 13, 2015

### HallsofIvy

Staff Emeritus
Using parentheses, (4x^2- x- 3)/(2x^2- 5x- 3)= (x+ 1)/(2x^2- 5x- 3)
Note that the denominator is 0 for x= 3 and x= -1/2. As long as the denominator is NOT 0, we can multiply on both sides by it to get
4x^2- x- 3= x+ 1 which is the same as the quadratic equation 4x^2- 2x- 4= 0

6. Oct 13, 2015

### jk22

From where you are you could put all on the same side since you have a common denominator you can put it away and remains a quadratic equation

7. Oct 24, 2015

### Weightofananvil

Thanks for the help everyone!