1. ## stupid me

I know the answer to the equation however i cant seem to get the math to work out. Here is the equation.

x-5/3 - x-4/3 = x-3/2 - (x-2)

2. Originally Posted by skittles900
I know the answer to the equation however i cant seem to get the math to work out. Here is the equation.

x-5/3 - x-4/3 = x-3/2 - (x-2)
I assume,
$\frac{x-5}{3}-\frac{x-4}{3}=\frac{x-3}{2}-(x-2)$
The common denominator exists and is 6.
Multiplication by six produces,
$2(x-5)-2(x-4)=3(x-3)-6(x-2)$
Open,
$2x-10-2x+8=3x-9-6x+12$
Combine,
$-2=-3x+3$
Subtract from both sides,
$-5=-3x$
Divide,
$x=5/3$

I did the same thing, but apparently the answer is x=5/2 !!! I've tried many ways to get this, but still nothing??? Can anyone get this answer???

5. Yes. Thank you. Since it's the original answer I had I'm going to assume I'm right until the instructor says otherwise. Thanks again for your help!

6. ## Stupid me!

Yesterday I posted a problem...and I have been typing it up wrong...here's the answer x=5/2

problem x-5/2 - x-4/3 = x-3/2 - (x-2) can anyone help me solve this. I know it's a simple step I keep missing....

7. Originally Posted by skittles900
Yesterday I posted a problem...and I have been typing it up wrong...here's the answer x=5/2

problem x-5/2 - x-4/3 = x-3/2 - (x-2) can anyone help me solve this. I know it's a simple step I keep missing....
I presume,
$\frac{x-5}{2}-\frac{x-4}{3}=\frac{x-3}{2}-(x-2)$
The Commond Denominator exists and is 6.
Multiplication upon the equation produces,
$3(x-5)-2(x-4)=3(x-3)-6(x-2)$
Thus,
$3x-15-2x+8=3x-9-6x+12$
Combine,
$x-7=-3x+3$
$x=-3x+10$
$4x=10$
Divide by 4,
$x=10/4=5/2=2.5$

8. Originally Posted by skittles900
Yesterday I posted a problem...and I have been typing it up wrong...here's the answer x=5/2

problem x-5/2 - x-4/3 = x-3/2 - (x-2) can anyone help me solve this. I know it's a simple step I keep missing....
A suggestion: Next time post the correction in the original thread. That way eveyone can see where the mistake was made and be able to follow it without having to jump threads.

-Dan

9. Originally Posted by topsquark
A suggestion: Next time post the correction in the original thread. That way eveyone can see where the mistake was made and be able to follow it without having to jump threads.

-Dan
Thank you I fixed it.

10. thank you very much. Problem solved