Can someone please explain in detail so I understand how x-5/2 - x-4/3 = x-3/2 - (x-2)

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- Sep 28th 2010, 07:50 PMquarinteenCan someone please explain
Can someone please explain in detail so I understand how x-5/2 - x-4/3 = x-3/2 - (x-2)

- Sep 28th 2010, 07:59 PMmr fantastic
- Sep 28th 2010, 08:03 PMquarinteen
Sorry im a programmer x-5 over 2 - x-4 over 2 = x-3 over 2 - (x-2) how does this = -5 over 2 so 5/2

- Sep 28th 2010, 08:10 PMEducated

Is this it? - Sep 29th 2010, 05:08 AMHallsofIvy
You're a programmer? Then you should understand even more how important parentheses are!

But you**still**refuse to use parentheses. "x- 4 over 2" could mean (x- 4)/2 or x- (4/2).

Also, originally, you had "x- 4/**3**" but now you say "x- 4 over [b]2[/sup]"!

In any case, if it were either x- (4/2) on the left or (x- 4)/2, the two sides**can't**be equal, the "x" would cancel on the left.

I suspect you mean (x- 5)/2- (x- 4)/3= (x- 3)/2- x- 2 but that is**still**not true! On the left, when you combine the fractions, you will have

but on the right

What you have written simply cannot be true. - Sep 29th 2010, 05:16 AMmr fantastic
It would appear that the question is meant to be:

Solve .

Most programmers understand the GIGO principle (Garbage in, garbage out) ....

My advice to the OP is to get each side over a common denominator of 6, simplify the resulting numerators and then equate. You end up with x - 7 = -3x + 3 and solving for x is simple. - Sep 29th 2010, 05:39 AMWilmer
Next time, use BRACKETS in order to show properly: (x - 5)/2 - (x - 4)/3 = (x - 3)/2 - (x - 2)

There is a huge difference (as example) between x - 5/2 and (x - 5)/2

Take this:

100 - 20/5 = 100 - 4 = 96

(100 - 20)/5 = 80/5 = 16 ; do you "get it" now?

I too am surprised that you're a programmer and didn't know this.

More so since you stated this on one of your previous posts:

" I am not in school I haven't been for a while. I was attempting to help a friend got to this problem and its bugging me."

Can you answer this:

(x - 5)/2 * 6 = ?

If not, you need to learn the Algebra basics; can't "teach" that here, sorry.