# Long-ish Equation: Need Help!

• Jun 25th 2010, 05:54 PM
Ingersoll
Long-ish Equation: Need Help!
Like the last problem I had with 0 of 0, I think this one has an easy fix. But there's not example in the textbook and it's Saturday, so obviously there's no teacher to turn to.

I'm close, but here it is:

2[4 -2(3-x)]-1= 4[2(4x-3) +7]-25

Here's what I did:

1st Step: 2[4 -6 +2x]-1 = 4[8x-6+7]-25

2nd: 2[-2 +2x]-1 = 4[8x +1]-25

3rd: -4 +4x -1 = 32x +4 -25

4th: -5 +4x = 32x -21

5th: ...

Oh. I just figured it out.

I subtract 4 from 4 and then 4 from 32x to get 28, which I then divide by itself.

Then I add 21 to -21 and add the same to the -5 to get 16...

So 16 over 28 reduces to 4 over 7. X= 4 over 7

So I guess now my question is this: why is this equation different in that I want X on the right side? Obviously there's a rule on this but what specifically is it? Is it that you always subtract the smaller number from the bigger number? Is that it?
• Jun 25th 2010, 06:28 PM
Sudharaka
Quote:

Originally Posted by Ingersoll
Like the last problem I had with 0 of 0, I think this one has an easy fix. But there's not example in the textbook and it's Saturday, so obviously there's no teacher to turn to.

I'm close, but here it is:

2[4 -2(3-x)]-1= 4[2(4x-3) +7]-25

Here's what I did:

1st Step: 2[4 -6 +2x]-1 = 4[8x-6+7]-25

2nd: 2[-2 +2x]-1 = 4[8x +1]-25

3rd: -4 +4x -1 = 32x +4 -25

4th: -5 +4x = 32x -21

5th: ...

Oh. I just figured it out.

I subtract 4 from 4 and then 4 from 32x to get 28, which I then divide by itself.

Then I add 21 to -21 and add the same to the -5 to get 16...

So 16 over 28 reduces to 4 over 7. X= 4 over 7

So I guess now my question is this: why is this equation different in that I want X on the right side? Obviously there's a rule on this but what specifically is it? Is it that you always subtract the smaller number from the bigger number? Is that it?

Dear Ingersoll,

You can do it either way you want.

$\displaystyle -5+4x=32x-21$

Substract 32x from both sides,

$\displaystyle -5+4x-32x=32x-21-32x$

$\displaystyle -5+(-28x)=-21$

$\displaystyle -5-28x=-21$

$\displaystyle -5-28x+5=-21+5$

$\displaystyle -28x=-16$

$\displaystyle x=\frac{-16}{-28}=\frac{16}{28}=\frac{4}{7}$

• Jun 25th 2010, 06:31 PM
Wilmer
Quote:

Originally Posted by Ingersoll
2[4 -2(3-x)]-1= 4[2(4x-3) +7]-25

Simplify as much as possible before "getting going!":
Move the -25:
2[4 - 2(3 - x)] + 24 = 4[2(4x - 3) + 7]
Divide by 2:
4 - 2(3 - x)] + 12 = 2[2(4x - 3) + 7]
carry on...
• Jun 25th 2010, 06:34 PM
Wilmer
Quote:

Originally Posted by Ingersoll
4th: -5 +4x = 32x -21
So I guess now my question is this: why is this equation different in that I want X on the right side? Obviously there's a rule on this but what specifically is it? Is it that you always subtract the smaller number from the bigger number? Is that it?

Nooooooo.......just rewrite like this:
32x -21 = -5 + 4x
Kapish?
• Jun 25th 2010, 08:34 PM
Ingersoll
Cool. Thank both of you very much. :)