Help with math (again!)

Page 1 of 1

6 Replies - 938 Views - Last Post: 17 September 2010 - 05:51 PM

#1 FrozenSnake

• En man från Sverige!

Reputation: 123
• Posts: 1,033
• Joined: 30-July 08

Help with math (again!)

Posted 17 September 2010 - 04:57 AM

I don't have my scanner here so I can't scan my notebook.

Here is the assignment: the X is ontop of the 2!
x - 3 = 6
-
2

Here I did:
step 1
x - 3 + 3 = 6 + 3
-
2

step 2

x = 9
-
2

Step 3
x*2 = 9*2
-
2

Step 4
2x = 18
------
2    2

x = 9
But I don't know if I did it correctly anyone that can help me verify ( I don't have a key for this one )

Is This A Good Question/Topic? 0

Replies To: Help with math (again!)

• Saucy!

Reputation: 6259
• Posts: 24,028
• Joined: 23-August 08

Re: Help with math (again!)

Posted 17 September 2010 - 05:05 AM

Not quite, and of course you can always check by plugging in the solution to the original equation and see if it makes sense. Obviously 9/2 - 3 != 6

In step three, you multiply both sides of the equation by 2, which is correct, but why are you dividing both sides by 2 in step 4? Multiplying the left side of x/2 by 2 results in x, not in 2x/2; the 2 removes the denominator, which is the whole reason for the multiplication in the first place.

#3 FrozenSnake

• En man från Sverige!

Reputation: 123
• Posts: 1,033
• Joined: 30-July 08

Re: Help with math (again!)

Posted 17 September 2010 - 05:08 AM

Oh, hmm... Is this more correct then?
x/2 = 9/1 and do a cross multiplication? x * 1 = 9 * 2
that would make x = 18, did I understand this now?

This post has been edited by FrozenSnake: 17 September 2010 - 05:09 AM

• Saucy!

Reputation: 6259
• Posts: 24,028
• Joined: 23-August 08

Re: Help with math (again!)

Posted 17 September 2010 - 05:13 AM

That is correct, if I remember by cross-multiplication rules correctly. Same end result in this case anyway.

You just always want to isolate the variable on one side of the equation, and you can do that by performing operations to reduce the variable to its base form, and you're free to do whatever you want to achieve that as long as you do the same operation of both sides of the equation.

#5 FrozenSnake

• En man från Sverige!

Reputation: 123
• Posts: 1,033
• Joined: 30-July 08

Re: Help with math (again!)

Posted 17 September 2010 - 05:22 AM

Ok, thanks a lot for the explanation

#6 irai

Reputation: -2
• Posts: 1
• Joined: 13-September 10

Re: Help with math (again!)

Posted 17 September 2010 - 11:22 AM

FrozenSnake, on 17 September 2010 - 03:57 AM, said:

I don't have my scanner here so I can't scan my notebook.

Here is the assignment: the X is ontop of the 2!
x - 3 = 6
-
2

Here I did:
step 1
x - 3 + 3 = 6 + 3
-
2

step 2

x = 9
-
2

Step 3
x*2 = 9*2
-
2

Step 4
2x = 18
------
2    2

x = 9
But I don't know if I did it correctly anyone that can help me verify ( I don't have a key for this one )

#7 IngeniousHax

• |>|20-514<|{3|2

Reputation: 84
• Posts: 1,385
• Joined: 28-March 09

Re: Help with math (again!)

Posted 17 September 2010 - 05:51 PM

Indeed, just simply add over the constant to remove it from the variable side, than multiply out the two, which will cancel it out on the variable side, than you have your x.

```1) (x/2) - 3 = 6 (add 3 to either side)
2) (x/2) = 9  (resulting answer from step 1. now multiply the two by either side)
3) x = 18  (the final answer)

```