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 )
Help with math (again!)
Page 1 of 16 Replies  504 Views  Last Post: 17 September 2010  05:51 PM
Replies To: Help with math (again!)
#2
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.
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
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?
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
#4
Re: Help with math (again!)
Posted 17 September 2010  05:13 AM
That is correct, if I remember by crossmultiplication 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.
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
Re: Help with math (again!)
Posted 17 September 2010  05:22 AM
Ok, thanks a lot for the explanation
#6
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 )
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
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)
Page 1 of 1
