Can someone help me understand how x-x1 = L*(x2 - x) equals x = (x1 + L * x2) / 1 + L

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GoldSpark

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Hi guys sorry I am new in this forum. I thought why not use some help. This is finding a coordinate for x, this formula is same for y and z. I am sorry if I posted in wrong forum.Here is my question.
I will mark lambda as L:

Can someone explain me how x - x1 = L * (x2 - x) is equal to x = (x1 + L * x2) / 1 + L

Thank you in advance.
 
GoldSpark said:
Hi guys sorry I am new in this forum. I thought why not use some help. This is finding a coordinate for x, this formula is same for y and z. I am sorry if I posted in wrong forum.Here is my question.
I will mark lambda as L:

Can someone explain me how x - x1 = L * (x2 - x) is equal to x = (x1 + L * x2) / 1 + L

Thank you in advance.
Click to expand...

First, it isn't. You've a notation problem. Perhaps you mean "(1 + L)" on the far right, rather than just "1 + L". Grouping matters.

Second, to solve for x, one must get all the 'x's together on one side of the "=". That "x" stuck inside the parentheses is the problem. Use the distributive property on the RHS to transform L * (x2 - x) = L*x2 - L*x, and this should get you on your way. There is more distributive property in your future.
 
tkhunny said:
First, it isn't. You've a notation problem. Perhaps you mean "(1 + L)" on the far right, rather than just "1 + L". Grouping matters.

Second, to solve for x, one must get all the 'x's together on one side of the "=". That "x" stuck inside the parentheses is the problem. Use the distributive property on the RHS to transform L * (x2 - x) = L*x2 - L*x, and this should get you on your way. There is more distributive property in your future.
Click to expand...


Hey thanks for replying so fast! Yeah I forgot to group that last part sorry. Thanks a lot man! :D I will try it out now.
 
Can't figure it out

I am trying and I get x = (x2 + x1) / L .. I dont know sorry..

Here is what I did:

1) x - x1 = L*(x2 - x)

2) x - x1 = L*x2 - L*x;

3) L*x + x - x1 = L*x2

4) 2*L*x - x1 = L*x2

5) x = (L*x2 + x1) / 2*L -Here i subtract denominator with nominator (?)

6) x = (x2 + x1) / L

:( ... I am sorry this is probably stupid easy to you but I have no idea whats going on why I cant get L*x2 in the nominator and 1+L in denominator
 
GoldSpark said:
I am trying and I get x = (x2 + x1) / L .. I dont know sorry..

Here is what I did:

1) x - x1 = L*(x2 - x)

2) x - x1 = L*x2 - L*x;

3) L*x + x - x1 = L*x2

4) 2*L*x - x1 = L*x2

5) x = (L*x2 + x1) / 2*L -Here i subtract denominator with nominator (?)

6) x = (x2 + x1) / L

:( ... I am sorry this is probably stupid easy to you but I have no idea whats going on why I cant get L*x2 in the nominator and 1+L in denominator
Click to expand...

3) is good

4) L*x + x = x * (L + 1), not 2*L*x -- You missed your second chance at the distributive property.
 
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