Gravitational Attraction Question

[wilson_chem90]

the { } is square root

 

[Redbelly98]

the { } is square root

Ah, okay. Then yes, this is correct:

alright so the new equation should be X/{Me} = D-X/{Mm}

Next is to solve that equation for X.

 

[wilson_chem90]

i got to X = D-X{Me} / {Mm}

 

[Doc Al]

i got to X = D-X{Me} / {Mm}

OK, write that as:
X = (D-X) {Me/Mm} (using {} for square root)

Now solve for X. (Get all the X terms on the same side.)

 

[wilson_chem90]

so would it just be X+X = D {Me/Mm}

 

[Doc Al]

so would it just be X+X = D {Me/Mm}

No. Realize that:

X = (D-X) {Me/Mm} →

X = D{Me/Mm} - X{Me/Mm}

 

[wilson_chem90]

oh okay, yeah i remember that now, is that the final equation though?

 

[Doc Al]

What do you mean by "final"? You still have to solve for X.

 

[wilson_chem90]

well i have to isolate the other x right, but the only think i can think of is it would be X^2 = D{Me/Mm} - {Me/Mm}

 

[Doc Al]

well i have to isolate the other x right, but the only think i can think of is it would be X^2 = D{Me/Mm} - {Me/Mm}

No. You must isolate X, but you don't end up with X^2.

How would you solve this one?

x = a -bx (a and b are just constants)

It's the same basic equation.

 

[wilson_chem90]

would it be x = a/b?

 

[Doc Al]

would it be x = a/b?

No. Do it step by step:

x = a - bx

group the x terms on the left:
x + bx = a

factor out the x:
x (1 + b) = a

isolate the x by dividing both sides by (1 + b):
x = a/(1 + b)

done!

 

[wilson_chem90]

YES! finally, thank you so much. i appreciate your time and patience with me