Question

How do I find the "a" value here??? sorry for the previous post, it was wrong equation. Need Help Please.
2a(L)^3 = 2h
-3a(L)^2= 0

Answer 1

You cannot really solve for a. It either equals 0, or it could equal anything.

Answer 2

I mean, we can sorta fake it, but there's a problem with that second equation.

Answer 3

here is the real equations through which i got these two. a(L)^3+b(L)^2=L and 3a(L)^2+2bL=0, how can do i find the value of either "b" or "a"?

Answer 4

Oh, that's better.

Answer 5

Divide the first equation by L.

Answer 6

but L has two two different powers there? how am i gona do that?

Answer 7

i got the answer, thanks for the hint.

Answer 8

2^2/2 = 2
2^3/2 =2^2
....

Answer 9

I divided the first equation by L and i got the answer as a=-2h/L^3, just one question. When I divide one equation by something, Isn't that the rule that I should apply same rule to the second equation too?

Answer 10

If you divide the first one by L, you will have something you can solve:

\(3aL^2 + 2bL = 0\)
- \(2(aL^2 + bL) = 2(1)\)
================
\(aL^2 + 0bL = -2\)
\[\implies a = -\frac{2}{L^2}\]

\[aL^2 + bL = 1\]\[\implies -2 + bL = 1\]\[\implies b = \frac{3}{L}\]

Answer 11

thanks for help.