How do you solve this algebraic equation?

M = (0.0333RW) + W
(0.0333RW) + W = M
0.0333RW + W = M
W(0.0333R+1) = M
[W(0.0333R+1)]/(0.0333R+1) = M/(0.0333R+1)
W = M/(0.0333R+1)

M = 0.0333RW + W
W(0.0333R + 1) = M
W = M/(0.0333R + 1)

Answer: W = M/(0.0333R + 1)

First step -- isolate among the variables. consequently, upload 3/y to the two edge of the decrease equation. This leaves you with that equation analyzing one million/x = 3/2 + 3/y. Invert the two edge, leaving x = one million/(3/2 + 3/y). Plug this cost for x into the suited equation, which then will turn into 4/(one million/(3/2 + 3/y)) + one million/y = 5/3. yet one million/(one million/(3/2 + 3/y)) = 3/2 + 3/y, so multiplying this by means of 4 provides 12/2 + 12/y + one million/y = 5/3. placing all the y's on one edge and all the numbers at the distinctive provides 12/y + one million/y = 5/3 - 12/2. including these provides thirteen/y = (-6+5/3) = -18/3 + 5/3 = -thirteen/3. consequently y = -3. Plugging this cost for y into the suited equation yields 4/x + one million/(-3) = 5/3 including one million/3 to each edge yields 4/x = 5/3 + one million/3 . or 6/3, or 2. on condition that 4/x = 2, x may desire to equivalent 2. Checking, applying the two values, 4/2 + one million/(-3) = 5/3, and 2 - one million/3 = 5/3 one million/2 -(3/-3) = 3/2, and one million/2 + one million = 2.

M = (0.0333RW) - W
M = 0.0333RW - W
M = W(0.0333R)
W = M/(0.0333R)

Weight = Maximum weight/(0.0333 * Repetition)

M=W(1+0.0333R)
W=M / (1+0.0333R)

combine like terms: M = W (0.0333R + 1)
transpose: W = M/(0.0333R +1)
just follow the pattern, its usually the same thing for all algebraic problems like this...

M = (0.0333RW) + W
(0.0333RW) + W = M
W(0.0333R+1) = M
W = M/(0.0333R+1)