abs(z-3i)+abs(z+3i)=10

Solution for abs(z-3i)+abs(z+3i)=10 equation:

Simplifying
abs(z + -3i) + abs(z + 3i) = 10

Reorder the terms:
abs(-3i + z) + abs(z + 3i) = 10
(-3i * abs + z * abs) + abs(z + 3i) = 10
(-3abis + absz) + abs(z + 3i) = 10

Reorder the terms:
-3abis + absz + abs(3i + z) = 10
-3abis + absz + (3i * abs + z * abs) = 10
-3abis + absz + (3abis + absz) = 10

Reorder the terms:
-3abis + 3abis + absz + absz = 10

Combine like terms: -3abis + 3abis = 0
0 + absz + absz = 10
absz + absz = 10

Combine like terms: absz + absz = 2absz
2absz = 10

Solving
2absz = 10

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Divide each side by '2bsz'.
a = 5b-1s-1z-1

Simplifying
a = 5b-1s-1z-1