One way to solve for the coefficients is to denote them with letters (x, y and z) for reactants and (a, b, c) for the products. Then the equation can be written as:

`x FeBr_2 + y HBrO_3 + z HI -> a FeI_3 + b IBr + c H_2O`

Step 2: is to calculate the number of atoms of each elements on both sides of the chemical reaction and write them in terms of these assumed coefficients. For example, the number of atoms of iron on reactant side is x, while those on the product side is 'a'.

That means, x = a (eq. 1)

Similarly, for bromine, the number of atoms on reactant side is 2x (from FeBr2) and y (from HBrO3), and on the product side, they are b (from IBr). Thus, 2x + y = b (eq. 2)

Similarly, for H: y + z = 2c (eq. 3)

for O: 3y = c (eq. 4)

and, for I: z = 3 a+ b (eq. 5)

Now we have 5 equations and 6 variables. This means that we will have one variable that is dependent on the other. ` `

Step 3: Solve the equations.

Looking at the equations, we can directly use the equations 1 and 3 to eliminate 'a' and 'c' from other equations.

For example, equation 2 can be written as:

y + z = 2c

or y + z = 2 (3y) = 6y

or, z = 5y (eq. 6)

Similarly, z = 3a + b

or, 5y = 3x + b

or. b = 5y -3x

Comparing this equation with equation (2), we get:

5y - 3x = 2x + y

or, 4y = 5x

Since, we do not have any more equations, we can assume one of the variables to be a constant and solve all other using it.

Let us assume, x = 4, then y = 5/4 x = 5/4 *4 = 5, z = 5y = 25, a = x = 4, b = 5y-3x = 5(5)-3(4) = 13 and c = 3y =12

Using all the variables, we can get the balanced equation as:

`4 FeBr_2 + 5 HBrO_3 + 25 HI -> 4 FeI_3 + 13 IBr + 15 H_2O`

The final step is to check the number of atoms of each elements on both sides of chemical equation.

For example, there are 5 + 25 = 30 atoms of hydrogen on reactant side, and there are 15 x 2 = 30 atoms on product side. Thus, hydrogen atoms are balanced. One can do the same check for all the atoms and verify the accuracy of the balancing steps.

Hope this helps.