Solution:

The equations in boxes are:

I: x + y = 36 ... (i)

II: \(\displaystyle y = \frac {5}{7} \times x\)
∴ 7y = 5x
i.e. 5x = 7y
∴ 5x − 7y = 0 ... (ii)

III:  x − y = 6 ... (iii)

IV:  2x − y = 27 ... (iv)

Now, let us solve equations (i) and (iii) to find the values of x and y.

Adding (i) and (iii), we get:

		x	+	y	=		36	... (i)
+		x	−	y	=		6	... (iii)
		2x			=		42

 \(\displaystyle x = \frac {42}{2} = 21\)

∴ x = 21 ... (v)

Substituting the value of x in equation (i), we get:
   x + y = 36 ... (i)
∴ 21 + y = 36
∴ y = 36 − 21
∴ y = 15 ... (vi)

∴ The length of the rectangle is 21 units and breadth is 15 units ... (vii)

Now, let us verify the solution using equation (iv).
Substituting the values of x = 21 and y = 15 in equation (iv),
we get:
  2x − y = 27 ... (iv)
∴ 2(21) − 15 = 42 − 15 = 27

∴ The solution is verified.

By taking one pair of equations at a time, 6 pairs can be formed from the 4 equations. Solving each pair will give the same solution, which can be verified by substituting the values of x and y in the remaining equations.