A false balance has equal arms and initially the balance beam is horizontal. An object weighs x when placed in one pan and y in the other pan. The true weight of the object is equal to

Since the balance has equal arms and it is a false balance, so the pans must be having different weights. Let w₁ and w₂ be the weights of left pan and right pan respectively. Let w be the true weight of object. When placed in left pan it weights x i.e. a total weight of w + w1 on left balances a total weight of w₂ + x on the right pan.

$\Rightarrow \mathrm{w}+{\mathrm{w}}_1=\mathrm{w}+\mathrm{x} ....\left(1\right)$
Similarly it weighs y when placed in right pan i.e.
 $\mathrm{y}+{\mathrm{w}}_1=\mathrm{w}+{\mathrm{w}}_2 ...\left(2\right)$
From (1)
${\mathrm{w}}_1-{\mathrm{w}}_2=\mathrm{x}-\mathrm{w} ....\left(3\right)$
From (2)
${\mathrm{w}}_1-{\mathrm{w}}_2=\mathrm{w}-\mathrm{y} ....\left(4\right)$
$\begin{array}{l}\Rightarrow \mathrm{x}-\mathrm{w}=\mathrm{w}-\mathrm{y}\\ \Rightarrow \mathrm{w}=\frac{\mathrm{x}+\mathrm{y}}{2}\end{array}$