In the figure
$$ AB = BC $$ and $$ AD = DC $$
$$ \angle ABD = 50^{\circ} , \angle ADB = y - 7^{\circ} $$
$$ \angle CBD = x + 5^{\circ} , \angle CDB = 38^{\circ} $$
In $$ \Delta ABD $$ and $$ \Delta CBD $$
$$ BD = BD $$ is common
It is given that
$$ AB = BC $$ and $$ AD = CD $$
Here $$ \Delta ABD \cong \Delta CBD $$ (SSS Axiom)
$$ \angle ABD = \angle CBD $$
So we get
$$ 50 = x + 5 $$
$$ x = 50 - 5 = 45^{\circ} $$
$$ \angle ADB = \angle CDB $$
$$ y - 7 = 38 $$
$$ y = 38 + 7 = 45^{\circ} $$
Therefore , $$ x = 45^{\circ} $$ and $$ y = 45^{\circ} $$