In a quadrilateral ABCD, AB = BC, AD = DC, ∠ABD = 68°, ∠ADB = (2y-7)°, ∠BDC = 33°, ∠DBC = (3x+2)°. Then the value of 2x+3y is:

AB = BC

AD = DC

​$$\angle ABD = 68^\circ \\\angle ADB = (2y - 7)^\circ \\\angle BDC = 33^\circ \\\angle DBC = (3x + 2)^\circ$$​​

Congruency property of triangle:

Side – Side – Side (SSS): if the respective sides of two triangle are equal than both the triangle are congruent to each other.

In a congruent triangle: the area, sides, angles of the respective triangle are equal.

In Quadrilaterals ABCD :

Two triangles $$\triangle ABD$$  and  $$\triangle CBD$$  are formed.

Now, in $$\triangle ABD$$​ and $$\triangle CBD$$ ​​

AB = CB (given}

AD = CD (given)

DB = BD (common side)

therefore,  $$\triangle ABD \cong \triangle CBD$$ 

So, $$\angle ADB = \angle CDB$$ 

 $$\implies 2y -7 = 33 \\ 2y = 40 \\ y = 20^\circ$$ 

Similarly, 

 $$\angle ABD = \angle CBD \\ \implies 68 = 3x +2 \\ 3x = 66 \\ x = 22^\circ$$ 

Thus , 2x + 3y = $$2 \times 22 + 3\times 20 = 44+60 = \bf104^\circ$$​​