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

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Step 1: Given AB = BC and AD = DC, triangles ABD and BCD are isosceles.

Step 2: In triangle ABD, AB = BC implies triangle ABC is isosceles with AB = BC, but the problem states AB = BC and AD = DC, so consider triangles ABD and BCD separately.

Step 3: Since AD = DC, triangle ADC is isosceles with AD = DC.

Step 4: Given angles:  
- ∠ABD = 68°  
- ∠ADB = (2y - 7)°  
- ∠BDC = 33°
- ∠DBC = (3x + 2)°

Step 5: Consider triangle ABD:
- Angles are ∠ABD = 68°, ∠ADB = (2y - 7)°, and ∠BAD.
- Since AB = BC, triangle ABC is isosceles with AB = BC, but we focus on ABD here.

Step 6: In triangle ABD, sum of angles = 180°  
=> ∠BAD + ∠ABD + ∠ADB = 180°  
=> ∠BAD + 68 + (2y - 7) = 180  
=> ∠BAD + 2y + 61 = 180  
=> ∠BAD = 119 - 2y

Step 7: Consider triangle BCD:
- AD = DC, so triangle ADC is isosceles, but we focus on BCD where angles ∠BDC = 33°, ∠DBC = (3x + 2)°, and ∠BCD.

Step 8: In triangle BCD, sum of angles = 180°   
=> ∠BDC + ∠DBC + ∠BCD = 180  
=> 33 + (3x + 2) + ∠BCD = 180  
=> ∠BCD = 145 - 3x

Step 9: Since AB = BC and AD = DC, quadrilateral ABCD has symmetry properties. Points B and D are connected to A and C respectively.

Step 10: Angles at B and D relate to each other.     
Specifically, ∠BAD and ∠BCD are equal because AB = BC and AD = DC.

Step 11: Equate ∠BAD and ∠BCD:    
=> 119 - 2y = 145 - 3x   
=> 3x - 2y = 145 - 119   
=> 3x - 2y = 26

Step 12: From triangle ABD, consider the isosceles property AB = BC, so angles opposite equal sides are equal.

Step 13: In triangle ABC, AB = BC implies ∠BAC = ∠BCA.

Step 14: Using the given angles at B and D, and the isosceles properties, solve for x and y using the equations:    
- 3x - 2y = 26    
- From triangle ABD, sum of angles gives another equation involving y.

Step 15: Using the given angles and properties, solve the system:   
From step 6: ∠BAD = 119 - 2y    
From step 8: ∠BCD = 145 - 3x   
Equate: 119 - 2y = 145 - 3x => 3x - 2y = 26

Step 16: Use the triangle ABD angles sum:    
∠ABD = 68°, ∠ADB = (2y - 7)°, ∠BAD = 119 - 2y    
Sum = 68 + (2y - 7) + (119 - 2y) = 180   
=> 68 - 7 + 119 + (2y - 2y) = 180    
=> 180 = 180 (valid, no new info)

Step 17: Use triangle BCD angles sum:   
∠BDC = 33°, ∠DBC = (3x + 2)°, ∠BCD = 145 - 3x   
Sum = 33 + (3x + 2) + (145 - 3x) = 180   
=> 33 + 2 + 145 + (3x - 3x) = 180    
=> 180 = 180 (valid, no new info)

Step 18: We have only one equation: 3x - 2y = 26

Step 19: Use the fact that in triangle ABD, AB = BC implies ∠ABD = ∠CBD, but ∠ABD is given as 68°, and ∠DBC is (3x + 2)°.

Step 20: Since AB = BC, ∠ABD = ∠CBD    
Given ∠ABD = 68°, so ∠DBC = 68°    
=> 3x + 2 = 68    
=> 3x = 66   
=> x = 22

Step 21: Substitute x = 22 into 3x - 2y = 26   
=> 3(22) - 2y = 26   
=> 66 - 2y = 26   
=> -2y = 26 - 66 = -40   
=> 2y = 40  
=> y = 20

Step 22: Calculate 2x + 3y    
=> 2(22) + 3(20) = 44 + 60 = 104

Answer: 104

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