Question

802:{572-[9×(4×(2a+3b)+1)+108]:3}=2

Answer 1

802:{572-[9×(4×(2a+3b)+1)+108]:3}=2
572-[9×(4×(2a+3b)+1)+108]:3=802:2
572-[9×(4×(2a+3b)+1)+108]:3=401
[9×(4×(2a+3b)+1)+108]:3=572-401
[9×(4×(2a+3b)+1)+108]:3=171
9×(4×(2a+3b)+1)+108=171×3
9×(4×(2a+3b)+1)+108=513
9×(4×(2a+3b)+1)=513-108
9×(4×(2a+3b)+1)=405
4×(2a+3b)+1=405:9
4×(2a+3b)+1=45
4×(2a+3b)=45-1
4×(2a+3b)=44
2a+3b=44:4
2a+3b=11

a=0
b=1
2a+3b=2×0+3×1=0+3=3

a=1
b=2
2a+3b=2×1+3×2=2+6=8


a=2
b=3
2a+3b=2×2+3×3=4+9=13


a=1
b=3
2a+3b=2×1+3×3=2+9=11 (soluția exercițiului)

a=8
b=1
2a+3b=2×4+3×1=8+3=11 (soluția exercițiului)