a). 5-3(b+2)=(2b+7)-(b-12) b). (2b+17)-(4b-5)=3b-7 c). (2b-6)-2(3b-5)=2(6-2b) svp aidez moi

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[tex]a)\ 5-3(b+2)=(2b+7)-(b-12)\\5-3b-6=2b+7-b+12\\-3b-1=b+19\\-3b-b=19+1\\-4b=20\\\\b=\dfrac{20}{-4b}\\\\\boxed{b=-5} [/tex]

[tex]b)\ (2b+17)-(4b-5)=3b-7\\2b+17-4b+5=3b-7\\-2b+22=3b-7\\-2b-3b=-7-22\\-5b=-29\\\\b=\dfrac{-29}{-5}\\\\\boxed{b=\dfrac{29}{5}=5,8}[/tex]

[tex]c)\ (2b-6)-2(3b-5)=2(6-2b)\\2b-6-6b+10=12-4b\\-4b+4=12-4b\\-4b+4b=12-4\\0\times b=8\\\\\boxed{\acute{e}quation\ impossible\Longrightarrow pas\ de\ solution }[/tex]