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32=4/5(b-5)
We move all terms to the left:
32-(4/5(b-5))=0
Domain of the equation: 5(b-5))!=0We multiply all the terms by the denominator
b∈R
-(4+32*5(b-5))=0
We calculate terms in parentheses: -(4+32*5(b-5)), so:
4+32*5(b-5)
determiningTheFunctionDomain 32*5(b-5)+4
Wy multiply elements
160b(b+4
Back to the equation:
-(160b(b+4)
We calculate terms in parentheses: -(160b(b+4), so:
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We calculate terms in parentheses: -(160b(b+4), so:
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We calculate terms in parentheses: -(160b(b+4), so:
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We calculate terms in parentheses: -(160b(b+4), so:
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We calculate terms in parentheses: -(160b(b+4), so:
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We calculate terms in parentheses: -(160b(b+4), so:
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We calculate terms in parentheses: -(160b(b+4), so:
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We calcu
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