2/5(25b)-40=1/4(16b+8

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Solution for 2/5(25b)-40=1/4(16b+8 equation: 



2/5(25b)-40=1/4(16b+8

We move all terms to the left:

2/5(25b)-40-(1/4(16b+8)=0



Domain of the equation: 525b!=0

b!=0/525

b!=0

b∈R





Domain of the equation: 4(16b+8)-40!=0

We move all terms containing b to the left, all other terms to the right

4(16b+8)!=40

b∈R



We calculate fractions


(8b(1-40)/(2100b^2(1-40)+(-(1*525b)/(2100b^2(1-40)=0



We calculate terms in parentheses: +(8b(1-40)/(2100b^2(1-40)+(-(1*525b)/(2100b^2(1-40), so:

8b(1-40)/(2100b^2(1-40)+(-(1*525b)/(2100b^2(1-40

We can not solve this equation

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