1/5(25-10b)+4=(9b-15)-6

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Solution for 1/5(25-10b)+4=(9b-15)-6 equation: 



1/5(25-10b)+4=(9b-15)-6

We move all terms to the left:

1/5(25-10b)+4-((9b-15)-6)=0



Domain of the equation: 5(25-10b)!=0

b∈R



We add all the numbers together, and all the variables

1/5(-10b+25)-((9b-15)-6)+4=0

We multiply all the terms by the denominator


-(((9b-15)-6))*5(-10b+25)+4*5(-10b+25)+1=0



We calculate terms in parentheses: -(((9b-15)-6))*5(-10b+25), so:

((9b-15)-6))*5(-10b+25

We add all the numbers together, and all the variables

-10b+((9b-15)-6))*5(+25

Back to the equation:

-(-10b+((9b-15)-6))*5(+25)



We add all the numbers together, and all the variables

-(-10b+((9b-15)-6))*525+4*5(-10b+25)+1=0

Wy multiply elements

-(-10b+((9b-15)-6))*525+20b(-+1=0

We use the square of the difference formula

-(-10b+((9b-15)-6))*525+20b(-1=0

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