4+3(2+5v)+7=7+(5v+2)3+4

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Solution for 4+3(2+5v)+7=7+(5v+2)3+4 equation: 



4+3(2+5v)+7=7+(5v+2)3+4

We move all terms to the left:

4+3(2+5v)+7-(7+(5v+2)3+4)=0

We add all the numbers together, and all the variables

3(5v+2)-(7+(5v+2)3+4)+4+7=0

We add all the numbers together, and all the variables

3(5v+2)-(7+(5v+2)3+4)+11=0

We multiply parentheses

15v-(7+(5v+2)3+4)+6+11=0



We calculate terms in parentheses: -(7+(5v+2)3+4), so:

7+(5v+2)3+4

determiningTheFunctionDomain
(5v+2)3+7+4

We add all the numbers together, and all the variables

(5v+2)3+11

We multiply parentheses

15v+6+11

We add all the numbers together, and all the variables

15v+17

Back to the equation:

-(15v+17)



We add all the numbers together, and all the variables

15v-(15v+17)+17=0

We get rid of parentheses

15v-15v-17+17=0

We add all the numbers together, and all the variables

=0

v=0/1

v=0

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