A(y+3)+5y=4(2y+1)+5

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Solution for A(y+3)+5y=4(2y+1)+5 equation: 



(A+3)+5A=4(2A+1)+5

We move all terms to the left:

(A+3)+5A-(4(2A+1)+5)=0

We add all the numbers together, and all the variables

5A+(A+3)-(4(2A+1)+5)=0

We get rid of parentheses

5A+A-(4(2A+1)+5)+3=0



We calculate terms in parentheses: -(4(2A+1)+5), so:

4(2A+1)+5

We multiply parentheses

8A+4+5

We add all the numbers together, and all the variables

8A+9

Back to the equation:

-(8A+9)



We add all the numbers together, and all the variables

6A-(8A+9)+3=0

We get rid of parentheses

6A-8A-9+3=0

We add all the numbers together, and all the variables

-2A-6=0

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

-2A=6

A=6/-2

A=-3

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