5(y+4)+y=4(y+4)-5

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Solution for 5(y+4)+y=4(y+4)-5 equation: 



5(y+4)+y=4(y+4)-5

We move all terms to the left:

5(y+4)+y-(4(y+4)-5)=0

We add all the numbers together, and all the variables

y+5(y+4)-(4(y+4)-5)=0

We multiply parentheses

y+5y-(4(y+4)-5)+20=0



We calculate terms in parentheses: -(4(y+4)-5), so:

4(y+4)-5

We multiply parentheses

4y+16-5

We add all the numbers together, and all the variables

4y+11

Back to the equation:

-(4y+11)



We add all the numbers together, and all the variables

6y-(4y+11)+20=0

We get rid of parentheses

6y-4y-11+20=0

We add all the numbers together, and all the variables

2y+9=0

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

2y=-9

y=-9/2

y=-4+1/2

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