-13(y+2)+6(3y-4)=4(y-5)+13

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Solution for -13(y+2)+6(3y-4)=4(y-5)+13 equation: 



-13(y+2)+6(3y-4)=4(y-5)+13

We move all terms to the left:

-13(y+2)+6(3y-4)-(4(y-5)+13)=0

We multiply parentheses

-13y+18y-(4(y-5)+13)-26-24=0



We calculate terms in parentheses: -(4(y-5)+13), so:

4(y-5)+13

We multiply parentheses

4y-20+13

We add all the numbers together, and all the variables

4y-7

Back to the equation:

-(4y-7)



We add all the numbers together, and all the variables

5y-(4y-7)-50=0

We get rid of parentheses

5y-4y+7-50=0

We add all the numbers together, and all the variables

y-43=0

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

y=43

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