5(n+3)-4(n-2)=2(n+1)-20

Solution for 5(n+3)-4(n-2)=2(n+1)-20 equation:

5(n+3)-4(n-2)=2(n+1)-20

We move all terms to the left:

5(n+3)-4(n-2)-(2(n+1)-20)=0

We multiply parentheses

5n-4n-(2(n+1)-20)+15+8=0


We calculate terms in parentheses: -(2(n+1)-20), so:

2(n+1)-20

We multiply parentheses

2n+2-20

We add all the numbers together, and all the variables

2n-18

Back to the equation:

-(2n-18)


We add all the numbers together, and all the variables

n-(2n-18)+23=0

We get rid of parentheses

n-2n+18+23=0

We add all the numbers together, and all the variables

-1n+41=0

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

-n=-41

n=-41/-1

n=+41