4(c-5)-c=2(c+4)+c

Solution for 4(c-5)-c=2(c+4)+c equation:

4(c-5)-c=2(c+4)+c

We move all terms to the left:

4(c-5)-c-(2(c+4)+c)=0

We add all the numbers together, and all the variables

-1c+4(c-5)-(2(c+4)+c)=0

We multiply parentheses

-1c+4c-(2(c+4)+c)-20=0


We calculate terms in parentheses: -(2(c+4)+c), so:

2(c+4)+c

We add all the numbers together, and all the variables

c+2(c+4)

We multiply parentheses

c+2c+8

We add all the numbers together, and all the variables

3c+8

Back to the equation:

-(3c+8)


We add all the numbers together, and all the variables

3c-(3c+8)-20=0

We get rid of parentheses

3c-3c-8-20=0

We add all the numbers together, and all the variables

-28!=0

There is no solution for this equation