23(-2c+3)=2(c+5)+c

Solution for 23(-2c+3)=2(c+5)+c equation:

23(-2c+3)=2(c+5)+c

We move all terms to the left:

23(-2c+3)-(2(c+5)+c)=0

We multiply parentheses

-46c-(2(c+5)+c)+69=0


We calculate terms in parentheses: -(2(c+5)+c), so:

2(c+5)+c

We add all the numbers together, and all the variables

c+2(c+5)

We multiply parentheses

c+2c+10

We add all the numbers together, and all the variables

3c+10

Back to the equation:

-(3c+10)


We get rid of parentheses

-46c-3c-10+69=0

We add all the numbers together, and all the variables

-49c+59=0

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

-49c=-59

c=-59/-49

c=1+10/49