1=(9/5c+32)

Solution for 1=(9/5c+32) equation:

1=(9/5c+32)

We move all terms to the left:

1-((9/5c+32))=0


Domain of the equation: 5c+32))!=0

c∈R


We multiply all the terms by the denominator


-((9+1*5c+32))=0


We calculate terms in parentheses: -((9+1*5c+32)), so:

(9+1*5c+32)

We add all the numbers together, and all the variables

(1*5c+41)

We get rid of parentheses

1*5c+41

Wy multiply elements

5c+41

Back to the equation:

-(5c+41)


We get rid of parentheses

-5c-41=0

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

-5c=41

c=41/-5

c=-8+1/5