Question

8c+32=5c+20
walk through please

Answer 1

so we have:
\[8c+32=5c+20\]
Now, in order to solve a first degree equation, we must leave the variables ("c") on one side of the equality and the constants in other.
so we'll substract 5c and 32 on either sides of the equation:
\[8c+32-5c-32=5c+20-32-5c\]
Good, now let's take out the terms that simplify:
\[8c-5c=20-32\]
We know how to operate the right side of the equiation, it's -12.
But on the left side, we'll use a little trick called, common factor:
\[c(8-5)=-12\]
So we now transformed those two numbers that were multiplying into a operable sustraction:
\[c(3)=-12\]
Let's get rid of that parenthesis by multiplying:
\[3c=-12\]
Now, I want to leave the variable "c" alone on the left side, so i'll divide both sides by 3:
\[\frac{ 3c }{ 3 }=-\frac{ 12 }{ 3 }\]
The 3's will simplify, so we'll end up with:
\[c=-\frac{ 12 }{ 3 }\]
Let's do that division:
\[c=-4\]
And there we have it :)