When solving an equation with two variables, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)
Solve 3c + 5c = c + 8y + 1 for c in terms of y.
| y + \(\frac{1}{2}\) | |
| 1\(\frac{1}{2}\)y + \(\frac{1}{2}\) | |
| -2\(\frac{3}{4}\)y - 1\(\frac{3}{4}\) | |
| -3\(\frac{2}{3}\)y - 1\(\frac{1}{3}\) |