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 6c + = -4c + z + 1 for c in terms of z.
| -1\(\frac{1}{4}\)z + \(\frac{1}{4}\) | |
| \(\frac{1}{10}\)z + \(\frac{1}{10}\) | |
| z - 9 | |
| -3z - 2\(\frac{1}{3}\) |