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 = 4c + 3z + 1 for c in terms of z.
| -2\(\frac{1}{9}\)z + \(\frac{2}{3}\) | |
| -\(\frac{1}{2}\)z + \(\frac{1}{2}\) | |
| -4z + 3 | |
| \(\frac{1}{2}\)z - \(\frac{7}{10}\) |