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 5b + 6b = 6b + 4z + 1 for b in terms of z.
| -\(\frac{1}{15}\)z - \(\frac{2}{5}\) | |
| 2z - 1 | |
| 4\(\frac{1}{3}\)z + 1\(\frac{1}{3}\) | |
| -5\(\frac{1}{2}\)z + 4 |