ASVAB Math Knowledge Practice Test 340535

Question 2 of 5
Two Variables

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 - 2c = c + 4z + 5 for c in terms of z.

\(\frac{15}{16}\)z - \(\frac{1}{16}\)
\(\frac{1}{12}\)z - \(\frac{2}{3}\)
-1\(\frac{1}{2}\)z - 1\(\frac{1}{4}\)
-\(\frac{1}{4}\)z + \(\frac{5}{12}\)