To solve the equation, Lorie applies the distributive property, combines like terms, then applies the addition and subtraction properties of equality to isolate the variable term on one side of the equation and the constant term on the other side. What are the possible coefficients of x after Lorie has completed these steps?

Question

10 (one-half x + 2) minus 5 = 3 (x minus 6) + 1
–32 and 2
–2 and 32
–2 and 2
–32 and 32

Answer 1

To solve the equation step by step, let's start with the given expression:

\[ 10 \left(\frac{1}{2}x + 2\right) - 5 = 3(x - 6) + 1 \]

1. Apply the Distributive Property:

   Left side: \[ 10 \left(\frac{1}{2}x + 2\right) = 10 \cdot \frac{1}{2}x + 10 \cdot 2 = 5x + 20 \]

   The left side becomes: \[ 5x + 20 - 5 = 5x + 15 \]

   Right side: \[ 3(x - 6) + 1 = 3x - 18 + 1 = 3x - 17 \]

   So the equation now is: \[ 5x + 15 = 3x - 17 \]

2. Combine Like Terms:

   To isolate \( x \), subtract \( 3x \) from both sides: \[ 5x - 3x + 15 = -17 \] This simplifies to: \[ 2x + 15 = -17 \]

3. Isolate the Variable Term:

   Next, subtract 15 from both sides: \[ 2x + 15 - 15 = -17 - 15 \] Which simplifies to: \[ 2x = -32 \]

4. Solve for x:

   Divide both sides by 2: \[ x = \frac{-32}{2} = -16 \]

Now, we have the solution \( x = -16 \).

5. Possible Coefficients of x:

Next, we look at the coefficients based on the steps that Lorie completed. Before the final operation where we divided by 2, we could have potential coefficients \(-32\) (from \(2x\) resulting in \(-32\)) and \(2\) (already present as the coefficient of \(x\) before moving through the steps).

Thus, the possible coefficients of \( x \) after Lorie has completed these steps are: \[ -32 \text{ and } 2 \]

From the options presented in your question, the correct answer is:

–32 and 2.