`{{:(a =b ^(2) + 4b -12),(a=-12 +b):}`
The ordered pair (a,b) satisfies the system of equatin above. What is one possible value of b ?

Options

A. `-6`
B. `-3`
C. 2
D. 3

Explanation

To solve the system of equations given by:

1. \( a = b^2 + 4b - 12 \)
2. \( a = -12 + b \)

we can set the two expressions for \( a \) equal to each other since they both represent \( a \):

\[ b^2 + 4b - 12 = -12 + b \]

### Step 1: Simplify the equation
First, we can eliminate \(-12\) from both sides:

\[ b^2 + 4b - 12 + 12 = b \]

This simplifies to:

\[ b^2 + 4b = b \]

### Step 2: Rearrange the equation
Next, we will move all terms to one side of the equation:

\[ b^2 + 4b - b = 0 \]

This simplifies to:

\[ b^2 + 3b = 0 \]

### Step 3: Factor the equation
Now we can factor the left-hand side:

\[ b(b + 3) = 0 \]

### Step 4: Solve for \( b \)
Setting each factor equal to zero gives us:

1. \( b = 0 \)
2. \( b + 3 = 0 \) which simplifies to \( b = -3 \)

### Conclusion
Thus, one possible value of \( b \) is:

\[ b = -3 \]