To determine the value of \( b \) that would allow the equations \( 3y + 12 = 6x \) and \( 2y = 4x + b \) to have infinitely many solutions, we need to express both equations in the slope-intercept form \( y = mx + c \), where \( m \) is the slope.
First, let's rearrange the first equation:
-
Start with the equation: \[ 3y + 12 = 6x \]
-
Subtract 12 from both sides: \[ 3y = 6x - 12 \]
-
Divide every term by 3: \[ y = 2x - 4 \]
Now we have the first equation in slope-intercept form: \[ y = 2x - 4 \]
Next, we'll rearrange the second equation:
-
Start with the equation: \[ 2y = 4x + b \]
-
Divide every term by 2: \[ y = 2x + \frac{b}{2} \]
Now the second equation in slope-intercept form is: \[ y = 2x + \frac{b}{2} \]
For these two equations to represent the same line (and hence have infinitely many solutions), their slopes and intercepts must be equal:
- The slopes are already equal (both are \( 2 \)).
- Set the y-intercepts equal: \[ -4 = \frac{b}{2} \]
Now, multiply both sides by 2: \[ -8 = b \]
Now, let's check if the options given contain \( b = -8 \). Since they are:
- \( b = -8 \)
- \( b = -4 \)
- \( b = 2 \)
- \( b = 6 \)
The correct answer is: \[ \boxed{-8} \]