Substituting the equation \( x=4 v-12 \) into the equation \( -2 v=x-6 \) will produce the equation a.) \( 2 y=-6 \) b.) \( -6 y=-6 \) c.) \( -6 y=-18 \) d.) \( 2 y=-18 \)

Let's substitute \( x = 4v - 12 \) into \( -2v = x - 6 \). Starting with the second equation: \[ -2v = (4v - 12) - 6 \] Now simplify the right side: \[ -2v = 4v - 12 - 6 \] \[ -2v = 4v - 18 \] Now, add \( 2v \) to both sides to eliminate the \( -2v \): \[ 0 = 6v - 18 \] Now, rearrange to isolate \( v \): \[ 6v = 18 \] This leads to: \[ v = 3 \] So, the equation produced after substitution simplifies to \( 6v - 18 = 0 \), but we need to express it as one of the provided options. Let's check what this means by rearranging it: \[ 6v = 18 \implies 2v = 6 \] So rewriting gives \( 2y = -18 \) suggests the adjustment to the format, essentially aligning with option d.) \( 2y = -18 \). Thus, the answer is d.) \( 2y = -18 \).