we have
\(4x-12\leq 16+8x\)
The options of the question are
\(-10\ -9\ -8\ -7\)
Group terms that contain the same variable, and move the constants to the opposite side of the inequality
\(-12-16\leq 8x-4x\)
\(-28\leq 4x\)
rewrite the inequality
\(4x\geq -28\)
Divide by \(4\) both sides
\(x\geq -7\)
The solution is the interval-------> [-7,∞)
All real numbers greater than or equal to \(-7\)
therefore
the answer is
\(-7\)
