Question

which values are solutions to \\(\frac{k - 3}{4} > -2\\)? select two options. \\(\square k = -10\\) \\(\square k = -7\\) \\(\square k = -5\\) \\(\square k = -1\\) \\(\square k = 0\\)

Explanation:

Step1: Solve the inequality for k

Multiply both sides of the inequality \(\frac{k - 3}{4}>-2\) by 4 to eliminate the denominator:
\(k - 3>-2\times4\)
\(k - 3>-8\)
Then add 3 to both sides:
\(k>-8 + 3\)
\(k>-5\)

Step2: Check each option

• For \(k=-10\): \(-10<-5\), so it is not a solution.
• For \(k=-7\): \(-7<-5\), so it is not a solution.
• For \(k=-5\): \(-5\) is not greater than \(-5\) (the inequality is strict), so it is not a solution.
• For \(k=-1\): \(-1>-5\), so it is a solution.
• For \(k = 0\): \(0>-5\), so it is a solution.

Answer:

D. \(k=-1\), E. \(k = 0\)