Question

which of the following is a solution to the inequality below?
$-103 geq \frac{99}{h} - 4$
$h = -3$
$h = -1$
$h = -9$
$h = -11$

Explanation:

Step1: Substitute \( h = -3 \)

Substitute \( h = -3 \) into \( \frac{99}{h}-4 \): \( \frac{99}{-3}-4=-33 - 4=-37 \). Check if \( - 103\geq - 37 \): No.

Step2: Substitute \( h=-1 \)

Substitute \( h = - 1\) into \( \frac{99}{h}-4 \): \( \frac{99}{-1}-4=-99 - 4=-103 \). Check if \( - 103\geq - 103 \): Yes, but let's check other options.

Step3: Substitute \( h = - 9\)

Substitute \( h=-9 \) into \( \frac{99}{h}-4 \): \( \frac{99}{-9}-4=-11 - 4=-15 \). Check if \( - 103\geq - 15 \): No.

Step4: Substitute \( h=-11 \)

Substitute \( h = - 11\) into \( \frac{99}{h}-4 \): \( \frac{99}{-11}-4=-9 - 4=-13 \). Check if \( - 103\geq - 13 \): No. Wait, but when \( h=-1 \), we have \( \frac{99}{-1}-4=-103 \), and \( - 103\geq - 103 \) is true. But wait, let's re - check the inequality manipulation.

Wait, maybe I made a mistake in the first approach. Let's solve the inequality \( - 103\geq\frac{99}{h}-4 \)

Step1: Add 4 to both sides

\( - 103 + 4\geq\frac{99}{h}\)
\( - 99\geq\frac{99}{h}\)

Step2: Consider the sign of \( h \)

Since the right - hand side has \( h \) in the denominator, if \( h>0 \), when we multiply both sides by \( h \) (a positive number), the inequality sign remains the same: \( - 99h\geq99\), then \( h\leq - 1 \) (but \( h>0 \), no solution in positive numbers). If \( h < 0 \), when we multiply both sides by \( h \) (a negative number), the inequality sign flips: \( - 99h\leq99\), then \( h\geq - 1 \). But \( h < 0 \), so \( - 1\leq h < 0 \). Now let's check the options:

• For \( h=-3 \): \( - 3<-1 \), not in the solution set.
• For \( h = - 1\): \( h=-1 \) satisfies \( - 1\leq h < 0 \). Substitute into the original inequality: \( \frac{99}{-1}-4=-103 \), and \( - 103\geq - 103 \) is true.
• For \( h=-9 \): \( - 9<-1 \), not in the solution set.
• For \( h=-11 \): \( - 11<-1 \), not in the solution set.

Answer:

B. \( h = - 1 \) (Wait, but in the initial substitution, when \( h=-1 \), the value of \( \frac{99}{h}-4=-103 \), and \( - 103\geq - 103 \) holds. The other options:

When \( h=-3 \): \( \frac{99}{-3}-4=-37 \), \( - 103\geq - 37 \) is false.

When \( h=-9 \): \( \frac{99}{-9}-4=-15 \), \( - 103\geq - 15 \) is false.

When \( h=-11 \): \( \frac{99}{-11}-4=-13 \), \( - 103\geq - 13 \) is false.

When \( h=-1 \): \( \frac{99}{-1}-4=-103 \), and \( - 103\geq - 103 \) is true. So the solution is \( h = - 1 \))