Given that  −10≤k≤10
and  k
is an integer. How many possible integers  k
so that the following inequality holds? \frac{k-3}{k+6}\le 1

Given the inequality \( \frac{k-3}{k+6} \leq 1 \) and the range \( -10 \leq k \leq 10 \) where \( k \) is an integer, we need to determine how many possible integer values of \( k \) satisfy this inequality.

### Step 1: Solve the inequality algebraically

We start by isolating the fraction:

\[
\frac{k-3}{k+6} \leq 1
\]

Subtract 1 from both sides:

\[
\frac{k-3}{k+6} - 1 \leq 0
\]

Rewrite the left side with a common denominator:

\[
\frac{k-3 - (k+6)}{k+6} \leq 0
\]

Simplify the numerator:

\[
\frac{k-3 - k - 6}{k+6} \leq 0 \implies \frac{-9}{k+6} \leq 0
\]

### Step 2: Analyze the inequality

The fraction \( \frac{-9}{k+6} \) will be non-positive (i.e., \( \leq 0 \)) when the denominator \( k+6 \) is positive or zero, because the numerator \(-9\) is negative. Therefore, we need:

\[
k+6 > 0 \implies k > -6
\]

Since \( k \) is an integer, \( k \geq -5 \).

### Step 3: Consider the range of \( k \)

We are given that \( -10 \leq k \leq 10 \). Combining this with the condition \( k \geq -5 \), we have:

\[
-5 \leq k \leq 10
\]

### Step 4: Count the number of integers

The integers \( k \) that satisfy \( -5 \leq k \leq 10 \) are:

\[
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
\]

This is a total of \( 10 - (-5) + 1 = 16 \) integers.

### Final Answer

There are **16 possible integers** for \( k \) that satisfy the given inequality.

Would you like more details on any part of the solution, or do you have any additional questions?

---

**Related Questions:**
1. How would the solution change if \( k \) were not restricted to integers?
2. What if the inequality was strict, \( \frac{k-3}{k+6} < 1 \)?
3. How would the problem change if the inequality were reversed to \( \frac{k-3}{k+6} \geq 1 \)?
4. Can you solve similar inequalities with different expressions in the numerator and denominator?
5. How would you graph the solution on a number line?

**Tip:** When solving inequalities involving fractions, always consider the sign of both the numerator and the denominator separately before combining the results.

Explore how to solve the inequality \( \frac{k-3}{k+6} \leq 1 \) for integer values of \( k \) in the range \( -10 \leq k \leq 10 \). Follow a step-by-step solution to find how many integers satisfy the inequality. Suitable for high school students and advanced learners.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation