For how many integer values of b is `b+3gt0 and 1gt2b-9`?

Options

A. Four
B. Five
C. Six
D. Seven

Answer

To solve the inequalities \( b + 3 > 0 \) and \( 1 > 2b - 9 \), we will break it down step by step.

### Step 1: Solve the first inequality
The first inequality is:
\[ b + 3 > 0 \]
To isolate \( b \), we subtract 3 from both sides:
\[ b > -3 \]

### Step 2: Solve the second inequality
The second inequality is:
\[ 1 > 2b - 9 \]
We can rearrange this by adding 9 to both sides:
\[ 10 > 2b \]
Next, we divide both sides by 2:
\[ 5 > b \quad \text{or} \quad b < 5 \]

### Step 3: Combine the inequalities
Now we have two inequalities:
1. \( b > -3 \)
2. \( b < 5 \)

Combining these, we find:
\[ -3 < b < 5 \]

### Step 4: Determine the integer values of \( b \)
The integer values of \( b \) that satisfy this combined inequality are:
- \( -2 \)
- \( -1 \)
- \( 0 \)
- \( 1 \)
- \( 2 \)
- \( 3 \)
- \( 4 \)

### Step 5: Count the integer values
Now, we count the integer values:
- The integers are: \( -2, -1, 0, 1, 2, 3, 4 \)

Thus, there are a total of 7 integer values of \( b \) that satisfy both inequalities.

### Final Answer
The total number of integer values of \( b \) is **7**.

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