Question

307 < 11b + 32 < 318
progress: 1/3
part 2 of 3
(b) the solution set for the compound inequality is graphed as
number line with ticks at 21, 22, 23, 24, 25, 26, 27, 28, 29

Explanation:

Step1: Subtract 32 from all parts

To isolate the term with \( b \), we subtract 32 from each part of the compound inequality.
\[
307 - 32 < 11b + 32 - 32 < 318 - 32
\]
\[
275 < 11b < 286
\]

Step2: Divide all parts by 11

To solve for \( b \), we divide each part of the inequality by 11.
\[
\frac{275}{11} < \frac{11b}{11} < \frac{286}{11}
\]
\[
25 < b < 26
\]

Answer:

The solution to the compound inequality \( 307 < 11b + 32 < 318 \) is \( 25 < b < 26 \). On the number line, this would be represented by an open circle at 25 and an open circle at 26, with a line segment connecting them (since \( b \) is strictly between 25 and 26).