Certainly! To solve the compound inequality
[tex]\[ 4x - 3 \leq \frac{1}{2}(x + 8) < x + 5, \][/tex]
we need to break it up into two separate inequalities and solve them individually. They are:
1. [tex]\( 4x - 3 \leq \frac{1}{2}(x + 8) \)[/tex]
2. [tex]\( \frac{1}{2}(x + 8) < x + 5 \)[/tex]
### Step 1: Solve the first inequality
[tex]\[ 4x - 3 \leq \frac{1}{2}(x + 8) \][/tex]
First, eliminate the fraction by multiplying everything by 2:
[tex]\[ 2(4x - 3) \leq x + 8 \][/tex]
[tex]\[ 8x - 6 \leq x + 8 \][/tex]
Now, isolate [tex]\( x \)[/tex]:
[tex]\[ 8x - x \leq 8 + 6 \][/tex]
[tex]\[ 7x \leq 14 \][/tex]
[tex]\[ x \leq 2 \][/tex]
### Step 2: Solve the second inequality
[tex]\[ \frac{1}{2}(x + 8) < x + 5 \][/tex]
Again, eliminate the fraction by multiplying everything by 2:
[tex]\[ x + 8 < 2(x + 5) \][/tex]
[tex]\[ x + 8 < 2x + 10 \][/tex]
Now, isolate [tex]\( x \)[/tex]:
[tex]\[ 8 < x + 10 \][/tex]
[tex]\[ 8 - 10 < x \][/tex]
[tex]\[ -2 < x \][/tex]
[tex]\[ x > -2 \][/tex]
### Step 3: Combine the Solutions
The solutions to each individual inequality are:
[tex]\[ x \leq 2 \][/tex]
[tex]\[ x > -2 \][/tex]
To find the combined solution, we take the intersection of these two solutions:
[tex]\[ -2 < x \leq 2 \][/tex]
### Number Line Representation
On a number line, this interval would be represented as follows:
- An open circle at [tex]\( x = -2 \)[/tex] (indicating that [tex]\( -2 \)[/tex] itself is not included in the solution).
- A closed circle at [tex]\( x = 2 \)[/tex] (indicating that [tex]\( 2 \)[/tex] is included in the solution).
[tex]\[
\begin{array}{ccccccccccc}
\text{ } & \text{ } & \text{ } & \text{ } & -2 & \text{ }\xrightarrow{\text{ }}\text{ } & 0 & \text{-}& \text{ } & 2 & \text{-}
\end{array}
\][/tex]
Thus, the solution on a number line is the interval [tex]\( (-2, 2] \)[/tex].
