Sure! Let's solve the given compound inequality step-by-step:
[tex]\[ 58 + x < 2x + 6 \leq 12 \][/tex]
To solve this, we need to break it down into two separate inequalities and then find the intersection of the solutions for each inequality.
### Step 1: Solve [tex]\( 58 + x < 2x + 6 \)[/tex]
1. Subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ 58 < x + 6 \][/tex]
2. Subtract 6 from both sides:
[tex]\[ 52 < x \][/tex]
This simplifies to:
[tex]\[ x > 52 \][/tex]
### Step 2: Solve [tex]\( 2x + 6 \leq 12 \)[/tex]
1. Subtract 6 from both sides:
[tex]\[ 2x \leq 6 \][/tex]
2. Divide both sides by 2:
[tex]\[ x \leq 3 \][/tex]
### Step 3: Combine the solutions
From the above steps, we have two results:
[tex]\[ x > 52 \][/tex]
[tex]\[ x \leq 3 \][/tex]
Now, we need to find the values of [tex]\( x \)[/tex] that satisfy both inequalities simultaneously.
### Step 4: Find the intersection of the two inequalities
1. The inequality [tex]\( x > 52 \)[/tex] means [tex]\( x \)[/tex] must be greater than 52.
2. The inequality [tex]\( x \leq 3 \)[/tex] means [tex]\( x \)[/tex] must be less than or equal to 3.
However, there's no number [tex]\( x \)[/tex] that can be both greater than 52 and less than or equal to 3 at the same time.
### Conclusion
There is no value of [tex]\( x \)[/tex] that satisfies both inequalities together. Therefore, the compound inequality has no solution in the real numbers.
The final answer is:
[tex]\[
\text{No solution}
\][/tex]
