Certainly! Let's solve the compound inequality [tex]\(2(2x + 3) \leq 7x + 5 < 3x - 1\)[/tex] step-by-step. We will solve the inequalities separately and then determine if there is a value of [tex]\(x\)[/tex] that satisfies both inequalities simultaneously.
### Step 1: Solve the first part of the inequality
[tex]\[ 2(2x + 3) \leq 7x + 5 \][/tex]
1. Distribute the 2 on the left-hand side:
[tex]\[ 4x + 6 \leq 7x + 5 \][/tex]
2. Move all [tex]\(x\)[/tex] terms to one side of the inequality:
[tex]\[ 4x - 7x \leq 5 - 6 \][/tex]
3. Simplify:
[tex]\[ -3x \leq -1 \][/tex]
4. Divide both sides by [tex]\(-3\)[/tex]. Remember, when dividing by a negative number, the inequality sign reverses:
[tex]\[ x \geq \frac{1}{3} \][/tex]
### Step 2: Solve the second part of the inequality
[tex]\[ 7x + 5 < 3x - 1 \][/tex]
1. Move all [tex]\(x\)[/tex] terms to one side of the inequality:
[tex]\[ 7x - 3x < -1 - 5 \][/tex]
2. Simplify:
[tex]\[ 4x < -6 \][/tex]
3. Divide both sides by 4:
[tex]\[ x < -\frac{3}{2} \][/tex]
### Step 3: Combine the results
We have:
[tex]\[ x \geq \frac{1}{3} \][/tex]
and
[tex]\[ x < -\frac{3}{2} \][/tex]
To satisfy the compound inequality, [tex]\(x\)[/tex] must meet both conditions simultaneously. However, no single value of [tex]\(x\)[/tex] can be both greater than or equal to [tex]\(\frac{1}{3}\)[/tex] and less than [tex]\(-\frac{3}{2}\)[/tex] at the same time.
### Conclusion
There is no value of [tex]\(x\)[/tex] that satisfies both inequalities simultaneously.
Therefore, the solution to the compound inequality [tex]\( 2(2x + 3) \leq 7x + 5 < 3x - 1 \)[/tex] is:
[tex]\[ \text{No solution} \][/tex]
