Which one of the following compound inequalities has no solution?

A. [tex]2x - 1 \ \textless \ x - 4[/tex] and [tex]-4x + 3 \ \textgreater \ 2x - 9[/tex]
B. [tex]2x - 5 \leq 3x + 5[/tex] and [tex]-4x - 1 \ \textgreater \ 2x + 3[/tex]
C. [tex]-3x - 1 \leq 2x + 9[/tex] and [tex]-4x + 3 \ \textgreater \ -2x + 11[/tex]
D. [tex]3(x + 1) \geq 2(x + 5)[/tex] and [tex]-2(x + 1) \leq 3(x + 1)[/tex]



Answer :

Let's analyze each set of compound inequalities step-by-step to determine which one has no solution.

### Set 1:
1. [tex]\(2 x - 1 < x - 4\)[/tex]
2. [tex]\(-4 x + 3 > 2 x - 9\)[/tex]

Solution for Inequality 1:
[tex]\[ 2 x - 1 < x - 4 \implies x < -3 \][/tex]

Solution for Inequality 2:
[tex]\[ -4 x + 3 > 2 x - 9 \implies -6 x > -12 \implies x < 2 \][/tex]

Combined Solution:
Combining [tex]\(x < -3\)[/tex] and [tex]\(x < 2\)[/tex] gives:
[tex]\[ x < -3 \][/tex]

### Set 2:
1. [tex]\(2 x - 5 \leq 3 x + 5\)[/tex]
2. [tex]\(-4 x - 1 > 2 x + 3\)[/tex]

Solution for Inequality 1:
[tex]\[ 2 x - 5 \leq 3 x + 5 \implies -x \leq 10 \implies x \geq -10 \][/tex]

Solution for Inequality 2:
[tex]\[ -4 x - 1 > 2 x + 3 \implies -6 x > 4 \implies x < -\frac{2}{3} \][/tex]

Combined Solution:
Combining [tex]\(x \geq -10\)[/tex] and [tex]\(x < -\frac{2}{3}\)[/tex] gives:
[tex]\[ -10 \leq x < -\frac{2}{3} \][/tex]

### Set 3:
1. [tex]\(-3 x - 1 \leq 2 x + 9\)[/tex]
2. [tex]\(-4 x + 3 > -2 x + 11\)[/tex]

Solution for Inequality 1:
[tex]\[ -3 x - 1 \leq 2 x + 9 \implies -5 x \leq 10 \implies x \geq -2 \][/tex]

Solution for Inequality 2:
[tex]\[ -4 x + 3 > -2 x + 11 \implies -2 x > 8 \implies x < -4 \][/tex]

Combined Solution:
Combining [tex]\(x \geq -2\)[/tex] and [tex]\(x < -4\)[/tex] gives:
There is no overlap between [tex]\(x \geq -2\)[/tex] and [tex]\(x < -4\)[/tex], hence, this set has no solution.

### Set 4:
1. [tex]\(3(x + 1) \geq 2(x + 5)\)[/tex]
2. [tex]\(-2(x + 1) \leq 3(x + 1)\)[/tex]

Solution for Inequality 1:
[tex]\[ 3(x + 1) \geq 2(x + 5) \implies 3x + 3 \geq 2x + 10 \implies x \geq 7 \][/tex]

Solution for Inequality 2:
[tex]\[ -2(x + 1) \leq 3(x + 1) \implies -2x - 2 \leq 3x + 3 \implies -5x \leq 5 \implies x \geq -1 \][/tex]

Combined Solution:
Combining [tex]\(x \geq 7\)[/tex] and [tex]\(x \geq -1\)[/tex] gives:
[tex]\[ x \geq 7 \][/tex]

### Conclusion:
The third set of compound inequalities has no solution, as there is no value of [tex]\(x\)[/tex] that satisfies both inequalities simultaneously.

So, the correct answer is:
[tex]\[ \boxed{3} \][/tex]