To solve the system of inequalities [tex]\(-\frac{4}{3} x + 2 \leq 6\)[/tex] and [tex]\(-2(x + 2) \geq 4\)[/tex], we should solve each inequality separately and then find the intersection of their solutions.
### Solving the First Inequality: [tex]\(-\frac{4}{3} x + 2 \leq 6\)[/tex]
1. Subtract 2 from both sides:
[tex]\[
-\frac{4}{3} x + 2 - 2 \leq 6 - 2
\][/tex]
[tex]\[
-\frac{4}{3} x \leq 4
\][/tex]
2. Multiply both sides by [tex]\(-\frac{3}{4}\)[/tex] to solve for [tex]\(x\)[/tex]. Remember that multiplying by a negative number reverses the inequality:
[tex]\[
x \geq -3
\][/tex]
So, our solution to the first inequality is:
[tex]\[
x \geq -3
\][/tex]
### Solving the Second Inequality: [tex]\(-2(x + 2) \geq 4\)[/tex]
1. Distribute the [tex]\(-2\)[/tex]:
[tex]\[
-2x - 4 \geq 4
\][/tex]
2. Add 4 to both sides:
[tex]\[
-2x - 4 + 4 \geq 4 + 4
\][/tex]
[tex]\[
-2x \geq 8
\][/tex]
3. Divide both sides by [tex]\(-2\)[/tex], remembering to reverse the inequality:
[tex]\[
x \leq -4
\][/tex]
So, our solution to the second inequality is:
[tex]\[
x \leq -4
\][/tex]
### Finding the Intersection of the Solutions
We have:
1. [tex]\(x \geq -3\)[/tex]
2. [tex]\(x \leq -4\)[/tex]
We need to find the values of [tex]\(x\)[/tex] that satisfy both inequalities. Looking at the regions defined by these inequalities, there are no values of [tex]\(x\)[/tex] that can satisfy both [tex]\(x \geq -3\)[/tex] and [tex]\(x \leq -4\)[/tex] simultaneously.
Thus, there is no solution to the system of inequalities. The correct answer is:
[tex]\[
\text{No solution}
\][/tex]
