Sure! Let's solve each inequality step-by-step to determine the correct answers.
### 1. Solve [tex]\( -2(2x + 5) \leq -7(x + 4) \)[/tex]
First, expand both sides:
[tex]\[
-2(2x + 5) = -4x - 10
\][/tex]
[tex]\[
-7(x + 4) = -7x - 28
\][/tex]
So the inequality becomes:
[tex]\[
-4x - 10 \leq -7x - 28
\][/tex]
Next, add [tex]\( 7x \)[/tex] to both sides:
[tex]\[
3x - 10 \leq -28
\][/tex]
Then, add 10 to both sides:
[tex]\[
3x \leq -18
\][/tex]
Finally, divide both sides by 3:
[tex]\[
x \leq -6
\][/tex]
Thus, [tex]\[ x \leq -6 \][/tex].
### 2. Solve [tex]\( \frac{4x - 6}{2} \geq 3x - 1 \)[/tex]
Simplify the fraction:
[tex]\[
2x - 3 \geq 3x - 1
\][/tex]
Subtract [tex]\( 3x \)[/tex] from both sides:
[tex]\[
- x - 3 \geq -1
\][/tex]
Add 3 to both sides:
[tex]\[
- x \geq 2
\][/tex]
Multiply both sides by -1, and reverse the inequality sign:
[tex]\[
x \leq -2
\][/tex]
Thus, [tex]\[ x \leq -2 \][/tex].
### 3. Solve [tex]\( 5(x - 3) \leq 9(x + 1) \)[/tex]
First, expand both sides:
[tex]\[
5(x - 3) = 5x - 15
\][/tex]
[tex]\[
9(x + 1) = 9x + 9
\][/tex]
So the inequality becomes:
[tex]\[
5x - 15 \leq 9x + 9
\][/tex]
Next, subtract [tex]\( 5x \)[/tex] from both sides:
[tex]\[
-15 \leq 4x + 9
\][/tex]
Then, subtract 9 from both sides:
[tex]\[
-24 \leq 4x
\][/tex]
Finally, divide both sides by 4:
[tex]\[
-6 \leq x
\][/tex]
or equivalently
[tex]\[
x \geq -6
\][/tex]
Thus, [tex]\[ x \geq -6 \][/tex].
### 4. Solve [tex]\( 3(x - 4) + 5 \geq 2x - 9 \)[/tex]
Expand:
[tex]\[
3(x - 4) + 5 = 3x - 12 + 5 = 3x - 7
\][/tex]
So the inequality becomes:
[tex]\[
3x - 7 \geq 2x - 9
\][/tex]
Subtract [tex]\( 2x \)[/tex] from both sides:
[tex]\[
x - 7 \geq -9
\][/tex]
Add 7 to both sides:
[tex]\[
x \geq -2
\][/tex]
Thus, [tex]\[ x \geq -2 \][/tex].
### Matching the inequalities to the solutions:
[tex]\[
\begin{array}{ll}
-2(2 x+5) \leq-7(x+4) & x \leq-6 \\
\frac{4 x-6}{2} \geq 3 x-1 & x \leq-2 \\
5(x-3) \leq 9(x+1) & x \geq-6 \\
3(x-4)+5 \geq 2 x-9 & x \geq-2 \\
\end{array}
\][/tex]
Thus, the correct matches are:
[tex]\[
\begin{array}{ll}
-2(2 x+5) \leq-7(x+4) & x \leq-6 \\
\frac{4 x-6}{2} \geq 3 x-1 & x \leq-2 \\
5(x-3) \leq 9(x+1) & x \geq-6 \\
3(x-4)+5 \geq 2 x-9 & x \geq-2 \\
\end{array}
\][/tex]
