18. Fill in the blank with an appropriate inequality a. If $$ x5 x-3 $$, then $$ 3 x $$

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18. Fill in the blank with an appropriate inequality a. If $$ x<-2 $$, then $$ 1-\mathrm{x} $$ b. If $$ \frac{1}{2}\left(3 x-\frac{2}{3}\right) \leq x+5 $$, then $$ x+\frac{7}{3} $$ c. If $$ 3 x-1>5 x-3 $$, then $$ 3 x $$

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Sure, let's solve each part step by step. ### a. If $$ x < -2 $$, then $$ 1 - x $$ ___ **Solution:** Given $$ x < -2 $$. To find the inequality for $$ 1 - x $$: $$ 1 - x = 1 + (-x) $$ Since $$ x < -2 $$, multiplying by $$-1$$ (and reversing the inequality) gives: $$ -x > 2 $$ Adding 1 to both sides: $$ 1 - x > 3 $$ **Answer:** $$ \text{If } x < -2, \text{ then } 1 - x > 3. $$ --- ### b. If $$ \frac{1}{2}\left(3x - \frac{2}{3}\right) \leq x + 5 $$, then $$ x + \frac{7}{3} $$ ___ **Solution:** Start with the inequality: $$ \frac{1}{2}\left(3x - \frac{2}{3}\right) \leq x + 5 $$ Multiply both sides by 2 to eliminate the fraction: $$ 3x - \frac{2}{3} \leq 2x + 10 $$ Subtract $$ 2x $$ from both sides: $$ x - \frac{2}{3} \leq 10 $$ Add $$ \frac{2}{3} $$ to both sides: $$ x \leq \frac{32}{3} $$ Now, add $$ \frac{7}{3} $$ to both sides: $$ x + \frac{7}{3} \leq \frac{32}{3} + \frac{7}{3} = \frac{39}{3} = 13 $$ **Answer:** $$ \text{If } \frac{1}{2}\left(3x - \frac{2}{3}\right) \leq x + 5, \text{ then } x + \frac{7}{3} \leq 13. $$ --- ### c. If $$ 3x - 1 > 5x - 3 $$, then $$ 3x $$ ___ **Solution:** Start with the inequality: $$ 3x - 1 > 5x - 3 $$ Subtract $$ 3x $$ from both sides: $$ -1 > 2x - 3 $$ Add 3 to both sides: $$ 2 > 2x $$ Divide both sides by 2: $$ 1 > x \quad \text{or} \quad x < 1 $$ Multiply both sides by 3 (since 3 is positive, the inequality direction remains the same): $$ 3x < 3 $$ **Answer:** $$ \text{If } 3x - 1 > 5x - 3, \text{ then } 3x < 3. $$ --- ### Summary of All Parts a. **If** $$ x < -2 $$, **then** $$ 1 - x > 3 $$. b. **If** $$ \frac{1}{2}\left(3x - \frac{2}{3}\right) \leq x + 5 $$, **then** $$ x + \frac{7}{3} \leq 13 $$. c. **If** $$ 3x - 1 > 5x - 3 $$, **then** $$ 3x < 3 $$. a. If $$ x < -2 $$, then $$ 1 - x > 3 $$. b. If $$ \frac{1}{2}\left(3x - \frac{2}{3}\right) \leq x + 5 $$, then $$ x + \frac{7}{3} \leq 13 $$. c. If $$ 3x - 1 > 5x - 3 $$, then $$ 3x < 3 $$.

Fill in the blank with an appropriate inequality
a. If , then
b. If , then
c. If , then

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