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## 1.6 Linear inequalities.
* Rules for solving similar to those of linear equations, with one exception: if you multiply or divide an inequality by a negative number, the inequality reverses.

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**Example.** If
$$ 2x < 5 $$
then
$$ x < \frac{5}{2}. $$

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**Example.** If
$$ -2x < 5 $$
then
$$ x > -\frac{5}{2}. $$

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### Example 1. Solve $$4x-1>7x-8$$

| **Step** | **Reason** |
|:------------------------------------------:|:---------------------------------------:|
| $4x - 1 > 7x - 8$ | Original inequality |
| $4x-1+1>7x-8+1$ | Adding 1 to both sides |
| $4x > 7x - 7$ | Simplify $-8+1=-7$ |
| $4x-7x > 7x -7x- 7$ | Subtract $7x$ from both sides |
| $-3x > -7$ | Simplify $4x-7x=-3x$ |
| $\dfrac{-3x}{-3} <\dfrac{-7}{-3}$ | Dividing by $-3$, flip inequality |
| $x < \frac{7}{3}$ | Final result |

Interval Notation: $\left(-\infty,\dfrac{7}{3}\right)$
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### Example 2. Solve $$3x + 5 \geq 9x + 3$$

| **Step** | **Reason** |
|:------------------------------------------:|:----------------------------------------:|
| $3x + 5 \geq 9x + 3$ | Original inequality |
| $3x + 5 - 3 \geq 9x + 3 - 3$ | Subtracting 3 from both sides |
| $3x + 2 \geq 9x$ | Simplify $5 - 3 = 2$ |
| $3x - 9x + 2 \geq 9x - 9x$ | Subtract $9x$ from both sides |
| $-6x + 2 \geq 0$ | Simplify $3x - 9x = -6x$ |
| $-6x + 2 - 2 \geq 0 - 2$ | Subtract 2 from both sides |
| $-6x \geq -2$ | Simplify $2 - 2 = 0$; $0-2=-2$ |
| $\dfrac{-6x}{-6} \leq \dfrac{-2}{-6}$ | Divide by $-6$, flip inequality |
| $x \leq \dfrac{1}{3}$ | Final result |

Interval Notation: $\left(-\infty,\dfrac{1}{3}\right]$

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### Example 3. Solve $$4 + 3(2x - 1) < 7 - 2(4x + 5)$$

| **Step** | **Reason** |
|:------------------------------------------:|:----------------------------------------:|
| $4 + 3(2x - 1) < 7 - 2(4x + 5)$ | Original inequality |
| $4 + 3 \cdot 2x - 3 \cdot 1 < 7 - 2 \cdot 4x - 2 \cdot 5$ | Detailed distribution |
| $4 + 6x - 3 < 7 - 8x - 10$ | Simplify after distribution |
| $6x + 1 < -8x - 3$ | Combine like terms: $4 - 3 = 1$ and $7 - 10 = -3$ |
| $6x + 8x + 1 < -8x + 8x - 3$ | Add $8x$ to both sides |
| $14x + 1 < -3$ | Simplify $6x + 8x = 14x$ |
| $14x + 1 - 1 < -3 - 1$ | Subtract $1$ from both sides |
| $14x < -4$ | Simplify $-3 - 1 = -4$ |
| $\frac{14x}{14} < \frac{-4}{14}$ | Divide both sides by $14$ |
| $x < -\frac{2}{7}$ | Final result |

Interval Notation: $\left(-\infty,-\dfrac{2}{7}\right)$

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### Example 4. Solve $$3x + 2(7x - 2) \geq 2 + 2(x - 5)$$

| **Step** | **Reason** |
|:------------------------------------------:|:----------------------------------------:|
| $3x + 2(7x - 2) \geq 2 + 2(x - 5)$ | Original inequality |
| $3x + 2 \cdot 7x - 2 \cdot 2 \geq 2 + 2 \cdot x - 2 \cdot 5$ | Detailed distribution |
| $3x + 14x - 4 \geq 2 + 2x - 10$ | Simplify after distribution |
| $17x - 4 \geq 2x - 8$ | Combine like terms $3x+14x=17x$; $2-10=-8$ |
| $17x - 2x - 4 \geq 2x - 2x - 8$ | Subtract $2x$ from both sides |
| $15x - 4 \geq -8$ | Simplify $17x - 2x = 15x$ |
| $15x - 4 + 4 \geq -8 + 4$ | Add $4$ to both sides |
| $15x \geq -4$ | Simplify $-8 + 4 = -4$ |
| $\frac{15x}{15} \geq \frac{-4}{15}$ | Divide both sides by $15$ |
| $x \geq -\frac{4}{15}$ | Final result |

Interval Notation: $\left[ -\dfrac{4}{15},\infty\right)$

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## Domain of Square Root Functions.

- The domain of functions involving square roots often involve solving an
inequality.

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### Example 5. Find the domain of $f(x) = \sqrt{3x - 8}$

**Solution:** We can only take the square root of $3x - 8$ if it is positive or zero, that is, if it is $\geq 0$:

$$3x - 8 \geq 0$$

$$3x \geq 8$$

$$x \geq \frac{8}{3}$$

The interval associated with these $x$s is $\left[\dfrac{8}{3}, \infty\right)$.

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### Example 6. Find the domain of $$f(x) = \frac{5}{\sqrt{7 - 4x}}$$

**Solution:** For the function to be defined, $7 - 4x$ must be positive and not equal to zero, as division by zero is undefined and the square root of a negative number is not a real number. We proceed with the following steps:

| **Step** | **Reason** |
|:-------------------------------------:|:----------------------------------------------------------:|
| $7 - 4x > 0$ | Requirement for the square root to be defined |
| $7 - 7 - 4x > -7$ | Subtract $7$ from both sides |
| $-4x > -7$ | Simplify $7 - 7 = 0$; $0-7=-7$ |
| $\dfrac{-4x}{-4} < \dfrac{-7}{-4}$ | Divide both sides by $-4$ and reverse the inequality |
| $x < \dfrac{7}{4}$ | Simplify the inequality $\dfrac{-4x}{-4}=x$; $\dfrac{-7}{-4}=\dfrac{7}{4}$ |

**Conclusion:** The domain of $f(x)$, in interval notation, is given by $\left(-\infty, \dfrac{7}{4}\right)$

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### Example 7. Find the domain of $$f(x) = \sqrt{5 - 3x}$$

| **Step** | **Reason** |
|:-------------------------------------:|:----------------------------------------------------------:|
| $5 - 3x \geq 0$ | The expression inside the square root must be non-negative (i.e., $\geq 0$) |
| $5 - 5 - 3x \geq -5$ | Subtract $5$ from both sides |
| $-3x \geq -5$ | Simplify $5 - 5 = 0$ |
| $\dfrac{-3x}{-3} \leq \dfrac{-5}{-3}$ | Divide both sides by $-3$ and reverse the inequality |
| $x \leq \dfrac{5}{3}$ | Simplify the inequality |

**Conclusion:** The domain of $f(x)$, in interval notation, is $\left(-\infty, \dfrac{5}{3}\right]$.

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### Example 8. Find the domain of $$f(x) = \frac{6}{\sqrt{4x + 15}}$$

| **Step** | **Reason** |
|:-------------------------------------:|:----------------------------------------------------------:|
| $4x + 15 > 0$ | The expression inside the square root must be positive (it cannot be zero, as division by zero is undefined) |
| $4x + 15 - 15 > 0 - 15$ | Subtract $15$ from both sides |
| $4x > -15$ | Simplify $15 - 15 = 0$ |
| $\dfrac{4x}{4} > \dfrac{-15}{4}$ | Divide both sides by $4$ |
| $x > \dfrac{-15}{4}$ | Simplify the inequality |

**Conclusion:** The domain of $f(x)$, in interval notation, is $\left(-\dfrac{15}{4}, \infty\right)$.