1.3 Solve the following linear inequalities [solve, plot \& write the soloution in an interva) 1.3.1 $$ \frac{1}{6} x+\frac{1}{4}(x-4) \leq \frac{2}{3} $$ $$ 1.3 .23 x-29 $$ or $$ -3 x+5 \geq-2 $$ $$ 1.3 .33 x+27 $$ and $$ -4 x+6 \geq-1 $$ $$ 1.3 .4-3 x+5 \geq-2 $$.

Question

1.3 Solve the following linear inequalities [solve, plot \& write the soloution in an interva) 1.3.1 $$ \frac{1}{6} x+\frac{1}{4}(x-4) \leq \frac{2}{3} $$ $$ 1.3 .23 x-2>9 $$ or $$ -3 x+5 \geq-2 $$ $$ 1.3 .33 x+2>7 $$ and $$ -4 x+6 \geq-1 $$ $$ 1.3 .4-3 x+5 \geq-2 $$.

UpStudy AI · Accepted Answer

Let's solve each of the linear inequalities step by step.

### 1.3.1: Solve $$ \frac{1}{6} x + \frac{1}{4}(x - 4) \leq \frac{2}{3} $$

First, we will simplify the left side of the inequality:

1. Distribute $$ \frac{1}{4} $$ in $$ \frac{1}{4}(x - 4) $$:
   $$
   \frac{1}{4}x - 1
   $$

2. Combine the terms:
   $$
   \frac{1}{6}x + \frac{1}{4}x - 1 \leq \frac{2}{3}
   $$

3. To combine $$ \frac{1}{6}x $$ and $$ \frac{1}{4}x $$, we need a common denominator (which is 12):
   $$
   \frac{2}{12}x + \frac{3}{12}x - 1 \leq \frac{2}{3}
   $$
   $$
   \frac{5}{12}x - 1 \leq \frac{2}{3}
   $$

4. Add 1 to both sides:
   $$
   \frac{5}{12}x \leq \frac{2}{3} + 1
   $$
   $$
   \frac{5}{12}x \leq \frac{5}{3}
   $$

5. Multiply both sides by $$ \frac{12}{5} $$:
   $$
   x \leq 4
   $$

### 1.3.2: Solve $$ 3x - 2 > 9 $$ or $$ -3x + 5 \geq -2 $$

1. For $$ 3x - 2 > 9 $$:
   $$
   3x > 11 \implies x > \frac{11}{3}
   $$

2. For $$ -3x + 5 \geq -2 $$:
   $$
   -3x \geq -7 \implies x \leq \frac{7}{3}
   $$

Thus, the solution for this inequality is:
$$
x > \frac{11}{3} \quad \text{or} \quad x \leq \frac{7}{3}
$$

### 1.3.3: Solve $$ x + 2 > 7 $$ and $$ -4x + 6 \geq -1 $$

1. For $$ x + 2 > 7 $$:
   $$
   x > 5
   $$

2. For $$ -4x + 6 \geq -1 $$:
   $$
   -4x \geq -7 \implies x \leq \frac{7}{4}
   $$

Since this is an "and" condition, there is no solution because $$ x $$ cannot be both greater than 5 and less than or equal to $$ \frac{7}{4} $$.

### 1.3.4: Solve $$ -3x + 5 \geq -2 $$

1. Rearranging gives:
   $$
   -3x \geq -7 \implies x \leq \frac{7}{3}
   $$

### Summary of Solutions

1. **1.3.1**: $$ x \leq 4 $$
2. **1.3.2**: $$ x > \frac{11}{3} $$ or $$ x \leq \frac{7}{3} $$
3. **1.3.3**: No solution
4. **1.3.4**: $$ x \leq \frac{7}{3} $$

Now, let's plot the inequalities and write the solutions in interval notation.

### Plotting and Interval Notation

1. **1.3.1**: $$ (-\infty, 4] $$
2. **1.3.2**: $$ \left( \frac{11}{3}, \infty \right) \cup \left( -\infty, \frac{7}{3} \right] $$
3. **1.3.3**: No solution
4. **1.3.4**: $$ (-\infty, \frac{7}{3}] $$

Now, I will plot the inequalities.
It seems that the system does not currently support plotting. However, I can summarize the solutions in interval notation for each inequality:

### Summary of Solutions in Interval Notation

1. **1.3.1**: $$ (-\infty, 4] $$
2. **1.3.2**: $$ \left( \frac{11}{3}, \infty \right) \cup \left( -\infty, \frac{7}{3} \right] $$
3. **1.3.3**: No solution
4. **1.3.4**: $$ (-\infty, \frac{7}{3}] $$

If you need any further assistance or specific details about any of the inequalities, feel free to ask!
Solutions in interval notation: 1. **1.3.1**: $$ (-\infty, 4] $$ 2. **1.3.2**: $$ \left( \frac{11}{3}, \infty \right) \cup \left( -\infty, \frac{7}{3} \right] $$ 3. **1.3.3**: No solution 4. **1.3.4**: $$ (-\infty, \frac{7}{3}] $$

1.3 Solve the following linear inequalities [solve, plot & write the soloution in an interva)
1.3.1
or
and
.

Solución de inteligencia artificial de Upstudy

Responder

Solución

1.3.1: Solve

1.3.2: Solve or

1.3.3: Solve and

1.3.4: Solve

Summary of Solutions

Plotting and Interval Notation

Summary of Solutions in Interval Notation

Beyond the Answer

preguntas relacionadas

Latest Algebra Questions

1.3 Solve the following linear inequalities [solve, plot & write the soloution in an interva) 1.3.1 or and .