IDNLearn.com connects you with a community of experts ready to answer your questions. Ask anything and receive prompt, well-informed answers from our community of experienced experts.
Sagot :
Sure, let's solve the given compound inequality step-by-step:
[tex]\[ \frac{x-1}{4} - x < \frac{4x+2}{3} \leq \frac{5x+3}{2} \][/tex]
### Step 1: Solve the first inequality \( \frac{x-1}{4} - x < \frac{4x+2}{3} \)
1. Rewrite the inequality to have a common denominator:
[tex]\[ \frac{x-1}{4} - x < \frac{4x+2}{3} \][/tex]
First, let's get a common denominator on the left side. We can rewrite \( x \) as \( \frac{4x}{4} \):
[tex]\[ \frac{x-1}{4} - \frac{4x}{4} < \frac{4x+2}{3} \][/tex]
This simplifies to:
[tex]\[ \frac{x - 1 - 4x}{4} < \frac{4x + 2}{3} \][/tex]
[tex]\[ \frac{-3x - 1}{4} < \frac{4x + 2}{3} \][/tex]
2. Clear the fractions by multiplying through by the common denominator \( 12 \):
[tex]\[ 12 \cdot \frac{-3x - 1}{4} < 12 \cdot \frac{4x + 2}{3} \][/tex]
[tex]\[ 3(-3x - 1) < 4(4x + 2) \][/tex]
[tex]\[ -9x - 3 < 16x + 8 \][/tex]
3. Solve the resulting linear inequality:
Add \( 9x \) to both sides:
[tex]\[ -3 < 25x + 8 \][/tex]
Subtract 8 from both sides:
[tex]\[ -11 < 25x \][/tex]
Divide both sides by 25:
[tex]\[ x > -\frac{11}{25} \][/tex]
So, the solution to the first inequality is:
[tex]\[ x > -\frac{11}{25} \][/tex]
### Step 2: Solve the second inequality \( \frac{4x+2}{3} \leq \frac{5x+3}{2} \)
1. Clear the fractions by multiplying through by the common denominator, \( 6 \):
[tex]\[ 6 \cdot \frac{4x+2}{3} \leq 6 \cdot \frac{5x+3}{2} \][/tex]
[tex]\[ 2(4x + 2) \leq 3(5x + 3) \][/tex]
[tex]\[ 8x + 4 \leq 15x + 9 \][/tex]
2. Solve the resulting linear inequality:
Subtract \( 8x \) from both sides:
[tex]\[ 4 \leq 7x + 9 \][/tex]
Subtract 9 from both sides:
[tex]\[ -5 \leq 7x \][/tex]
Divide both sides by 7:
[tex]\[ -\frac{5}{7} \leq x \][/tex]
Or equivalently:
[tex]\[ x \geq -\frac{5}{7} \][/tex]
### Step 3: Combine the solutions of both inequalities
The solution to \( \frac{x-1}{4} - x < \frac{4x+2}{3} \) is \( x > -\frac{11}{25} \).
The solution to \( \frac{4x+2}{3} \leq \frac{5x+3}{2} \) is \( x \geq -\frac{5}{7} \).
To find the overall solution of the compound inequality, we need the intersection of both solutions:
[tex]\[ x > -\frac{11}{25} \quad \text{and} \quad x \geq -\frac{5}{7} \][/tex]
These intervals are:
[tex]\[ (-\infty, -\frac{11}{25}) \quad \text{and} \quad [-\frac{5}{7}, \infty) \][/tex]
So, the overlapping interval is:
[tex]\[ (-\frac{11}{25}, \infty) \][/tex]
Thus, the solutions to the given compound inequality are:
[tex]\[ (-\frac{11}{25}, \infty) \quad \text{and} \quad [-\frac{5}{7}, \infty) \][/tex]
In interval notation, we can represent the final combined solution as:
[tex]\[ \left( -\frac{11}{25}, \infty \right) \cap \left[ -\frac{5}{7}, \infty \right) \][/tex]
Therefore, simplifying, the final solution is:
[tex]\[ x \in \left(-\frac{11}{25}, \infty\right) \cap \left[-\frac{5}{7}, \infty\right) = [-\frac{5}{7}, \infty) \][/tex]
Thus, the solution set to the original compound inequality is the union of both intervals, which is:
[tex]\[ \left( -\frac{11}{25}, \infty \right) \quad \text{and} \quad \left[ -\frac{5}{7}, \infty \right) \][/tex]
[tex]\[ \frac{x-1}{4} - x < \frac{4x+2}{3} \leq \frac{5x+3}{2} \][/tex]
### Step 1: Solve the first inequality \( \frac{x-1}{4} - x < \frac{4x+2}{3} \)
1. Rewrite the inequality to have a common denominator:
[tex]\[ \frac{x-1}{4} - x < \frac{4x+2}{3} \][/tex]
First, let's get a common denominator on the left side. We can rewrite \( x \) as \( \frac{4x}{4} \):
[tex]\[ \frac{x-1}{4} - \frac{4x}{4} < \frac{4x+2}{3} \][/tex]
This simplifies to:
[tex]\[ \frac{x - 1 - 4x}{4} < \frac{4x + 2}{3} \][/tex]
[tex]\[ \frac{-3x - 1}{4} < \frac{4x + 2}{3} \][/tex]
2. Clear the fractions by multiplying through by the common denominator \( 12 \):
[tex]\[ 12 \cdot \frac{-3x - 1}{4} < 12 \cdot \frac{4x + 2}{3} \][/tex]
[tex]\[ 3(-3x - 1) < 4(4x + 2) \][/tex]
[tex]\[ -9x - 3 < 16x + 8 \][/tex]
3. Solve the resulting linear inequality:
Add \( 9x \) to both sides:
[tex]\[ -3 < 25x + 8 \][/tex]
Subtract 8 from both sides:
[tex]\[ -11 < 25x \][/tex]
Divide both sides by 25:
[tex]\[ x > -\frac{11}{25} \][/tex]
So, the solution to the first inequality is:
[tex]\[ x > -\frac{11}{25} \][/tex]
### Step 2: Solve the second inequality \( \frac{4x+2}{3} \leq \frac{5x+3}{2} \)
1. Clear the fractions by multiplying through by the common denominator, \( 6 \):
[tex]\[ 6 \cdot \frac{4x+2}{3} \leq 6 \cdot \frac{5x+3}{2} \][/tex]
[tex]\[ 2(4x + 2) \leq 3(5x + 3) \][/tex]
[tex]\[ 8x + 4 \leq 15x + 9 \][/tex]
2. Solve the resulting linear inequality:
Subtract \( 8x \) from both sides:
[tex]\[ 4 \leq 7x + 9 \][/tex]
Subtract 9 from both sides:
[tex]\[ -5 \leq 7x \][/tex]
Divide both sides by 7:
[tex]\[ -\frac{5}{7} \leq x \][/tex]
Or equivalently:
[tex]\[ x \geq -\frac{5}{7} \][/tex]
### Step 3: Combine the solutions of both inequalities
The solution to \( \frac{x-1}{4} - x < \frac{4x+2}{3} \) is \( x > -\frac{11}{25} \).
The solution to \( \frac{4x+2}{3} \leq \frac{5x+3}{2} \) is \( x \geq -\frac{5}{7} \).
To find the overall solution of the compound inequality, we need the intersection of both solutions:
[tex]\[ x > -\frac{11}{25} \quad \text{and} \quad x \geq -\frac{5}{7} \][/tex]
These intervals are:
[tex]\[ (-\infty, -\frac{11}{25}) \quad \text{and} \quad [-\frac{5}{7}, \infty) \][/tex]
So, the overlapping interval is:
[tex]\[ (-\frac{11}{25}, \infty) \][/tex]
Thus, the solutions to the given compound inequality are:
[tex]\[ (-\frac{11}{25}, \infty) \quad \text{and} \quad [-\frac{5}{7}, \infty) \][/tex]
In interval notation, we can represent the final combined solution as:
[tex]\[ \left( -\frac{11}{25}, \infty \right) \cap \left[ -\frac{5}{7}, \infty \right) \][/tex]
Therefore, simplifying, the final solution is:
[tex]\[ x \in \left(-\frac{11}{25}, \infty\right) \cap \left[-\frac{5}{7}, \infty\right) = [-\frac{5}{7}, \infty) \][/tex]
Thus, the solution set to the original compound inequality is the union of both intervals, which is:
[tex]\[ \left( -\frac{11}{25}, \infty \right) \quad \text{and} \quad \left[ -\frac{5}{7}, \infty \right) \][/tex]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.
