Answer :
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]
