Answer :
Let's solve the inequality step by step:
Given the inequality is
[tex]\[ \frac{2 - x}{2} + 3x \leq \frac{x}{3} - 1 \][/tex]
1. Simplify the left-hand side (LHS) and right-hand side (RHS):
First, let's clear the fraction on the LHS:
[tex]\[ \frac{2 - x}{2} + 3x \][/tex]
This can be rewritten by distributing the division on the first term:
[tex]\[ \frac{2}{2} - \frac{x}{2} + 3x = 1 - \frac{x}{2} + 3x \][/tex]
2. Combine like terms on the LHS:
Combine [tex]\(-\frac{x}{2}\)[/tex] and [tex]\(3x\)[/tex]:
[tex]\[ 1 - \frac{x}{2} + 3x = 1 - \frac{x}{2} + \frac{6x}{2} = 1 + \frac{5x}{2} \][/tex]
So, the LHS simplifies to:
[tex]\[ 1 + \frac{5x}{2} \][/tex]
The RHS remains:
[tex]\[ \frac{x}{3} - 1 \][/tex]
3. Combine LHS and RHS into a single inequality:
Now we have:
[tex]\[ 1 + \frac{5x}{2} \leq \frac{x}{3} - 1 \][/tex]
4. Eliminate the fractions by finding a common denominator:
The common denominator for [tex]\(2\)[/tex] and [tex]\(3\)[/tex] is [tex]\(6\)[/tex]:
[tex]\[ \left(1 + \frac{5x}{2}\right) \leq \frac{x}{3} - 1 \][/tex]
To eliminate the fractions, multiply every term by [tex]\(6\)[/tex]:
[tex]\[ 6 \left(1 + \frac{5x}{2}\right) \leq 6 \left(\frac{x}{3} - 1\right) \][/tex]
Expand these terms:
[tex]\[ 6 \cdot 1 + 6 \cdot \frac{5x}{2} \leq 6 \cdot \frac{x}{3} - 6 \cdot 1 \][/tex]
Simplify further:
[tex]\[ 6 + 15x \leq 2x - 6 \][/tex]
5. Isolate the variable [tex]\(x\)[/tex] on one side:
Move all terms involving [tex]\(x\)[/tex] to one side of the inequality and constants to the other:
[tex]\[ 15x - 2x \leq -6 - 6 \][/tex]
Simplify:
[tex]\[ 13x \leq -12 \][/tex]
6. Solve for [tex]\(x\)[/tex]:
Divide both sides by 13 to isolate [tex]\(x\)[/tex]:
[tex]\[ x \leq -\frac{12}{13} \][/tex]
Therefore, the solution to the inequality is:
[tex]\[ x \leq -\frac{12}{13} \][/tex]
Given the inequality is
[tex]\[ \frac{2 - x}{2} + 3x \leq \frac{x}{3} - 1 \][/tex]
1. Simplify the left-hand side (LHS) and right-hand side (RHS):
First, let's clear the fraction on the LHS:
[tex]\[ \frac{2 - x}{2} + 3x \][/tex]
This can be rewritten by distributing the division on the first term:
[tex]\[ \frac{2}{2} - \frac{x}{2} + 3x = 1 - \frac{x}{2} + 3x \][/tex]
2. Combine like terms on the LHS:
Combine [tex]\(-\frac{x}{2}\)[/tex] and [tex]\(3x\)[/tex]:
[tex]\[ 1 - \frac{x}{2} + 3x = 1 - \frac{x}{2} + \frac{6x}{2} = 1 + \frac{5x}{2} \][/tex]
So, the LHS simplifies to:
[tex]\[ 1 + \frac{5x}{2} \][/tex]
The RHS remains:
[tex]\[ \frac{x}{3} - 1 \][/tex]
3. Combine LHS and RHS into a single inequality:
Now we have:
[tex]\[ 1 + \frac{5x}{2} \leq \frac{x}{3} - 1 \][/tex]
4. Eliminate the fractions by finding a common denominator:
The common denominator for [tex]\(2\)[/tex] and [tex]\(3\)[/tex] is [tex]\(6\)[/tex]:
[tex]\[ \left(1 + \frac{5x}{2}\right) \leq \frac{x}{3} - 1 \][/tex]
To eliminate the fractions, multiply every term by [tex]\(6\)[/tex]:
[tex]\[ 6 \left(1 + \frac{5x}{2}\right) \leq 6 \left(\frac{x}{3} - 1\right) \][/tex]
Expand these terms:
[tex]\[ 6 \cdot 1 + 6 \cdot \frac{5x}{2} \leq 6 \cdot \frac{x}{3} - 6 \cdot 1 \][/tex]
Simplify further:
[tex]\[ 6 + 15x \leq 2x - 6 \][/tex]
5. Isolate the variable [tex]\(x\)[/tex] on one side:
Move all terms involving [tex]\(x\)[/tex] to one side of the inequality and constants to the other:
[tex]\[ 15x - 2x \leq -6 - 6 \][/tex]
Simplify:
[tex]\[ 13x \leq -12 \][/tex]
6. Solve for [tex]\(x\)[/tex]:
Divide both sides by 13 to isolate [tex]\(x\)[/tex]:
[tex]\[ x \leq -\frac{12}{13} \][/tex]
Therefore, the solution to the inequality is:
[tex]\[ x \leq -\frac{12}{13} \][/tex]
