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Difficulty: Very hardSolving Linear Inequalities

What is the maximum integer value of xx that satisfies the inequality 3(2−3x)4−2(2x+5)3≥1−x2+76\frac{3(2 - 3x)}{4} - \frac{2(2x + 5)}{3} \geq \frac{1 - x}{2} + \frac{7}{6}?

Answer: -2

Answer

The maximum integer value of xx that satisfies the inequality is −2-2.
Multiplying the inequality by the common denominator 12 and simplifying yields the inequality −37x≥42-37x \ge 42. Dividing by −37-37 requires reversing the inequality sign, which gives x≤−4237x \le -\frac{42}{37}. The value of −4237-\frac{42}{37} is approximately −1.135-1.135. The largest integer less than or equal to −1.135-1.135 is −2-2.

Step-by-Step Solution

1
Multiply both sides of the inequality by the least common multiple of the denominators (12).
9(2−3x)−8(2x+5)≥6(1−x)+149(2 - 3x) - 8(2x + 5) \geq 6(1 - x) + 14
This eliminates the fractions and simplifies the algebraic manipulation.
2
Expand the terms on both sides of the inequality.
18−27x−16x−40≥6−6x+1418 - 27x - 16x - 40 \geq 6 - 6x + 14
Expanding the terms allows us to combine like terms.
3
Combine the constant and variable terms on each side.
−43x−22≥20−6x-43x - 22 \geq 20 - 6x
This simplifies the inequality to a standard linear form.
4
Add 6x6x and 2222 to both sides to isolate the variable term on the left.
−37x≥42-37x \geq 42
Grouping variable terms on one side and constant terms on the other prepares for the final division.
5
Divide both sides by −37-37 and reverse the direction of the inequality sign.
x≤−4237x \leq -\frac{42}{37}
Dividing an inequality by a negative number requires flipping the inequality sign.
6
Find the largest integer that is less than or equal to −4237-\frac{42}{37}.
−2-2
Since −4237≈−1.135-\frac{42}{37} \approx -1.135, the integers less than or equal to this value are −2,−3,−4,…-2, -3, -4, \dots, of which −2-2 is the greatest.

Key Concept

Solving multi-step linear inequalities with rational coefficients, applying the inequality sign-flip rule, and finding boundary integer conditions.

Alternative Method

Instead of clearing the fractions first, you can group all terms containing xx on one side and the constant terms on the other side by finding a common denominator for only the variables and only the constants. However, clearing the fractions first is generally less prone to errors.
Estimated Time:2m 0s
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Topics

Solving Linear Inequalities

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