Question

The set of all x satisfying the inequality |3x+4| ≤ 7 is

• A. \([-1,\frac{11}{3}]\)
• B. \([\frac{4}{3},\frac{7}{3}]\)
• C. \([\frac{-11}{3},1]\)
• D. \([\frac{-4}{3},\frac{7}{3}]\)
• E. \([\frac{-4}{3},\frac{11}{3}]\)

Answer 1

The Correct Option is C

We are given the inequality: \[ |3x + 4| \leq 7 \] This can be rewritten as two inequalities: \[ -7 \leq 3x + 4 \leq 7 \] Now, subtract 4 from all parts of the inequality: \[ -7 - 4 \leq 3x \leq 7 - 4 \] \[ -11 \leq 3x \leq 3 \] Next, divide through by 3: \[ \frac{-11}{3} \leq x \leq 1 \] Thus, the set of all \( x \) satisfying the inequality is: \[{\left[ -\frac{11}{3}, 1 \right]} \]

The correct option is (C) : \([\frac{-11}{3},1]\)

Answer 2

We are given the inequality \(|3x+4| \le 7\).

This inequality is equivalent to:

\(-7 \le 3x + 4 \le 7\)

Subtract 4 from all parts of the inequality:

\(-7 - 4 \le 3x \le 7 - 4\)

\(-11 \le 3x \le 3\)

Divide all parts of the inequality by 3:

\(\frac{-11}{3} \le x \le \frac{3}{3}\)

Which simplifies to:

\(\frac{-11}{3} \le x \le 1\)

Thus, the set of all x satisfying the inequality is \([\frac{-11}{3}, 1]\).