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Exercise 5.1 · Q7

Q.3(x−1)≤2(x−3)3(x - 1) \le 2(x - 3)

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The inequality 3(x−1)≤2(x−3)3(x - 1) \le 2(x - 3) simplifies to x≤−3x \le -3, so the solution set is all real numbers less than or equal to −3-3, i.e., (−∞,−3](-\infty, -3].

The Core Idea: Linear Inequality Solutions

A linear inequality is like a linear equation, but instead of an equals sign, we have ≤\le, ≥\ge, <<, or >>. The goal is the same: isolate the variable xx on one side. The only extra rule is that if you multiply or divide both sides by a negative number, you must flip the inequality sign. Here, we only need addition and subtraction, so no sign-flipping is needed — it’s as straightforward as solving an equation.

We want to find all xx that make 3(x−1)3(x - 1) less than or equal to 2(x−3)2(x - 3). Let’s work through it.

Step-by-Step Solution

  1. Expand both sides Distribute the constants inside the parentheses:

3(x−1)=3x−33(x - 1) = 3x - 3

2(x−3)=2x−62(x - 3) = 2x - 6

So the inequality becomes:

3x−3≤2x−63x - 3 \le 2x - 6

  1. Move variable terms to one side Subtract 2x2x from both sides to bring all xx terms to the left:

3x−3−2x≤2x−6−2x3x - 3 - 2x \le 2x - 6 - 2x

This simplifies to:

x−3≤−6x - 3 \le -6

  1. Isolate xx Add 33 to both sides to get xx alone:

x−3+3≤−6+3x - 3 + 3 \le -6 + 3

Which gives:

x≤−3x \le -3

  1. Interpret the result …

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