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Exercise 5.1 · Q13

Q.2(2x+3)−10<6(x−2)2(2x + 3) - 10 < 6(x - 2)

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This is a linear inequality in one variable. The key idea is to simplify both sides using distribution, collect like terms, isolate the variable, and then solve for xx. The final solution is x>4x > 4, meaning all real numbers greater than 4 satisfy the inequality.

Understanding Linear Inequalities

A linear inequality is like a linear equation, but instead of an equals sign, we have a relationship like <<, >>, ≤\le, or ≥\ge. The goal is the same: find the set of values for the variable that make the statement true. The core difference is that when you multiply or divide both sides by a negative number, the inequality sign flips direction. Here, we have a straightforward case with no such flip.

We start with:

2(2x+3)−10<6(x−2)2(2x + 3) - 10 < 6(x - 2)

The approach is to simplify each side independently, then bring terms together.

Step-by-Step Solution

  1. Distribute the constants On the left side, multiply 22 by each term inside (2x+3)(2x + 3):

2(2x+3)=4x+62(2x + 3) = 4x + 6

On the right side, multiply 66 by each term inside (x−2)(x - 2):

6(x−2)=6x−126(x - 2) = 6x - 12

So the inequality becomes:

4x+6−10<6x−124x + 6 - 10 < 6x - 12

  1. Combine like terms on the left The constants 66 and −10-10 combine to −4-4:

4x−4<6x−124x - 4 < 6x - 12

  1. Move variable terms to one side To isolate xx, subtract 4x4x from both sides. This keeps the coefficient of xx positive on the right, which is often easier:

4x−4−4x<6x−12−4x4x - 4 - 4x < 6x - 12 - 4x

Simplifying:

−4<2x−12-4 < 2x - 12

  1. Isolate the term with xx Add 1212 to both sides to move the constant away from the xx term:

−4+12<2x−12+12-4 + 12 < 2x - 12 + 12

Which gives:

8<2x8 < 2x

  1. Solve for xx …

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