Question

Solve the inequality 3(x – 2) + 1 ≥ x + 2(x + 2).

Answer 1

Answer:

no solution

Step-by-step explanation:

3(x – 2) + 1 ≥ x + 2(x + 2).

Distribute

3x – 6 + 1 ≥ x + 2x + 4

Combine like terms

3x -5 ≥ 3x +4

Subtract 3x from each side

-5 ≥ 4

This is never true so there is no solution

Answer 2

By solving the given inequality "$3(x – 2) + 1 \geq x + 2(x + 2)$", we get "$-5 \geq 4$".

The given inequality is:

• $3(x - 2) + 1 \geq x + 2(x + 2)$

By solving the brackets, we get

→  $3x-6+1 \geq x+2x+4$

→        $3x-5 \geq 3x+4$

By subtracting "3x" from both sides, we get

→ $3x-5-3x \geq 3x+4-3x$

→               $-5 \geq 4$

Thus the above answer is right.

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