(2-y) (3+y) ≤ (4+y) (6-y)

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вопрос

Let's expand both sides of the inequality to simplify it:

Left side:
(2-y) (3+y) = 6 - y^2

Right side:
(4+y) (6-y) = 24 - 4y + 6y - y^2 = 24 + 2y - y^2

Now we have:
6 - y^2 ≤ 24 + 2y - y^2

To simplify further, we can cancel out the -y^2 terms:
6 ≤ 24 + 2y

Subtract 24 from both sides:
-18 ≤ 2y

Divide by 2:
-9 ≤ y

Therefore, the solution to the inequality is y ≥ -9.