: {(x+2)(2−x)<(x+3)(4−x),

4

5+x

​+6

1−2x

​≥1.

​

Let's solve the first inequality:

$(x+2)(2-x) < (x+3)(4-x)$

Expanding both sides:

$2x-x^2+4-2x < 4x-x^2+9-3x$

Simplifying:

$x^2-x-3 < 0$

Factorizing:

$(x-3)(x+1) < 0$

The solution is $-1 < x < 3$.

Now let's solve the second inequality:

$45+\frac{x}{3}+\frac{61-2x}{3} \geq 1$

Simplifying:

$35+\frac{x}{3} \geq 0$

Multiplying both sides by 3:

$x \geq -105$

The solution is $x \geq -105$.

Therefore, the solution to the system of inequalities is $-1 < x < 3$ and $x \geq -105$.