Question

Решить уравнение 6 - I x + 3 I >2

Answer 1

-1х >2-6-31
-1х>-35 /(-1)
х<35

Answer 2

To solve the inequality $6 - |Ix + 3| > 2$, we need to consider two cases based on the absolute value expression:

Case 1: $Ix + 3$ is positive or zero ($Ix + 3 \geq 0$): In this case, the absolute value becomes $|Ix + 3| = Ix + 3$. So the inequality becomes: $6 - (Ix + 3) > 2$

Now, we can solve for $x$:

$6 - Ix - 3 > 2$

Subtract 3 from both sides:

$6 - 3 - Ix > 2$

Simplify:

$3 - Ix > 2$

Subtract 3 from both sides:

$-Ix > -1$

Divide by -1, remembering to reverse the inequality when dividing by a negative number:

$Ix < 1$

Case 2: $Ix + 3$ is negative ($Ix + 3 < 0$): In this case, the absolute value becomes $|Ix + 3| = -(Ix + 3)$. So the inequality becomes: $6 - (-(Ix + 3)) > 2$

Now, we can solve for $x$:

$6 + Ix + 3 > 2$

Subtract 9 from both sides:

$Ix > 2 - 9$

Simplify:

$Ix > -7$

In summary, we have two solutions for $x$ based on the two cases:

1. $Ix < 1$ (when $Ix + 3 \geq 0$)
2. $Ix > -7$ (when $Ix + 3 < 0$)

Thus, the solution to the inequality is $x \in (-7, 1)$.