Question

3)(4(x+1) < 9-2x,2(x+5) < 1–x;​

Answer 1

Ответ:

хзхзхзхз я не знаю

Объяснение:

хзхз

Answer 2

It seems like you have a system of inequalities that you'd like to solve. Let's tackle each inequality separately and find the solutions:

1. $4(x+1) < 9-2x$:

First, distribute the $4$ on the left side:

$4x + 4 < 9 - 2x$

Now, add $2x$ to both sides:

$6x + 4 < 9$

Subtract $4$ from both sides:

$6x < 5$

Finally, divide by $6$ (since $6$ is positive, the inequality direction doesn't change):

$x < \frac{5}{6}$

So, the solution for this inequality is: $x < \frac{5}{6}$.

2. $2(x+5) < 1 - x$:

First, distribute the $2$ on the left side:

$2x + 10 < 1 - x$

Now, add $x$ to both sides:

$3x + 10 < 1$

Subtract $10$ from both sides:

$3x < -9$

Finally, divide by $3$ (since $3$ is positive, the inequality direction doesn't change):

$x < -3$

So, the solution for this inequality is: $x < -3$.

To find the solution for the system of inequalities, we need to find the intersection of the individual solutions. Since both inequalities have $x$ values less than a certain threshold, the common solution is the intersection of the two ranges:

The solutions for the system of inequalities are: $x < \frac{5}{6}$ and $x < -3$.

This means any value of $x$ that satisfies both of these conditions will be a solution to the system of inequalities. In other words, the solution set is all values of $x$ that are less than both $\frac{5}{6}$ and $-3$.