Question

952. 1) 3 - 2(u - 1) > 8 +u; 2) 5(u + 2) + 14 < 6 - u;3) (3 + 8u) > 6,25+u ; 4) 4(u + 3) < 3(u + 2); 5) 3 (2u+1) > 5(u - 1); 6) ​

Answer 1

Ответ:

5(u +2) +14<6-u

5u+10+14<6-u

5u+u<6+10-14

6u>2

u>3

3-2(u-1)>8+u

3-2u-2>8+u

-2u-u>8-3+2

-3u>7

u<- 3/7

(3+8u)>6,25+u

3+8u>6,25+u

8u-u>6,25-3

7u>3,25

u< 2целыз 2/13

4(u+3)<3(u+2)

4u+12<3u+6

4u-3u<6-12

1u<-6

u> - 1/6

3(2u+1)>5(u-1)

6u+3>5u-5

6u-5u>-5-3

1u>-8

u> -1/8

пж дайте лучший ответ пожалуйста

Answer 2

Let's solve each of these inequalities step by step:

1. $3 - 2(u - 1) > 8 + u$

First, distribute the -2 on the left side:

$3 - 2u + 2 > 8 + u$

Combine like terms:

$5 - 2u > 8 + u$

Now, add 2u to both sides to isolate the u terms on one side:

$5 > 8 + 3u$

Subtract 8 from both sides:

$-3 > 3u$

Finally, divide by 3 (remember to reverse the inequality sign when dividing by a negative number):

$u < -1$

So the solution is $u < -1$.

2. $5(u + 2) + 14 < 6 - u$

First, distribute the 5 on the left side:

$5u + 10 + 14 < 6 - u$

Combine like terms:

$5u + 24 < 6 - u$

Add u to both sides to isolate the u terms on one side:

$6u + 24 < 6$

Subtract 24 from both sides:

$6u < -18$

Finally, divide by 6:

$u < -3$

So the solution is $u < -3$.

3. $(3 + 8u) > (6.25 + u)$

First, subtract u from both sides:

$3 + 7u > 6.25$

Next, subtract 3 from both sides:

$7u > 3.25$

Finally, divide by 7:

$u > \frac{3.25}{7}$

So the solution is $u > \frac{13}{28}$.

4. $4(u + 3) < 3(u + 2)$

First, distribute both sides:

$4u + 12 < 3u + 6$

Subtract 3u from both sides to isolate the u terms on one side:

$u + 12 < 6$

Subtract 12 from both sides:

$u < -6$

So the solution is $u < -6$.

5. $3(2u + 1) > 5(u - 1)$

First, distribute both sides:

$6u + 3 > 5u - 5$

Subtract 5u from both sides to isolate the u terms on one side:

$u + 3 > -5$

Subtract 3 from both sides:

$u > -8$

So the solution is $u > -8$.

If you have any more inequalities to solve or need further clarification, feel free to ask!