Question

Решите неравенства 1)7х+4(х-2)>6(1+3х) 2)7(у+3)-2(у+2)》2(5у+1) 3)5(u +2)+14<6-u4) 4(u+3)<3(u+2)​

Answer 1

Ответ:

1)7x+4x-8>6+18x

7x+4x-18x>6+8

-7x>14

x<-2

2)7y+21-2y-4>10y+2

7y-2y-10y>2-21+4

-15y>-15

y<1

3)5u+10+14<6-u

5u+u<6-14-10

6u<-18

u<-3

4)4u+12<3u+6

4u-3u<6-12

u<-6

Answer 2

1. 7x + 4(x - 2) > 6(1 + 3x)

Expanding the equation: 7x + 4x - 8 > 6 + 18x

Combining like terms: 11x - 8 > 6 + 18x

Moving all x terms to one side and constants to the other: 11x - 18x > 6 + 8 -7x > 14

Dividing both sides by -7 (and flipping the inequality): x < -2

So the solution to the inequality is x < -2.

2. 7(y + 3) - 2(y + 2) > 2(5y + 1)

Expanding the equation: 7y + 21 - 2y - 4 > 10y + 2

Combining like terms: 5y + 17 > 10y + 2

Moving all y terms to one side and constants to the other: 5y - 10y > 2 - 17 -5y > -15

Dividing both sides by -5 (and flipping the inequality): y < 3

So the solution to the inequality is y < 3.

3. 5(u + 2) + 14 < 6 - u

Expanding the equation: 5u + 10 + 14 < 6 - u

Combining like terms: 5u + 24 < 6 - u

Moving all u terms to one side and constants to the other: 5u + u < 6 - 24 6u < -18

Dividing both sides by 6: u < -3

So the solution to the inequality is u < -3.

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

Expanding the equation: 4u + 12 < 3u + 6

Moving all u terms to one side and constants to the other: 4u - 3u < 6 - 12 u < -6

So the solution to the inequality is u < -6.