Question

1) (2x + 7>1,x-3<1;2) 3y < 21,4-y >0;​

Answer 1

Ответ:1)

{2x+7 >1

{x - 3<1

{2x > 1 - 7

{x < 1 + 3

{x > (- 6)/2

{x < 4

{ x> - 3

{ x< 4

x∈ (-3;4)

2)

{3y<21

{4-y>0

{y < 21/3

{4 > y

{y<7

{y< 4

y∈(-∞,4 )

3)

{4х +9 > -15

{2-x<5

{4x> -15 -9

{2-5 < x

{4x > -24

{x >-3

{x>-24/4

{x>-3

{x>-6

{x>-3

x∈(-3; ∞ )

4)

{2x+3>x-1

{5x-22<x+2

{2x -x > -1-3

{5x-x< 2+22

{x>-4

{4x<24

{x>-4

{x<6

x∈(-4 , 6 )

5)

{7x+9<2x-1

{4+11x>9x-14

{7x-2x<-1-9

{11x-9x>-14-4

{5x<-10

{2x> - 18

{x<-2        

{x> - 9

x∈ (-9, - 2)

6)

{x>0

{x-5>2x+1           ⇒  -5-1>2x-x  

{x>0

{x< -6

Пошаговое объяснение:

Answer 2

It looks like you've provided two sets of inequalities. Let's solve each set step by step:

Set 1:

1. $2x + 7 > 1$
2. $x - 3 < 1$

For the first inequality: $2x + 7 > 1$ Subtract 7 from both sides: $2x > -6$ Divide both sides by 2 (remember to reverse the inequality since you're dividing by a negative number): $x > -3$

For the second inequality: $x - 3 < 1$ Add 3 to both sides: $x < 4$

So, the solution for the first set of inequalities is: $-3 < x < 4$.

Set 2:

1. $3y < 21$
2. $4 - y > 0$

For the first inequality: $3y < 21$ Divide both sides by 3: $y < 7$

For the second inequality: $4 - y > 0$ Subtract 4 from both sides: $-y > -4$ Multiply both sides by -1 (remember to reverse the inequality since you're multiplying by a negative number): $y < 4$

So, the solution for the second set of inequalities is: $y < 7$ and $y < 4$. Since the second inequality is more restrictive, the overall solution is $y < 4$.

To summarize, the solutions are:

1. Set 1: $-3 < x < 4$
2. Set 2: $y < 4$