Question

Помогите решить зарание спасибо вам огромное. 4){ 2х + 3 ≥ х - 1 5х - 22 ≤ х + 2 6){х ≥ 0 х - 5 > 2х + 1

Answer 1

2х-х》-1-3
х》-4
Ж: [-4;+шексіздік)

5x-22《x+2
5x-x《2+22
4x《24
x《24:4
x《6
Ж: (-шексіз;6]

х》0
Ж: [0;+шексіз)

х-5>2х+1
х-2х>1+5
-х>6
х<-6

Ж:(-шексіз;-6)

Answer 2

4) 2x + 3 ≥ x - 1 To solve this inequality, we can start by subtracting x from both sides: 2x - x + 3 ≥ -1 This simplifies to: x + 3 ≥ -1 Next, we can subtract 3 from both sides: x + 3 - 3 ≥ -1 - 3 This simplifies to: x ≥ -4

5x - 22 ≤ x + 2 To solve this inequality, we can start by subtracting x from both sides: 5x - x - 22 ≤ 2 This simplifies to: 4x - 22 ≤ 2 Next, we can add 22 to both sides: 4x - 22 + 22 ≤ 2 + 22 This simplifies to: 4x ≤ 24 Finally, we can divide both sides by 4 to solve for x: x ≤ 6

6) x ≥ 0 This inequality is already solved for x.

x - 5 > 2x + 1 To solve this inequality, we can start by subtracting x from both sides: x - x - 5 > 2x - x + 1 This simplifies to: -5 > x + 1 Next, we can subtract 1 from both sides: -5 - 1 > x + 1 - 1 This simplifies to: -6 > x or x < -6

So, the solution to the system of inequalities is: x ≥ -4 and x ≤ 6 and x < -6

I hope this helps! Let me know if you have any further questions.