Question

1) (3(x - 1) < x - 3,5(x+3) > 2x + 3;2) 2(y – 2)>3y +1,5(y +1)<4y + 3;​

Answer 1

Пошаговое объяснение:

1) Система неравенств:

3(x-1)<x-3; 3x-3<x-3; 3x-x<3-3; 2x<0; x<0

5(x+3)>2x+3; 5x+15>2x+3; 5x-2x>3-15; x>-12/3; x>-4

x принадлежит (-4; 0).

2) Система неравенств:

2(y-2)>3y+1; 2y-4>3y+1; 2y-3y>4+1; -y>5; y<-5

5(y+1)<4y+3; 5y+5<4y+3; 5y-4y<3-5; y<-2

y принадлежит (-∞; -5)

Answer 2

To solve these inequalities, you'll need to simplify and isolate the variable, just like you would with equations. Here's how to solve each inequality:

1. 3(x - 1) < x - 3 and 5(x + 3) > 2x + 3:

Let's start with the first inequality: 3(x - 1) < x - 3

First, distribute the 3 on the left side: 3x - 3 < x - 3

Now, subtract x from both sides: 2x - 3 < -3

Add 3 to both sides: 2x < 0

Divide by 2 (since it's a positive number, we don't need to change the direction of the inequality): x < 0

Now, let's move on to the second inequality: 5(x + 3) > 2x + 3

First, distribute the 5 on the left side: 5x + 15 > 2x + 3

Subtract 2x from both sides: 3x + 15 > 3

Subtract 15 from both sides: 3x > -12

Divide by 3 (since it's a positive number, we don't need to change the direction of the inequality): x > -4

So, the solution to the system of inequalities is: x < 0 and x > -4 (or -4 < x < 0)

2. 2(y - 2) > 3y + 1 and 5(y + 1) < 4y + 3:

Let's start with the first inequality: 2(y - 2) > 3y + 1

First, distribute the 2 on the left side: 2y - 4 > 3y + 1

Subtract 3y from both sides: -y - 4 > 1

Add 4 to both sides: -y > 5

Multiply both sides by -1 (since we want to isolate y and reverse the inequality): y < -5

Now, let's move on to the second inequality: 5(y + 1) < 4y + 3

First, distribute the 5 on the left side: 5y + 5 < 4y + 3

Subtract 4y from both sides: y + 5 < 3

Subtract 5 from both sides: y < -2

So, the solution to the system of inequalities is: y < -5 and y < -2 (or y < -5, since that is the more restrictive condition)