Question

943. Решите неравенство: 1) 3х – 7 < x+ 1; 3) 1- x < 2x – 5; 2) 2 + x > 8 – х; 4) 2x +1 > x+6; 5) 4x + 2 > 3x + 1; 6) 6 x + 1 < 2x + 9. ​

Answer 1

Ответ:

решение на фото

Пошаговое объяснение:

там второй и третий пример местами поменялись..

Answer 2

1) 3x - 7 < x + 1 First, let's get all the x terms on one side of the inequality: 3x - x < 7 + 1 2x < 8 Now, divide both sides by 2: x < 4

2) 1 - x < 2x - 5 First, let's get all the x terms on one side of the inequality: 1 + 5 < 2x + x 6 < 3x Now, divide both sides by 3: x > 2

3) 2 + x > 8 - x First, let's get all the x terms on one side of the inequality: x + x > 8 - 2 2x > 6 Now, divide both sides by 2: x > 3

4) 2x + 1 > x + 6 First, let's get all the x terms on one side of the inequality: 2x - x > 6 - 1 x > 5

5) 4x + 2 > 3x + 1 First, let's get all the x terms on one side of the inequality: 4x - 3x > 1 - 2 x > -1

6) 6 + x < 2x + 9 First, let's get all the x terms on one side of the inequality: x - 2x < 9 - 6 -x < 3 Now, multiply both sides by -1 (don't forget to reverse the inequality sign when multiplying or dividing by a negative number): x > -3

So, the solutions to the inequalities are: 1) x < 4 2) x > 2 3) x > 3 4) x > 5 5) x > -1 6) x > -3