Question

A) 2 |x-2| =2;b) 12|3x-5|=-20.​

Answer 1

Ответ:

Пошаговое объяснение:

2|x-2|=2; |x-2|=2/2; |x-2|=1

1) x-2=1; x=1+2; x₁=3

2) x-2=-1; x=-1+2; x₂=1

Ответ: 1; 3.

12|3x-5|=-20; |3x-5|=-20/12 - решений нет, модуль не может быть отрицательным.

Answer 2

To solve these absolute value equations, you need to consider two cases for each equation: one where the expression inside the absolute value is positive and another where it is negative. Then, you solve for the variable in each case.

A) 2 |x - 2| = 2

Case 1: x - 2 is positive 2(x - 2) = 2 2x - 4 = 2 2x = 2 + 4 2x = 6 x = 6/2 x = 3

Case 2: x - 2 is negative 2(-x + 2) = 2 -2x + 4 = 2 -2x = 2 - 4 -2x = -2 x = (-2)/(-2) x = 1

So, for equation A, the solutions are x = 3 and x = 1.

B) 12 |3x - 5| = -20

Case 1: 3x - 5 is positive 12(3x - 5) = -20 36x - 60 = -20 36x = -20 + 60 36x = 40 x = 40/36 x = 10/9

Case 2: 3x - 5 is negative

In this case, the absolute value of a negative number cannot equal a negative number. So there are no solutions in this case.

So, for equation B, the only solution is x = 10/9.