Question

|1-2x|+|3x+2|+|x|=5

Answer 1

Ответ:

x=1/3

Объяснение:

По правилам модуля

1+2x+3x+2+x=5

6x+3=5

x=2/6

x=1/3

Answer 2

To solve the equation |1-2x| + |3x+2| + |x| = 5, we need to consider different cases based on the absolute value expressions.

Case 1: (1-2x) ≥ 0, (3x+2) ≥ 0, and (x) ≥ 0 In this case, all three absolute value expressions are positive, so we can rewrite the equation without the absolute value signs: (1 - 2x) + (3x + 2) + x = 5 Now, solve for x: 1 - 2x + 3x + 2 + x = 5 4x + 3 = 5 4x = 5 - 3 4x = 2 x = 2/4 x = 1/2

However, we need to check if this solution satisfies the conditions for Case 1: 1 - 2x ≥ 0 1 - 2(1/2) ≥ 0 1 - 1 ≥ 0 0 ≥ 0 (True)

3x + 2 ≥ 0 3(1/2) + 2 ≥ 0 3/2 + 2 ≥ 0 7/2 ≥ 0 (True)

x ≥ 0 (True)

Since all three conditions are satisfied, x = 1/2 is a valid solution for Case 1.

Case 2: (1-2x) < 0, (3x+2) ≥ 0, and (x) ≥ 0 In this case, the first absolute value expression is negative, and the other two are positive, so we can rewrite the equation as follows: -(1 - 2x) + (3x + 2) + x = 5 Now, solve for x: -1 + 2x + 3x + 2 + x = 5 6x + 1 = 5 6x = 5 - 1 6x = 4 x = 4/6 x = 2/3

However, we need to check if this solution satisfies the conditions for Case 2: 1 - 2x < 0 1 - 2(2/3) < 0 1 - 4/3 < 0 -1/3 < 0 (True)

3x + 2 ≥ 0 3(2/3) + 2 ≥ 0 2 + 2 ≥ 0 4 ≥ 0 (True)

x ≥ 0 (True)

Since all three conditions are satisfied, x = 2/3 is a valid solution for Case 2.

Case 3: (1-2x) < 0, (3x+2) < 0, and (x) ≥ 0 In this case, all three absolute value expressions are negative, so we can rewrite the equation as follows: -(1 - 2x) - (3x + 2) + x = 5 Now, solve for x: -1 + 2x - 3x - 2 + x = 5 -x - 3 = 5 -x = 5 + 3 -x = 8 x = -8 (but x ≥ 0 condition is violated)

Since x cannot be negative in this case, there are no valid solutions for Case 3.

Case 4: (1-2x) < 0, (3x+2) < 0, and (x) < 0 In this case, all three absolute value expressions are negative, so we can rewrite the equation as follows: -(1 - 2x) - (3x + 2) - x = 5 Now, solve for x: -1 + 2x - 3x - 2 - x = 5 -2x - 3 = 5 -2x = 5 + 3 -2x = 8 x = -8/2 x = -4 (but x < 0 condition is violated)

Since x cannot be negative in this case, there are no valid solutions for Case 4.

In summary, the valid solutions to the equation |1-2x| + |3x+2| + |x| = 5 are x = 1/2 and x = 2/3.