Number of real solutions of |x - 3| + 2|x + 1| = 4 is:

(1) 0
(2) 1
(3) 2
(4) infinite

The equation ∣ x − 3∣ + 2∣ x + 1∣ = 4 has one real solution at x = − 1 . Therefore, the answer is option (2) 1. Further analysis shows that both other intervals yield no valid solutions.
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Answered by Anonymous | 2025-07-04

To solve the equation ∣ x − 3∣ + 2∣ x + 1∣ = 4 , we should consider the different cases that arise due to the absolute value expressions. Absolute values create piecewise functions that change based on the sign of the expression inside them. For this equation, we'll consider the behavior of ∣ x − 3∣ and ∣ x + 1∣ over different intervals:

Case 1: When x ≥ 3


Both ∣ x − 3∣ = x − 3 and ∣ x + 1∣ = x + 1 .

The equation becomes ( x − 3 ) + 2 ( x + 1 ) = 4 .

Simplifying this gives: x − 3 + 2 x + 2 = 4 or 3 x − 1 = 4 .

Solving for x , we get 3 x = 5 , so x = 3 5 ​ .

However, x = 3 5 ​ does not fall within this interval ≥ 3 , so there are no solutions here.



Case 2: When − 1 ≤ x < 3


Here, ∣ x − 3∣ = 3 − x and ∣ x + 1∣ = x + 1 .

The equation becomes ( 3 − x ) + 2 ( x + 1 ) = 4 .

Simplifying, 3 − x + 2 x + 2 = 4 or x + 5 = 4 .

Solving for x , we get x = − 1 .

x = − 1 lies within the interval − 1 ≤ x < 3 , so this is a valid solution.



Case 3: When x < − 1


Both ∣ x − 3∣ = 3 − x and ∣ x + 1∣ = − ( x + 1 ) = − x − 1 .

The equation becomes ( 3 − x ) + 2 ( − x − 1 ) = 4 .

Simplifying gives 3 − x − 2 x − 2 = 4 or − 3 x + 1 = 4 .

Solving for x , − 3 x = 3 , thus x = − 1 .

x = − 1 does not fall within this interval x < − 1 , so there are no solutions here.


Thus, the only valid solution is x = − 1 , which is from Case 2.
Therefore, the number of real solutions is 1 .
The correct multiple-choice option is (2) 1 .

Answered by EmmaGraceJohnson | 2025-07-07