Algebra-2

We can solve ∣4x − 1∣+∣2x + 3∣ = 7 by considering the critical points of each absolute value function, which are x = 1/4 and x = − 3/2 .

Case 1: x < − 3/2 (both terms negative):

−(4x − 1) − (2x + 3) = 7 ⟹ −4x + 1 − 2x − 3 = 7 ⟹ −6x − 2 = 7 ⟹ −6x = 9 ⟹ x = −3/2

Case 2: -3/2 ≤ x < 1/4 (first term negative, second term positive):

−(4x − 1) + (2x + 3) = 7 ⟹ −4x + 1 + 2x + 3 = 7 ⟹ −2x + 4 = 7 ⟹ −2x = 3 ⟹ x = −3/2

Case 3: x ≥ 1/4 (both terms positive):

(4x − 1) + (2x + 3) = 7 ⟹ 4x − 1 + 2x ...