Solve the inequality x < 4 - |2x + 1|.

In order to deal with the modulus sign, we must take account of 2 possible cases:

Case 1: |2x + 1| = (2x +1). In this case we can solve algebraicly, preserving the inequality sign, to get that x < 4 -(2x + 1) = 3 - 2x. Then by adding 2x to each side and dividing both sides by 3 we get x < 1.

Case 2: |2x + 1| = -(2x +1). In this case we solve algebraicly again so that x < 4 + (2x +1) = 2x + 5. Hence by subtracting a 5 and an x from each side we get x > -5.

Finally we combine the results of each case, namely that x < 1 and x > -5 to get -5 < x < 1 as our final solution.

Answered by Joe C. Maths tutor

5662 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Solve 2sec^2(x) = 3 + tan(x) for 0 < x <pi/2


What is the definite integral of 2x^2 + 4x + 1 with a lower limit of 3 and a higher limit of 6?


Rationalise the fraction : 5/(3-sqrt(2))


Find the total area enclosed between y = x^3 - x, the x axis and the lines x = 1 and x= -1 . (Why do i get 0 as an answer?)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy