Question

Which point is NOT part of the solution of the inequality y < |3x| + 1? (2, –9) (–4, 12) (3, 15) (0, 0)

Answer 1

All options satisfies the given inequality

Solution:

Given that,

$y < | 3x | + 1$

Here, modulus means that, the absolute value of a number is never negative

Option A

$(2, -9)$

Substitute x = 2 and y = -9 in given

$-9 < | 3(2)| + 1\\\\-9 < 6 + 1\\\\-9 < 7$

-9 is less than 7 is true

This forms a solution to given inequality

Option B

(-4, 12)

Substitute x = -4 and y = 12 in given

$12 < | 3(12)| + 1\\\\12 < 36 + 1\\\\\12 < 37$

12 is less than 37 is true

This forms a solution to given inequality

Option C

(3, 15)

Substitute x = 3 and y = 15 in given

$15 < | 3(15)| + 1\\\\15 < 45+1\\\\15< 46$

15 less than 46 is true

This forms a solution to given inequality

Option D

(0, 0)

Substitute x = 0 and y = 0 in given

$0 < | 3(0) | + 1\\\\0 < 0 + 1\\\\0 < 1$

0 is less than 1

This forms a solution to given inequality