I don't get this question. (1 Viewer)

[Dragonmaster262]

2 < l2x + 3l
2^2 < (2x + 3)^2
4 < 4x^2 + 12x + 9
0 < 4x^2 + 12x + 5
0 < (2x + 5)(2x + 1)
x < -5/2 and x > -1/2

l2x + 3l < 11
(2x + 3)^2 < 11^2
4x^2 + 12x + 9 < 121
4x^2 + 12x - 112 < 0
4 (x^2 + 3x - 24) < 0
4 (x + 5)(x - 3) < 0
-5 < x < 3

So when they intersect those are your solutions:
-5 < x < -5/2 AND -1/2 < x < 3 hope it helped : )

Yeah, I already knew that method and how to do that but I htought you were supposed to use some other method because it said solve simultaneously. I thought 'simultaneously' always involved elimination/substitution.

 

[drowning_pocoyo]

Yeah LOL for simulateously equating inequalities its kinda hard to do that =="

 

[Dragonmaster262]

Yeah LOL for simulateously equating inequalities its kinda hard to do that =="

Is it possible to do it though? And if so, how? This question is part of an assignment for school and I don't want to get it wrong.

 

[gurmies]

The solutions that have been provided ARE simultaneously solving. They are taking one case and combining with another case SIMULTANEOUSLY in order for both to be true for the original question...