# Inequalities / Interval notation

• Apr 27th 2012, 03:58 AM
jase03
Inequalities / Interval notation
Find all values of x satisfying the inequalities. Present your answer using the interval notation and illustrate it with the help of the number axis.

Hey guys.
I just wanted to know if I'm on the right track with these questions.

a) (x - 1) (2x + 3) > 0

answer: x > 1 , x > - 1.5

b) |7 + 2x| >/_ 3

answer: x>/_ 3 (x is greater than or equal to 3)

c) |2x - 7| </_ 3x - 10

answer: x >/_ -2 or x >/_ -5 (x is greater than or equal to -2 or x is greater than or equal to -5)

(Apologies for the greater than / equal to signs - I didn't know how to in put them. The underscore (_) represents the equal to sign)
• Apr 27th 2012, 04:01 AM
jase03
Re: Inequalities / Interval notation
I'm just not sure about the interval notation and how you represent it on the number axis.
• Apr 27th 2012, 04:08 AM
princeps
Re: Inequalities / Interval notation
type \leq for : $\displaystyle \leq$

and

type \geq for : $\displaystyle \geq$

• Apr 27th 2012, 04:12 AM
biffboy
Re: Inequalities / Interval notation
Two comments about part a. If x>1 it will be greater than -3/2 so that part of the answer is just x>1

Now use the fact that if both brackets are negative (that is <0) the product will be >0
• Apr 27th 2012, 04:19 AM
jase03
Re: Inequalities / Interval notation
So its not x>-1.5 just x>1 ?
• Apr 27th 2012, 04:32 AM
biffboy
Re: Inequalities / Interval notation
Yes. Any value that is greater than 1 will also be greater than -1.5 so we don't need to write x greater than -1.5 as well as x greater than 1
• Apr 28th 2012, 02:50 PM
TheIntegrator
Re: Inequalities / Interval notation
a) seems to be resolved

b) $\displaystyle |7 + 2x| \geq 3$ for this one please remember the rule $\displaystyle |x-a| \geq d => x-a \geq d$ or $\displaystyle x-a \leq -d$

following the rule

$\displaystyle => |7+2x| \geq 3$

$\displaystyle => 7+2x \geq 3$ or $\displaystyle 7+2x \leq -3$

$\displaystyle => 2x \geq 3-7$ or $\displaystyle 2x \leq -3-7$

$\displaystyle => x \geq -2$ or $\displaystyle x \leq -5$ are your solutions

c) similar situation but remember the rule $\displaystyle |x-a| \leq d => -d \leq x-a \leq d$