1. ## inequality help

lx+10l + 3 > 6

do i subtract 3 to both sides or subtract 10? thanks

2. Originally Posted by rj2001
lx+10l + 3 > 6

do i subtract 3 to both sides or subtract 10? thanks
subtract 3 first to get |x + 10| > 3

now we must have x + 10 > 3 or -(x + 10) > 3 depending on the sign of (x + 10) with the absolute values

3. Hi rj2001,

We have $|x+10|+3>6$

Firstly as you suggested subtract $3$ from both sides which gives,

$|x+10|>3$

Since the modulus sign implies that the "stuff" inside it is "turned" positive when negative and remains positive when positive this equation means two things. The first thing it means is that,

$x+10>3$

Subtracting $10$ from both sides gives

$x>-7$

The second thing it means is

$-(x+10)>3$

Now divide through by $-1$ . When you divide through by $-1$ in an inequality the inequality sign switches round (e.g. if $-x>3$ then $x<-3$). Since this is the case we are left with,

$
x+10<-3 \implies x<-13
$

Thus we are left with

$x>-7~~,~~x<-13$