EIV Chapter 1 q.4

[Guest]

Q. 4 |6+x| = 2x+1
The question says solve for all unknows.

The explnations says solve this for two cases 1. which the expression is positive and one in which it is negative. Notice that the solution arrived in Case 2 is not valid since it violates the condition that x be less that -6.

I do not understand where is the condition stated or from where do i derive the condition that x be less than -6.

Can you please help me how did i get to this condition x be less than -6

regards
chandni

[rfernandez]

Q. 4 |6+x| = 2x+1

Here's another way to go about this problem. In general, we know that |a| = a if a is positive. Also, we know that |a| = -a if a is negative. (Another way to say the latter sentence is that |a| equals the opposite of a if a is negative.)

So, what this tells us is that we can essentially "drop" the absolute value bars in an absolute value equation if we account for two cases: 1) when the expression within the absolute value bars is positive and 2) when the expression within the absolute value bars is negative. Let's use this approach for this problem:

|6+x| = 2x+1

Case 1
We know that we can rewrite |6+x| as (6+x) as long as (6+x)>0. When is that true? Solve the inequality:
(6+x)>0
x>-6

So, assuming that x>-6, |6+x| = 2x+1 can be rewritten as 6+x = 2x+1.
6+x = 2x+1
5+x = 2x
5 = x

This is a valid solution because x=5 holds under the initial condition above that x>-6.

Case 2
We know that we can rewrite |6+x| as -(6+x) as long as (6+x)<0. When is that true? Solve the inequality:
-(6+x)>0
-6-x>0
-x>6
x<-6

So, assuming that x<-6, |6+x| = 2x+1 can be rewritten as -(6+x) = 2x+1.
-(6+x) = 2x+1
-6-x = 2x+1
-7-x = 2x
-7 = 3x
-7/3 = x

This is NOT a valid solution, however, because x=-7/3 violates the initial condition that x<-6.

[johnnyboa]

This is NOT a valid solution, however, because x=-7/3 violates the initial condition that x<-6.

I was stuck on this same problem, too. Glad to find the explanation here!

[JonathanSchneider]

: )