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## Main Question or Discussion Point

I was hoping someone could give a little more insight, or perhaps enlighten me to a better way of approaching solving these seemingly simple Algebra 2 inequalities.

I did some google searching but I was not able to find the answers I seek.

The problem came up when a friend of mine had an equation that looked like this:

|x-3| > -2 + |2x + 8|

My method of solving involved setting up 4 equalities:

1) x-3 = -2 + 2x + 8

2) -(x-3) = -2 + 2x + 8

3) x-3 = -2 - (2x + 8)

4) -(x-3) = -2 - (2x + 8)

These yielded the following values for x:

1) x= -9

2) x = -1

3) x = -7/3

4) x = -13

With each of these x values I then went back and plugged into the original inequality. My goal was to check for equality.

Only 2) and 4) resulted in equality so I threw the other answers out.

Once I had these two solutions I tested the points 0, -5, and -20 in the original inequality. And since -5 was the only value that yielded a true statement, I declared the solution:

(technically I had checked -7/3 and -9, checking -5 was unnecessary I realize now)

-13 < x < -1

As I understand things, when you have:

|x| = A

You need to solve x = A and x = -A

When you have:

|x| = |y|

You need to solve x = y, x = -y, -x = y and -x = -y, but because x = -y, -x = y are the same equation and x = y, -x = -y are the same equation you only need to solve two equations.

When you have:

|x| = A + |y|

You need to solve x = A + y, x = A - y, -x = A + y, and -x = A - y, since these are all different, you need to solve all four. Essentially the prior case (|x| = |y|) is a special case of this, just two of the equations you need to solve happen to be the same.

Thank you in advance.

I did some google searching but I was not able to find the answers I seek.

The problem came up when a friend of mine had an equation that looked like this:

|x-3| > -2 + |2x + 8|

**Question 1**: What is the generally accepted best way to solve this inequality.My method of solving involved setting up 4 equalities:

1) x-3 = -2 + 2x + 8

2) -(x-3) = -2 + 2x + 8

3) x-3 = -2 - (2x + 8)

4) -(x-3) = -2 - (2x + 8)

These yielded the following values for x:

1) x= -9

2) x = -1

3) x = -7/3

4) x = -13

With each of these x values I then went back and plugged into the original inequality. My goal was to check for equality.

Only 2) and 4) resulted in equality so I threw the other answers out.

Once I had these two solutions I tested the points 0, -5, and -20 in the original inequality. And since -5 was the only value that yielded a true statement, I declared the solution:

(technically I had checked -7/3 and -9, checking -5 was unnecessary I realize now)

-13 < x < -1

**Question 2**: Can someone give me more information on why there are extraneous solutions?As I understand things, when you have:

|x| = A

You need to solve x = A and x = -A

When you have:

|x| = |y|

You need to solve x = y, x = -y, -x = y and -x = -y, but because x = -y, -x = y are the same equation and x = y, -x = -y are the same equation you only need to solve two equations.

When you have:

|x| = A + |y|

You need to solve x = A + y, x = A - y, -x = A + y, and -x = A - y, since these are all different, you need to solve all four. Essentially the prior case (|x| = |y|) is a special case of this, just two of the equations you need to solve happen to be the same.

**Question 3**: Is there a more rigorous way of handling these equations? Should I be using |x| = Sqrt(x^2) here?Thank you in advance.

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