Solving Inequalities with One Absolute Value (II)

Solving Inequalities with One Absolute Value (II)

 

These steps explain how to solve an inequality with one absolute value and one variable when there are variable terms both inside and outside the absolute value sign.

 

Step 1: Isolate the absolute value

 

Exactly like Part I of this lesson.

 

3 - | x – 3 | > - x           è - | x – 3 | > - x – 3               è | x – 3 | < x + 3

2 | y² – 4 | > 6y è | y² – 4 | > 3y

 

Step 2: Is the expression in the absolute value negative or nonnegative?

 

This can be a difficult step for complicated expressions, but for the simple polynomials we are usually working with, there is an easy way.  Solve the related equation where the expression is equal to zero.  This will tell you where the graph crosses zero.  Mark the variable in this related equation so you don’t confuse it with the original problem.

 

x₀ – 3 = 0         è x₀ = 3

y₀² – 4 = 0        è y₀ = -2 or 2

 

Divide the domain into sections using the roots above and mark each section as positive or negative by plugging in a test value.

 

x – 3    è negative where x < 3, nonnegative where x >= 3

y² – 4   è nonnegative where y < -2, negative where -2 < y < 2, nonnegative where y > 2

 

Step 3: Split the problem into sections

 

Write a separate inequality for the negative section and the positive sections.  Mark the variables so you don’t mix the two up.  To make the positive inequality, just drop the absolute value sign.  To make the negative inequality, multiply the contents of the absolute value sign by -1.  If step 2 showed that your expression was always positive or always negative, you only need to make one inequality.

 

x_p - 3 < x_p + 3

3 – x_n < x_n + 3

y_p² – 4 > 3y_p

4 – y_n² > 3y_n

 

Step 4: Solve using basic algebra

 

x_p - 3 < x_p + 3              è -3 < 3                     è true (everything)

3 – x_n < x_n + 3              è 0 < 2x_n                    è x_n > 0

 

y_p² – 4 > 3y_p                 è y_p² – 3y_p – 4 > 0     è (y_p – 4) (y_p + 1) > 0  è y_p > 4 or y_p < -1

4 – y_n² > 3y_n                 è - y_n² – 3y_n + 4 > 0   è (1 – y_n) (y_n + 4) > 0  è -4 < y_n < 1

 

Step 5: Take the intersection for each section

 

Now we combine the domain notes from step 2 with the partial solutions from step 4 by taking the intersection for each section.  If you got an always-true result in step 4, take the entire corresponding section.  If you got an always-false result, take nothing for that section.  At this point we can drop our special markings for the sections.  Be extra careful when the section has an OR in it --- you may want to sketch a graph in two colors.

 

x_p >=3 AND true         è x >= 3

x_n < 3 AND x_n > 0       è 0 < x < 3

 

(y_p <= -2 OR y_p >= 2) AND (y_p > 4 OR y_p < -1)         è y <= -2 OR y > 4

(-2 < y < 2) AND (-4 < y_n < 1)                                    è -2 < y < 1

 

Step 6: Take the union of the two sections

 

0 < x < 3 OR x>=3                              è x > 0

y <= -2 OR y > 4 OR -2 < y < 1          è y < 1 or y > 4