Warm UpWarm Up
Solve.Solve.
1. 1. yy + 7 < –11 + 7 < –11
2. 42. 4mm ≥ –12 ≥ –12
3. 5 – 23. 5 – 2xx ≤ 17 ≤ 17
yy < –18< –18
mm ≥ –3≥ –3
xx ≥ –6 ≥ –6
Use interval notation to indicate the graphed numbers.Use interval notation to indicate the graphed numbers.
4.4.
5.5.
(-2, 3](-2, 3]
(-(-, 1], 1]
Absolute Value Equations and Inequalities
College Algebra
Absolute Value (of x)
Symbol lxl The distance x is from 0 on the number line. Always positive Ex: l-3l = 3
-4 -3 -2 -1 0 1 2-4 -3 -2 -1 0 1 2
Ex: x = 5 What are the possible values of x?
x = 5 or x = -5
To solve an absolute value equation:ax+b = c, where c > 0
To solve, set up 2 new equations, then solve each equation.
ax + b = c or ax + b = -c
** make sure the absolute value is by itself before you split to solve.
Ex: Solve 6x - 3 = 156x-3 = 15 or 6x-3 = -15
6x = 18 or 6x = -12
x = 3 or x = -2
* Plug in answers to check your solutions!
Ex: Solve 2x + 7 - 3 = 8Get the abs. value part by itself first!
2x+7 = 11
Now split into 2 parts.
2x+7 = 11 or 2x+7 = -11
2x = 4 or 2x = -18
x = 2 or x = -9
Check the solutions.
Solving Absolute Value Inequalities
1. ax+b < c, where c > 0
Becomes an “and” problem
Changes to: ax+b < c and ax+b > -c
2. ax+b > c, where c > 0
Becomes an “or” problem
Changes to: ax+b > c or ax+b < -c
““less thless thANDAND””
““greatgreatOROR””
Ex: Solve & graph.
Becomes an “and” problem
2194 x
2
153 x
-3 7 8-3 7 8
Solve & graph.
Get absolute value by itself first.
Becomes an “or” problem
11323 x
823 x
823or 823 xx63or 103 xx
2or 3
10 xx
-2 3 4-2 3 4
Solving an Absolute Value Equation
Solve 952 x
x=7 or x=−2x=7 or x=−2
Solving with less than
Solve .1172 x
29 x
Solving with greater than
Solve 823 x
3
102 xorx
Example 1:
● |2x + 1| > 7
● 2x + 1 > 7 or 2x + 1 >7
● 2x + 1 >7 or 2x + 1 <-7
● x > 3 or x < -4
This is an ‘This is an ‘or’or’ statement. statement. (Great(Greatoror). Rewrite.). Rewrite.
In the 2In the 2ndnd inequality, inequality, reversereverse the the inequality sign and inequality sign and negatenegate the the
right side value.right side value.
Solve each inequality.Solve each inequality.
Graph the solution.Graph the solution.
33-4-4
Example 2:
● |x -5|< 3
● x -5< 3 and x -5< 3
● x -5< 3 and x -5> -3
● x < 8 and x > 2
● 2 < x < 8
This is an ‘This is an ‘and’and’ statement. statement. (Less th(Less thandand).).
Rewrite.Rewrite.
In the 2nd inequality, In the 2nd inequality, reversereverse the the inequality sign and inequality sign and negatenegate the the
right side value.right side value.
Solve each inequality.Solve each inequality.
Graph the solutionGraph the solution..
88 22
Solve the equation.Solve the equation.
Rewrite the absolute Rewrite the absolute value as a value as a
disjunction.disjunction.
This can be read as This can be read as ““the the distance from k to –3 is 10.distance from k to –3 is 10.””
Add 3 to both sides of Add 3 to both sides of each equation.each equation.
|–3 + |–3 + kk| = 10| = 10
––3 + 3 + kk = 10 or –3 + = 10 or –3 + kk = –10 = –10
k k = 13 or = 13 or k k = –7= –7
Solve the equation.Solve the equation.
x x = 16 or = 16 or xx = –16 = –16
Isolate the absolute-value Isolate the absolute-value expression.expression.
Rewrite the absolute value as a Rewrite the absolute value as a disjunction.disjunction.
Multiply both sides of each equation Multiply both sides of each equation by 4.by 4.
Solve the inequality. Then graph the solution.Solve the inequality. Then graph the solution.
Rewrite the absolute Rewrite the absolute value as a disjunction.value as a disjunction.
|–4|–4qq + 2| ≥ 10 + 2| ≥ 10
––44q q + 2 ≥ 10 or –4+ 2 ≥ 10 or –4q q + 2 ≤ –10 + 2 ≤ –10
––44q q ≥ 8 or –4≥ 8 or –4q q ≤ –12 ≤ –12
Divide both sides of Divide both sides of each inequality by each inequality by –4 –4
and reverse the and reverse the inequality symbols.inequality symbols.
Subtract 2 from both Subtract 2 from both sides of each inequality.sides of each inequality.
q q ≤ –2 or ≤ –2 or q q ≥ 3 ≥ 3
Solve the inequality. Then graph the solution.Solve the inequality. Then graph the solution.
|3|3xx| + 36 > 12| + 36 > 12
Divide both sides of each Divide both sides of each inequality by 3inequality by 3..
Isolate the absolute value Isolate the absolute value as a disjunction.as a disjunction.
Rewrite the absolute Rewrite the absolute value as a disjunction.value as a disjunction.
33xx > –24 or 3 > –24 or 3x x < 24< 24
xx > –8 or > –8 or x x < 8< 8
|3|3xx| > –24| > –24
––3 –2 3 –2 –1 0 1 2 3 4 5 6–1 0 1 2 3 4 5 6
(–∞, ∞)(–∞, ∞)
The solution is The solution is allall real numbers, real numbers, R.R.
Solve the compound inequality. Then graph the solution set.Solve the compound inequality. Then graph the solution set.
|2|2x +x +7| ≤ 37| ≤ 3 Multiply both sides by 3.Multiply both sides by 3.
Subtract 7 from both Subtract 7 from both sides of each inequality.sides of each inequality.
Divide both sides of Divide both sides of each inequality by 2.each inequality by 2.
Rewrite the absolute Rewrite the absolute value as a conjunction.value as a conjunction.
22x + x + 7 ≤ 3 and 27 ≤ 3 and 2x + x + 7 ≥ –3 7 ≥ –3
22x x ≤ –4 and 2≤ –4 and 2x x ≥ –10 ≥ –10
x x ≤ –2 and ≤ –2 and x x ≥ –5≥ –5
Solve the compound inequality. Then graph the solution set.Solve the compound inequality. Then graph the solution set.
||p p – 2| ≤ –6– 2| ≤ –6Multiply both sides by Multiply both sides by –2, and –2, and reverse the inequality symbol.reverse the inequality symbol.
Add 2 to both sides of Add 2 to both sides of each inequality.each inequality.
Rewrite the absolute value Rewrite the absolute value as a conjunction.as a conjunction.
|p – |p – 2| ≤ –6 and 2| ≤ –6 and p – p – 2 ≥ 6 2 ≥ 6
p p ≤ –4 and ≤ –4 and p p ≥ 8 ≥ 8
Because no real number satisfies both Because no real number satisfies both p p ≤ –4 and≤ –4 andp ≥ p ≥ 8, there is 8, there is no solutionno solution. The solution set is ø.. The solution set is ø.
