Transcript of x - Nutley Public Schools

Warm Up
1. y + 7 < –11
Solve.
y < –18
2. 4m ≥ –12
m ≥ –3
3. 5 – 2x ≤ 17
x ≥ –6
Use interval notation to indicate the graphed numbers.
4.
(-2, 3]
5.
(-, 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
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 = 15
6x-3 = 15 or
6x = 18 or
x = 3 or
6x-3 = -15
6x = -12
x = -2
* Plug in answers to check your solutions!
Ex: Solve 2x + 7 - 3 = 8
Get 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 thAND”
“greatOR”
Ex: Solve & graph.
4 x  9  21
 Becomes an “and” problem
4x - 9 £ 21 and 4x - 9 ³ -21
4x £ 30 and 4x ³ -12
30
15
x£
and x ³ -3
3 x 
4
2
-3
7
8
Solve & graph.
3x  2  3  11
 Get absolute value by itself first.
3x  2  8
 Becomes an “or” problem
3x  2  8 or 3x  2  8
3x  10 or
3x  6
10
x
or x  2
3
-2
3
4
Solving an Absolute Value Equation
Solve
2x  5  9
x=7 or x=−2
Solving with less than
Solve 2 x  7  11.
9  x  2
Solving with greater than
Solve
3x  2  8
10
x  2 or x 
3
Example 1:
● |2x + 1| > 7
● 2x + 1 > 7 or 2x + 1 >7
● 2x + 1 >7 or 2x + 1 <-7
●
x > 3 or
This is an ‘or’ statement.
(Greator). Rewrite.
In the 2nd inequality, reverse the
inequality sign and negate the
right side value.
Solve each inequality.
x < -4
Graph the solution.
-4
3
Example 2:
This is an ‘and’ statement.
(Less thand).
● |x -5|< 3
● x -5< 3 and x -5< 3
● x -5< 3 and x -5> -3
●
●
Rewrite.
In the 2nd inequality, reverse the
inequality sign and negate the
right side value.
x < 8 and x > 2
2<x<8
Solve each inequality.
Graph the solution.
2
8
Solve the equation.
|–3 + k| = 10
This can be read as “the
distance from k to –3 is 10.”
–3 + k = 10 or –3 + k = –10
Rewrite the absolute
value as a
disjunction.
k = 13 or k = –7
Add 3 to both sides of
each equation.
Solve the equation.
Isolate the absolute-value
expression.
Rewrite the absolute value as a
disjunction.
x = 16 or x = –16
Multiply both sides of each equation
by 4.
Solve the inequality. Then graph the solution.
|–4q + 2| ≥ 10
–4q + 2 ≥ 10 or –4q + 2 ≤ –10
Rewrite the absolute
value as a disjunction.
–4q ≥ 8 or –4q ≤ –12
Subtract 2 from both
sides of each inequality.
q ≤ –2
or q ≥ 3
Divide both sides of
each inequality by –4
and reverse the
inequality symbols.
Solve the inequality. Then graph the solution.
|3x| + 36 > 12
Isolate the absolute value
as a disjunction.
|3x| > –24
Rewrite the absolute
value as a disjunction.
3x > –24 or 3x < 24
Divide both sides of each
inequality by 3.
x > –8 or x < 8
The solution is all real numbers, R.
(–∞, ∞)
–3 –2 –1
0
1
2
3
4
5
6
Solve the compound inequality. Then graph the solution set.
|2x +7| ≤ 3
2x + 7 ≤ 3 and 2x + 7 ≥ –3
Multiply both sides by 3.
Rewrite the absolute
value as a conjunction.
2x ≤ –4 and 2x ≥ –10
Subtract 7 from both
sides of each inequality.
x ≤ –2 and x ≥ –5
Divide both sides of
each inequality by 2.
Solve the compound inequality. Then graph the solution set.
|p – 2| ≤ –6
|p – 2| ≤ –6 and p – 2 ≥ 6
p ≤ –4 and p ≥ 8
Multiply both sides by –2, and
reverse the inequality symbol.
Rewrite the absolute value
as a conjunction.
Add 2 to both sides of
each inequality.
Because no real number satisfies both p ≤ –4 and
p ≥ 8, there is no solution. The solution set is ø.