Transcript 1.6 Solving Absolute-Value Equations and Inequalities

I can write and solve absolute-value equations and
inequalities.
Do Now (Turn on laptop to my calendar)
Write the inequality used to solve.
Success Criteria:

I can interpret complicated expressions by
viewing one or more of their parts as a
single entity

Be able to create equations and
inequalities in one variable and use them
to solve problems
Today
1. Do Now
2. Check HW #6
3. Lesson 1.6
4. HW #7
5. Complete iReady
Recall that the absolute value of a number x,
written |x|, is the distance from x to zero on the
number line. Because absolute value represents
distance without regard to direction, the absolute
value of any real number is nonnegative.
Absolute-value equations and inequalities can be
represented by compound statements. Consider
the equation |x| = 3.
The solutions of |x| = 3 are the two points that
are 3 units from zero. The solution is a
disjunction: x = –3 or x = 3.
Example 2A: Solving Absolute-Value Equations
Solve the equation.
|–3 + k| = 10
–3 + k = 10 or –3 + k = –10
k = 13 or k = –7
Solve the equation.
|6x| – 8 = 22
|6x| = 30
6x = 30 or 6x = –30
x=5
or
x = –5
Example 2B: Solving Absolute-Value Equations
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.
Example 3A: Solving Absolute-Value Inequalities
Solve the inequality. Then graph the solution.
|–4q + 2| ≥ 10
–4q + 2 ≥ 10 or –4q + 2 ≤ –10
–4q ≥ 8
q ≤ –2
–3 –2 –1
0
Rewrite the absolute
value as or.
or –4q ≤ –12
Subtract 2 from both
sides of each inequality.
or q ≥ 3
Divide both sides of
each inequality by –4
and reverse the
inequality symbols.
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2
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5
6
Check It Out! Example 4b
Solve the compound inequality. Check for
Extraneous Solutions
|3x +2| = 4x + 5
x = –3 or x =-1
Assignment #7
Pg 46 #12-24 x 3, 57- 66x3
HW#7 Pg 46 #12-24 x 3, 57- 66x3
Example 3B: Solving Absolute-Value Inequalities
Solve the inequality. Then graph the solution.
|0.5r| – 3 ≥ –3
Isolate the absolute value as
or.
|0.5r| ≥ 0
0.5r ≥ 0 or 0.5r ≤ 0
r≤0
Rewrite the absolute
value as or.
Divide both sides of each
inequality by 0.5.
or r ≥ 0
The solution is all real numbers, R.
(–∞, ∞)
–3 –2 –1
0
1
2
3
4
5
6
The solutions of |x| < 3 are the points that are
less than 3 units from zero. The solution is an
“and” statement: –3 < x < 3.
The solutions of |x| > 3 are the points that are more
than 3 units from zero. The solution is an “or”
statement:
x < –3 or x > 3.
Note: The symbol ≤ can replace <, and the rules
still apply. The symbol ≥ can replace >, and the
rules still apply.
Helpful Hint
Think: Greator inequalities involving > or ≥
symbols are disjunctions.
Think: Less thand inequalities involving < or
≤ symbols are conjunctions.
Check It Out! Example 4b
Solve the compound inequality.
–2|x +5| > 10
Divide both sides by –2, and
reverse the inequality symbol.
|x + 5| < –5
x + 5 < –5
x+5>5
x < –10 or x > 0
Rewrite the absolute value
as a conjunction.
Subtract 5 from both
sides of each inequality.
Lesson Quiz: Part I
Solve. Then graph the solution.
1. y – 4 ≤ –6 or 2y >8
–4 –3 –2 –1 0
{y|y ≤ –2 ≤ or y > 4}
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2
3
4
5
2. –7x < 21 and x + 7 ≤ 6 {x|–3 < x ≤ –1}
–4 –3 –2
–1 0
1
2
3
4
5
Solve each equation.
3. |2v + 5| = 9
2 or –7
4. |5b| – 7 = 13
+4
Lesson Quiz: Part II
Solve. Then graph the solution.
5. |1 – 2x| > 7 {x|x < –3 or x > 4}
–4 –3 –2
–1 0
1
2
3
4
5
6. |3k| + 11 > 8 R
–4 –3 –2 –1
7. –2|u + 7| ≥ 16
0
ø
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2
3
4
5
Put your answers into socrative student. Read
carefully and make sure you answer the question.
Do Now:
What’s the ERROR?
Do Now – Pick the correct
answer