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DO NOW
10/13/2015
Solve.
1. y + 7 < –11
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]
Objective
Solve equations in one variable that contain
absolute-value expressions.
The absolute-value of a number is that numbers
distance from zero on a number line. For
example, |–5| = 5.
5 units
6 5 4 3 2 1
0
1
2
3
4
5
6
Both 5 and –5 are a distance of 5 units from 0, so
both 5 and –5 have an absolute value of 5.
To write this using algebra, you would write |x| = 5.
This equation asks, “What values of x have an
absolute value of 5?” The solutions are 5 and –5.
Notice this equation has two solutions.
Example 1A: Solving Absolute-Value Equations
Solve each equation. Check your answer.
|x| = 12
|x| = 12
Case 1
x = 12
Think: What numbers are 12 units
from 0?
Case 2
x = –12
Rewrite the equation as two
cases.
The solutions are 12 and –12.
Check
|x| = 12
|12|
12
12
12
|x| = 12
|12| 12
12
12
Example 1B: Solving Absolute-Value Equations
3|x + 7| = 24
Since |x + 7| is multiplied by 3,
divide both sides by 3 to undo
the multiplication.
Think: What numbers are 8
|x + 7| = 8
units from 0?
Rewrite the equations as two
Case 2
Case 1
cases. Since 7 is added to x
x + 7 = 8 x + 7 = –8
subtract 7 from both sides
– 7 –7
–7 –7
of each equation.
x
=1 x
= –15
The solutions are 1 and –15.
2-Ext
Solving Absolute-Value Equations
Example 1B Continued
3|x + 7| = 24
The solutions are 1 and –15.
Check
3|x + 7| = 24
3|1 + 7|
24
3|15 + 7|
24
3|8|
24
3|8|
24
3(8)
24
3(8)
24
Holt Algebra 1
3|x + 7| = 24
24
24
24
24
Example 2a
Solve each equation. Check your answer.
|x| – 3
|x| – 3
+3
|x|
Case 1
x =7
x=7
=4
=4
+3
=7
Since 3 is subtracted from |x|, add 3 to
both sides.
Think: what numbers are 7 units from 0?
Case 2
-x = 7
Rewrite the case 2 equation by
multiplying by –1 to change
–1(–x) = –1(7)
the minus x to a positive..
x = –7
The solutions are 7 and –7.
Example 2a Continued
Solve the equation. Check your answer.
|x|  3 = 4
The solutions are 7 and 7.
Check
|x| 3 = 4
|x| 3 = 4
|7|  3
4
| 7|  3
4
7 3
4
73
4
4
4
4
4
Try It Out!
Solve the equation. Check your answer.
|x  2| = 8
|x  2| = 8
Case 1
x2= 8
+2 +2
x = 10
Think: what numbers are 8
units from 0?
Case 2
Rewrite the equations as two
cases. Since 2 is subtracted
x  2 = 8
+2 +2
from x add 2 to both sides
of each equation.
x = 6
The solutions are 10 and 6.
Check It Out!
Solve the equation. Check your answer.
|x  2| = 8
The solutions are 10 and 6.
Check
|x 2| = 8
|x 2| = 8
|10 2|
8
| 6 + (2)|
8
10 2|
8
6+2
8
8
8
8
8
Not all absolute-value equations have two
solutions. If the absolute-value expression
equals 0, there is one solution. If an equation
states that an absolute-value is negative, there
are no solutions.
Example 3A: Special Cases of Absolute-Value
Equations
Solve the equation. Check your answer.
8 = |x + 2|  8
8 = |x + 2|  8
+8
+8
0 = |x +2|
Since 8 is subtracted from |x + 2|,
add 8 to both sides to undo the
subtraction.
0= x+2
2
2
2 = x
There is only one case. Since 2
is added to x, subtract 2 from
both sides to undo the
addition.
Example 3A Continued
Solve the equation. Check your answer.
8 = |x +2|  8
Solution is x = 2
Check
8 =|x + 2|  8
8
|2 + 2|  8
8
|0|  8
8
08
8
8 
To check your solution,
substitute 2 for x in
your original equation.
Example 2B: Special Cases of Absolute-Value
Equations
Solve the equation. Check your answer.
3 + |x + 4| = 0
3 + |x + 4| = 0
3
3
|x + 4| = 3
Since 3 is added to |x + 4|,
subtract 3 from both sides to
undo the addition.
Absolute values cannot be
negative.
This equation has no solution.
Remember!
Absolute value must be nonnegative because
it represents distance.
Check It Out! Example 2a
Solve the equation. Check your answer.
2  |2x  5| = 7
2  |2x  5| = 7
2
2
 |2x  5| = 5
|2x  5| = 5
Since 2 is added to |2x  5|,
subtract 2 from both sides to
undo the addition.
Since |2x  5| is multiplied by a
negative 1, divide both sides
by negative 1.
Absolute values cannot be
negative.
This equation has no solution.
Try It Out!
Solve the equation. Check your answer.
6 + |x  4| = 6
6 + |x  4| = 6
+6
+6
|x  4| = 0
x4 = 0
+ 4 +4
x
=4
Since 6 is subtracted from
|x  4|, add 6 to both
sides to undo the
subtraction.
There is only one case.
Since 4 is subtracted from
x, add 4 to both sides to
undo the addition.
Check It Out!
Solve the equation. Check your answer.
6 + |x  4| = 6
The solution is x = 4.
6 + |x  4| = 6
6 + |4  4|
6
6 +|0| 6
6 + 0 6
6 6
To check your solution,
substitute 4 for x in your
original equation.
Do Now
10/14/2015
Solve the equation.
|6x| –This
8 =can
22be read as “the
distance from x to +9 is 4.”
|x + 9| = 13
x + 9 = 13 or x + 9 = –13
x=4
or
x = –22
Rewrite the absolute
value as a disjunction.
Subtract 9 from both
sides of each equation.
Check It Out! Example 2b
Solve the equation.
|6x| – 8 = 22
Isolate the absolutevalue expression.
|6x| = 30
6x = 30 or 6x = –30
x=5
or
x = –5
Rewrite the absolute
value as a disjunction.
Divide both sides of each
equation by 6.