6.5 warm up warm up lesson quiz lesson quiz lesson presentation lesson presentation solve absolute...
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6.5
Warm UpWarm Up
Lesson QuizLesson Quiz
Lesson PresentationLesson Presentation
Solve Absolute Value Equations
6.5 Warm-Up
ANSWER 12, 12
ANSWER 1
2. Evaluate |x| – 2 when x = –3.
1. For a = –12, find, –a and |a|.
The change in elevation as a diver explored a reef was –0.5 foot, 1.5 feet, –2.5 feet, and 2.25 feet. Which change in elevation had the greatest absolute value?
ANSWER –2.5 ft
3.
6.5 Example 1
SOLUTION
The distance between x and 0 is 7. So, x = 7 or x = –7.
ANSWER
The solutions are 7 and –7.
Solve x = 7.
6.5 Guided Practice
Solve (a) |x| = 3 and (b) |x| = 15.
ANSWER 3, –3a.
15, –15b.
6.5 Example 2
SOLUTION
Rewrite the absolute value equation as two equations. Then solve each equation separately.
x – 3 = 8 Write original equation.
x – 3 = 8 or x – 3 = –8 Rewrite as two equations.
x = 11 or x = –5 Add 3 to each side.
ANSWER
The solutions are 11 and –5. Check your solutions.
x – 3 = 8.Solve
6.5 Example 2
|x – 3| = 8 |x – 3| = 8
CHECK
Substitute for x.
Subtract.
Simplify. The solution checks.
|11 – 3| = 8 |–5 – 3| = 8? ?
| 8| = 8 |–8| = 8??
Write original inequality.
8 = 8 8 = 8
6.5 Example 3
SOLUTION
First, rewrite the equation in the form ax + b = c.
3 2x – 7 – 5 = 4
3 2x – 7 = 9
2x – 7 = 3
Write original equation.
Add 5 to each side.
Divide each side by 3.
3 2x – 7 – 5 = 4.Solve
6.5 Example 3
Next, solve the absolute value equation.
2x – 7 = 3
2x – 7 = 3 or 2x – 7 = –3
2x = 10 or 2x = 4
x = 5 or x = 2
Write absolute value equation.
Rewrite as two equations.
Add 7 to each side.
Divide each side by 2.
ANSWER
The solutions are 5 and 2.
6.5 Guided Practice
Solve the equation.r – 7 = 92.
16, –2ANSWER
2 s + 4.1 = 18.93.
7.4, –7.4ANSWER
4 t + 9 – 5 = 194.
–3, –15ANSWER
6.5 Example 4
3x + 5 + 6 = –2 Write original equation.
3x + 5 = –8 Subtract 6 from each side.
ANSWER
The absolute value of a number is never negative. So, there are no solutions.
3x + 5 + 6 = –2, if possible.Solve
6.5 Guided Practice
ANSWER no solution
5. 2 m – 5 + 4 = 2
Solve the equation, if possible
6. –3 n +2 –7 = –10
ANSWER 1, 3
6.5
Absolute Deviation
The absolute deviation of a number x from a given value is the absolute value of the difference of x and the given value:
Absolute Deviation =
6.5 Example 5
BASKETBALLS
Before the start of a professional basketball game, a basketball must be inflated to an air pressure of 8 pounds per square inch (psi) with an absolute error of 0.5 psi.
Absolute error is the absolute deviation of a measured value from an accepted value.
Find the minimum and maximum acceptable air pressures for the basketball.
6.5 Example 5
SOLUTION
Let p be the air pressure (in psi) of a basketball. Write a verbal model. Then write and solve an absolute value equation.
0.5 = p – 8
6.5 Example 5
p – 80.5 =
p 80.5 = – or p 8–0.5 = –
p8.5 = or p7.5 =
Write original equation.
Rewrite as two equations.
Add 8 to each side.
ANSWER
The minimum and maximum acceptable pressures are 7.5 psi and 8.5 psi.
6.5 Guided Practice
7. A volleyball league is preparing a two minute radio ad to announce tryouts. The ad has an absolute deviation of 0.05 minute. Find the minimum and the maximum acceptable times the radio ad can run.
Minimum: 1.95 minMaximum: 2.05 min
ANSWER
6.5 Lesson Quiz
Solve the equation, if possible.
ANSWER –9, 17
ANSWER no solutions
1. 3| x – 4 | = 39
2. | x + 2 | + 7 = 3
ANSWER –5, 11
ANSWER no solutions
3. | 2x – 6 | – 18 = – 2
4. –2 | x – 5 | + 7 = 12
6.5 Lesson Quiz
A pattern for a 26-inch skirt has an absolute deviation of 1.5 inches. Find the minimum and maximum skirt lengths that can be made from the pattern.
5.
ANSWER minimum : 24.5 in.; maximum 27.5 in.