Chapter 1 Section 5 Absolute Value Equations and Inequalities.
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Transcript of Chapter 1 Section 5 Absolute Value Equations and Inequalities.
Chapter 1
Section 5Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5ALGEBRA 2 LESSON 1-5
Solve each equation.
1. 5(x – 6) = 40 2. 5b = 2(3b – 8) 3. 2y + 6y = 15 – 2y + 8
4. 4x + 8 > 20
(For help, go to Lessons 1-3 and 1-4.)
Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities
Solve each inequality.
6. 4(t – 1) < 3t + 55. 3a – 2 a + 6>–
1. 5(x – 6) = 40 2. 5b = 2(3b – 8)
= 5b = 6b – 16
x – 6 = 8 –b = –16x = 14 b = 16
Solutions
ALGEBRA 2 LESSON 1-5ALGEBRA 2 LESSON 1-5Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities
5(x – 6)5
405
5. 3a – 2 a + 6 6. 4(t – 1) < 3t + 53a – a 6 + 2 4t – 4 < 3t + 5
2a 8 4t – 3t < 5 + 4a 4 t < 9
3. 2y + 6y = 15 – 2y + 8 4. 4x + 8 > 202y + 6y + 2y = 15 + 8 4x > 12
10y = 23 x > 3y = 2.3
>–>–>–>–
Absolute Value Absolute Value – the distance from zero a number is
on the number line – it is always positive Symbol : │x│ Definition:
If x is positive ( x > 0) then │x│ = x If x is negative ( x < 0) then │x │ = -x
Absolute Value Equations have a possibility of two solutions This is because the value inside the │ │ can equal either the
negative or the positive of the value on the other side of the equal sign
Always isolate the absolute value expression on one side of the equal sign before breaking the problem into two pieces.
Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities
Solve |15 – 3x| = 6.
|15 – 3x| = 6
15 – 3x = 6 or15 – 3x = –6The value of 15 – 3x can
be 6 or –6 since |6| and |–6| both equal 6.
x = 3 or x = 7Divide each side of both
equations by –3.
–3x = –9 –3x = –21Subtract 15 from each
side of both equations.
Check: |15 – 3x| = 6 |15 – 3x| = 6|15 – 3(3)| 6 |15 – 3(7)| 6
|6| 6 |–6| 66 = 6 6 = 6
Try This Problem│3x + 2 │ = 7
3x + 2 = 7 or 3x + 2 = -7 3x = 5 or 3x = -9
x = 5/3 or x = -3
Check your answer by plugging it back into the equation.
.
77
725
72353
77
729
72)3(3
Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities
Solve 4 – 2|x + 9| = –5.
4 – 2|x + 9| = –5
–2|x + 9| = –9 Add –4 to each side.
x = –4.5 or x = –13.5Subtract 9 from
each side of both equations. Check: 4 – 2 |x + 9| = –5 4 – 2|x + 9| = –5
4 – 2 |–4.5 + 9| –5 4 – 2 |–13.5 + 9| –5
4 – 2 |4.5| –5 4 – 2 |–4.5| –5
4 – 2 (4.5) –5 4 – 2 (4.5) –5
–5 = –5 –5 = –5
|x + 9| = Divide each side by –2.92
x + 9 = or x + 9 = – Rewrite as two equations.92
92
Try This Problem
Solve 2│3x - 1 │ + 5 = 33.
2 │3x - 1 │ = 28
│3x – 1│ = 14
3x – 1 = 14 or 3x – 1 = -14
3x = 15 or 3x = -14
x = 5 or x = -8/3
Check the solutions by plugging them into the original problem. They both work.
Extraneous Solutions
Extraneous solution – a solution of an equation derived from an original equation that is not a solution of the original equation
This is why we MUST check all answers to see if they work in the original problem.
Solve │2x + 5 │ = 3x + 4
2x + 5 = 3x + 4 5 = x + 4 1 = x
Check:2(1) + 5 = 3(1) + 4 2 + 5 = 3 + 4 7 = 7
2x + 5 = -(3x + 4)2x + 5 = -3x – 4 5x + 5 = -4 5x = -9 x = -9/5Check:│2(-9/5) + 5 │ = 3(-9/5) + 4
│-18/5 + 25/5 │ = -27/5 + 20/5
│7/5 │ = -7/5
│7/5 │ ≠ -7/5
The only solution is x = 1.
Try These Problems
Solve and check for extraneous solutions.
a) │2x + 3 │ = 3x + 2
2x + 3 = 3x + 2 or 2x + 3 = -3x – 2 3 = x + 2 or 5x + 3 = -2 1 = x or 5x = -5 1 = x or x = -1
Check:
The solution is x = 1
b) │x │ = x – 1
x = x – 1 or x = -x + 10 = -1 or 2x = 1 x = ½
Check:
There is No Solution.
Homework
Practice 1.5 #13-24 all
Absolute Value Inequalities
Let k represent a positive real number │x │ ≥ k is equivalent to x ≤ -k or x ≥ k
│x │ ≤ k is equivalent to -k ≤ x ≤ k
Remember to isolate the absolute value before rewriting the problem with two inequalities
Solve |2x – 5| > 3. Graph the solution.
|2x – 5| > 3
2x – 5 < –3 or 2x – 5 > 3 Rewrite as a compound inequality.
x < 1 or x > 4
2x < 2 2x > 8 Solve for x.
Try This Problem
Solve │2x - 3 │ > 7
2x – 3 > 7 or 2x – 3 < -7
2x > 10 or 2x < -4
x > 5 or x < -2
Solve –2|x + 1| + 5 –3. Graph the solution.>–
|x + 1| 4 Divide each side by –2 and reverse the inequality.
<–
–2|x + 1| + 5 –3>–
–2|x + 1| –8Isolate the absolute value expression. Subtract 5 from each side.
>–
–4 x + 1 4Rewrite as a compound inequality.
<– <–
–5 x 3Solve for x.
<– <–
Try This Problem
Solve |5z + 3| - 7 < 34. Graph the solution. |5z + 3| -7 < 34 |5z + 3| < 41-41 < 5z + 3 < 41-44 < 5z < 38-44/5 < z < 38/5
-8 4/5 < z < 7 3/5
Ranges in Measurement
Absolute value inequalities and compound inequalities can be used to specify an allowable range in measurement.
Tolerance – the difference between a measurement and its maximum and minimum allowable values Equals one half (½) the difference
between the max and min values
ToleranceFor example, if a manufacturing specification calls
for a dimension d of 10 cm with a tolerance of 0.1 cm, then the allowable difference between d and 10 is less than or equal to 0.1
|d - 10 | ≤ 0.1 absolute value inequality
d – 10 ≤ 0.1 and d – 10 ≥-0.1 equivalent compound inequality
-0.1 ≤ d – 10 ≤ 0.1 equivalent compound inequality
9.9 ≤ d ≤ 10.1 simplified compound inequality
The area A in square inches of a square photo is required to
satisfy 8.5 ≤ A ≤ 8.9. Write this requirement as an absolute
value inequality.
Write an inequality.–0.2 A – 8.7 0.2<– <–
Rewrite as an absolute value inequality.|A – 8.7| 0.2<–
Find the tolerance.8.9 – 8.5
2 =0.42 = 0.2
Find the average of the maximum andminimum values.
8.9 + 8.52 =
17.42 = 8.7
Try This ProblemThe specification for the circumference C in inches of a
basketball for junior high school is 27.75 ≤ C ≤ 30. Write the specification as an absolute value inequality.
375.2
75.
2
75.275.28
125.282
25.56
2
75.275.28
375.125.28 C
Find the tolerance.
Find the average from min and max values.
Write the absolute value inequality.
Homework
Practice 1.5 # 1 – 12, 25 – 30
Watch the units on the back page Make sure all units are inches,
centimeters, meters, etc…