Section 5 Absolute Value Equations and Inequalities Chapter 1 Section 5 Absolute Value Equations and Inequalities

Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 (For help, go to Lessons 1-3 and 1-4.) Solve each equation. 1. 5(x – 6) = 40 2. 5b = 2(3b – 8) 3. 2y + 6y = 15 – 2y + 8 Solve each inequality. 4. 4x + 8 > 20 5. 3a – 2 a + 6 > – 6. 4(t – 1) < 3t + 5

Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solutions 1. 5(x – 6) = 40 2. 5b = 2(3b – 8) = 5b = 6b – 16 x – 6 = 8 –b = –16 x = 14 b = 16 5(x – 6) 5 40 5 3. 2y + 6y = 15 – 2y + 8 4. 4x + 8 > 20 2y + 6y + 2y = 15 + 8 4x > 12 10y = 23 x > 3 y = 2.3 5. 3a – 2 a + 6 6. 4(t – 1) < 3t + 5 3a – a 6 + 2 4t – 4 < 3t + 5 2a 8 4t – 3t < 5 + 4 a 4 t < 9 > – > – > – > –

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 Inequalities Solve |15 – 3x| = 6. |15 – 3x| = 6 15 – 3x = 6 or 15 – 3x = –6 The value of 15 – 3x can be 6 or –6 since |6| and |–6| both equal 6. –3x = –9 –3x = –21 Subtract 15 from each side of both equations. x = 3 or x = 7 Divide each side of both equations by –3. Check: |15 – 3x| = 6 |15 – 3x| = 6 |15 – 3(3)| 6 |15 – 3(7)| 6 |6| 6 |–6| 6 6 = 6 6 = 6

Try This Problem │3x + 2 │ = 7 3x + 2 = 7 or 3x + 2 = -7 Check your answer by plugging it back into the equation. .

Absolute 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 + 9| = Divide each side by –2. 9 2 x + 9 = or x + 9 = – Rewrite as two equations. 9 2 x = –4.5 or x = –13.5 Subtract 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

Try This Problem Solve 2│3x - 1 │ + 5 = 33. 2 │3x - 1 │ = 28 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 The only solution is x = 1. 2x + 5 = 3x + 4 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 = -9/5 Check: │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. │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 │x │ = x – 1 x = x – 1 or x = -x + 1 0 = -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

|2x – 5| > 3 2x < 2 2x > 8 Solve for x. x < 1 or x > 4 Solve |2x – 5| > 3. Graph the solution. |2x – 5| > 3 2x – 5 < –3 or 2x – 5 > 3 Rewrite as a compound inequality. 2x < 2 2x > 8 Solve for x. x < 1 or x > 4

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. > – –2|x + 1| + 5 –3 > – –2|x + 1| –8 Isolate the absolute value expression. Subtract 5 from each side. > – |x + 1| 4 Divide each side by –2 and reverse the inequality. < – –4 x + 1 4 Rewrite as a compound inequality. < – –5 x 3 Solve for x. < –

Try This Problem Solve |5z + 3| - 7 < 34. Graph the solution.

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

Tolerance For 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 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. Find the tolerance. 8.9 – 8.5 2 = 0.4 = 0.2 Find the average of the maximum and minimum values. 8.9 + 8.5 2 = 17.4 = 8.7 Write an inequality. –0.2 A – 8.7 0.2 < – Rewrite as an absolute value inequality. |A – 8.7| 0.2 < –

Try This Problem The 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. 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…