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Take out HW from last night. Text p. 32, #4-14 evens , 15-24 all Copy HW in your planner.

Take out HW from last night. Text p. 32, #4-14 evens , 15-24 all Copy HW in your planner. Text p. 33, #26, #35-45 & 47 In your notebook, describe the process of solving an absolute value equation . Then solve the following equations . Describe how they are different.

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Take out HW from last night. Text p. 32, #4-14 evens , 15-24 all Copy HW in your planner.

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  1. Take out HW from last night. • Text p. 32, #4-14 evens, 15-24 all • Copy HW in your planner. • Text p. 33, #26, #35-45 & 47 • In your notebook, describetheprocess of solvingan absolute valueequation. Thensolvethefollowingequations. Describehowthey are different. 3|x + 8| = 12 |x + 8| = |2x + 1|

  2. HomeworkText p. 32, #4-14 evens, 15-24 all

  3. HomeworkText p. 32, #4-14 evens, 15-24 all

  4. Learning Goal • SWBAT solve absolute value equations and equations with two absolute values SWBAT solve linear equations in one variable

  5. ABSOLUTE VALUE EQUATION– an equation that contains an absolute value expression. |x| = 4 means the distance ‘x’ is from zero. -4 -3 -2 -1 0 1 2 3 4 The solution to |x| = 4 is 4 and -4 because they are the only numbers whose distance from 0 is 4.

  6. 2x – 7 = 3 First, rewrite the equation in the form ax + b = c. 32x – 7 – 5 = 4 32x – 7 = 9 2x – 7 = 3 Solve an Absolute Value Equation Solve 3|2x – 7| - 5 = 4 Write original equation. Add 5 to each side. Divide each side by3. Next, solve the absolute value equation. Write absolute value equation. 2x – 7 = 3 or2x – 7 = –3 Rewrite as two equations. 2x = 10 or2x = 4 Add 7 to each side. x = 5 or x = 2 Divide each side by 2.

  7. The equation |ax + b| = c where c ≥ 0, is equivalent to the statement: ax + b = c or ax + b = -c The equation |ax + b| = c where c < 0, is NO SOLUTION.

  8. First, rewrite the equation in the form ax + b = c. Solve an Absolute Value Equation Solve |3x + 5| + 6 = -2, if possible. |3x + 5| + 6 = -2 Write original equation. |3x + 5| + 6 = -2 Subtract 6 from each side. - 6 - 6 |3x + 5| = -8 The equation |ax + b| = c where c < 0, is NO SOLUTION.

  9. In a cheerleading competition, the minimum length of the routine is 4 minutes. The maximum length of a routine is 5 minutes. Write an absolute value equation that represents the minimum and maximum lengths.

  10. Solving Equations with Two Absolute Values Solve |x + 8| = |2x + 1|, if possible. x + 8 = -(2x + 1) x + 8 = 2x + 1 or -x -x x + 8 = -2x – 1 +2x +2x 8 = x + 1 - 1 - 1 3x + 8 = -1 - 8 - 8 7 = x 3x = -9 x = -3 CHECK |x + 8| = |2x+1| |x + 8| = |2x+1| Substitute for x. |7+ 8| = |2(7) + 1| |-3 + 8| = |2(-3) +1| Simplify. 15 = 15 5 = 5.

  11. Solving Equations with Two Absolute Values Solve |2x + 5| = 3|x – 4|, if possible. 2x + 5 = 3(x – 4) 2x + 5 = -3(x - 4) or 2x + 5 = 3x – 12 2x + 5 = -3x + 12 -2x -2x +3x +3x 5x + 5 = 12 5 = x – 12 - 5 - 5 +12 +12 5x = 7 17 = x x = 7/5 CHECK |2x + 5| = 3|x - 4| |2x + 5| = 3|x - 4| Substitute for x. |2(17)+ 5| = 3|(17) - 4| |2(7/5) + 5| = 3|(7/5) - 4| Simplify. 39 = 39 39/5 = 39/5

  12. Solve the Absolute Value Equation. Check your solution. Solve |3x + 18| = 6x 3x + 18 = -6x 3x + 18 = 6x or -3x -3x -3x -3x 18 = -9x 18 = 3x -2 = x 6 = x Reject x = -2; Extraneous Solution CHECK |3x + 18| = 6x |3x + 18| = 6x |-6 +18| = 6(-2). |18 + 18| = 6(6). Substitute for x. Simplify. 12 = -12 36 = 36.

  13. Text p. 32, #26, #35-45 & 47

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