Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Signal-flow graph wikipedia , lookup
Quartic function wikipedia , lookup
Cubic function wikipedia , lookup
Quadratic equation wikipedia , lookup
Elementary algebra wikipedia , lookup
System of linear equations wikipedia , lookup
History of algebra wikipedia , lookup
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 73 4 4 4 4 4 Try It Out! Solve the equation. Check your answer. |x 2| = 8 |x 2| = 8 Case 1 x2= 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 08 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 x4 = 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.