 # Solving Absolute-Value Equations

## Presentation on theme: "Solving Absolute-Value Equations"— Presentation transcript:

Solving Absolute-Value Equations
2-Ext Solving Absolute-Value Equations Holt Algebra 1 Lesson Presentation

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 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.

To solve absolute-value equations, perform inverse operations to isolate the absolute-value expression on one side of the equation. Then you must consider two cases.

Example 1A: Solving Absolute-Value Equations
Solve each equation. Check your answer. |x| = 12 Think: What numbers are 12 units from 0? |x| = 12 Case 1 x = 12 Case 2 x = –12 Rewrite the equation as two cases. The solutions are 12 and –12. Check |x| = 12 |12| |12| 12

Example 1B: Solving Absolute-Value Equations
Since |x + 7| is multiplied by 3, divide both sides by 3 to undo the multiplication. Think: What numbers are 8 units from 0? |x + 7| = 8 Case 1 x + 7 = 8 Case 2 x + 7 = –8 – 7 –7 – 7 – 7 x = 1 x = –15 Rewrite the equations as two cases. Since 7 is added to x subtract 7 from both sides of each equation. The solutions are 1 and –15.

Example 1B Continued 3|x + 7| = 24 The solutions are 1 and –15. Check 3|x + 7| = 24 3|8| 24 3|1 + 7| 24 3(8) 24 3|15 + 7| 3|8| 3(8)

Check It Out! Example 1a Solve each equation. Check your answer. |x| – 3 = 4 Since 3 is subtracted from |x|, add 3 to both sides. |x| – 3 = 4 |x| = 7 Think: what numbers are 7 units from 0? Case 1 x = 7 Case 2 –x = 7 –1(–x) = –1(7) x = –7 x = 7 Rewrite the case 2 equation by multiplying by –1 to change the minus x to a positive.. The solutions are 7 and –7.

Check It Out! Example 1a Continued
Solve the equation. Check your answer. |x|  3 = 4 The solutions are 7 and 7. Check |x| 3 = 4 7  |7|  | 7|  7 

Check It Out! Example 1b Solve the equation. Check your answer. |x  2| = 8 Think: what numbers are 8 units from 0? |x  2| = 8 Case 1 x  2 = 8 x = 10 x = 6 Case 2 x  2 = 8 Rewrite the equations as two cases. Since 2 is subtracted from x add 2 to both sides of each equation. The solutions are 10 and 6.

Check It Out! Example 1b Continued
Solve the equation. Check your answer. |x  2| = 8 The solutions are 10 and 6. Check |x 2| = 8 10 2| 8 |10 2| 8 | 6 + (2)| 8

Not all absolute-value equations have two solutions
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 2A: Special Cases of Absolute-Value Equations
Solve the equation. Check your answer. 8 = |x + 2|  8 Since 8 is subtracted from |x + 2|, add 8 to both sides to undo the subtraction. 8 = |x + 2|  8 0 = |x +2| There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition. 0 = x + 2  2 2 = x

Example 2A Continued Solve the equation. Check your answer. 8 = |x +2|  8 Solution is x = 2 Check 8 =|x + 2|  8 To check your solution, substitute 2 for x in your original equation. 8 8 8 |2 + 2|  8 8 |0|  8   8

Example 2B: Special Cases of Absolute-Value Equations
Solve the equation. Check your answer. 3 + |x + 4| = 0 Since 3 is added to |x + 4|, subtract 3 from both sides to undo the addition. 3 + |x + 4| = 0  3 |x + 4| = 3 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 Since 2 is added to |2x  5|, subtract 2 from both sides to undo the addition. 2  |2x  5| = 7  2  |2x  5| = 5 Since |2x  5| is multiplied by a negative 1, divide both sides by negative 1. |2x  5| = 5 Absolute values cannot be negative. This equation has no solution.

Check It Out! Example 2b Solve the equation. Check your answer. 6 + |x  4| = 6 Since 6 is subtracted from |x  4|, add 6 to both sides to undo the subtraction. 6 + |x  4| = 6 |x  4| = 0 x  4 = 0 x = 4 There is only one case. Since 4 is subtracted from x, add 4 to both sides to undo the addition.

Check It Out! Example 2b Continued
Solve the equation. Check your answer. 6 + |x  4| = 6 The solution is x = 4. 6 + |x  4| = 6 To check your solution, substitute 4 for x in your original equation. 6 + |4  4| 6 6 +|0| 6  6 6 6