1. Teacher – Mrs. Volynskaya Algebra 2 Absolute Value Equations Absolute value A numbers distance from 0 on number line. ALWAYS POSITIVE!!! Extraneous Solution A solution that appears to be a solution, but doesn’t work in the original equation.

2. 1.5 Interpreting Absolute Value Equations Equation|x| = |x  0| = k MeaningThe distance between x and 0 is __k__. Graph Solutionsx  0 =  k or x  0 = k x = _  k_ or x = _k_

3. 1.5 Interpreting Absolute Value Equations Equation|x  b| = k MeaningThe distance between x and b is _k_. Graph Solutionsx  b =  k or x  b = k x = _b  k_ or x = _b + k_

4. Teacher – Mrs. Volynskaya Algebra 2 1.5 Solving an absolute value equation Use these steps to solve an absolute value equation | ax + b | = c where c > 0. Step 1 Write two equations: ax + b = _c_ or ax + b = _  c__. Step 2 Solve each equation. Step 3 Check each solution in the original _absolute value_ equation.

5. Example 1 Solve a simple absolute value equation Solve |x  3 | = 6. Graph the solution. x  3 = -6 or x  3 = 6 +3 +3 +3 +3 x = -3 x = 9 These are the values that are 6 from 3 on the number line.

6. Example 2 Solve an absolute value equation Solve |4x + 10 | = 6x. Check for extraneous solutions. 4x + 10 = 6x or4x + 10 = -6x -4x -4x -4x-4x 10 = 2x10 = -10x 10/2 = 2x/2 10/-10 = -10x/-10 5 = x-1 = x CHECK

7. Example 2 Solve an absolute value equation Check |4x + 10 | = 6x | 4(_5_) + 10| ≟ 6(_5_) |4(  1_ ) + 10 | ≟ 6(_  1_) |_30_| ≟ _30_|_6_| ≟ _  6_ _30_ = _30__6_  _  6_ X = 5 x = -1 is rejected, because it is an extraneous solution.

8. 1.5 Absolute Value Inequalities InequalityEquivalent Form |ax + b| _<_ c  c < ax + b < c Graph of Solution InequalityEquivalent Form |ax + b| _>_ c ax + b c Graph of Solution

9. Example 3 Solve an inequality Solve |2x + 5 | > 3. Then graph the solution. 2x + 5 3 -5 -5 -5 -5 2x -2 2x/2 -2/2 x -1

10. Example 4 Solve an inequality Solve |x  1.5|  4.5. Then graph the solution. x – 1.5 ≤ 4.5 or x – 1.5 ≥ -4.5 +1.5 +1.5+1.5 +1.5 x ≤ 6 x ≥ -3