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Published byNigel Atkins Modified over 7 years ago

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EXAMPLE 1 Solve a simple absolute value equation Solve |x – 5| = 7. Graph the solution. SOLUTION | x – 5 | = 7 x – 5 = – 7 or x – 5 = 7 x = 5 – 7 or x = 5 + 7 x = –2 or x = 12 Write original equation. Write equivalent equations. Solve for x. Simplify.

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EXAMPLE 1 The solutions are –2 and 12. These are the values of x that are 7 units away from 5 on a number line. The graph is shown below. ANSWER Solve a simple absolute value equation

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EXAMPLE 2 Solve an absolute value equation | 5x – 10 | = 45 5x – 10 = 45 or 5x –10 = – 45 5x = 55 or 5x = – 35 x = 11 or x = – 7 Write original equation. Expression can equal 45 or – 45. Add 10 to each side. Divide each side by 5. Solve |5x – 10 | = 45. SOLUTION

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EXAMPLE 2 Solve an absolute value equation The solutions are 11 and –7. Check these in the original equation. ANSWER Check: | 5x – 10 | = 45 | 5(11) – 10 | = 54 ? |45| = 45 ? 45 = 45 | 5x – 10 | = 45 | 5(– 7 ) – 10 | = 54 ? 45 = 45 | – 45| = 45 ?

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EXAMPLE 3 | 2x + 12 | = 4x 2x + 12 = 4x or 2x + 12 = – 4x 12 = 2x or 12 = – 6x 6 = x or –2 = x Write original equation. Expression can equal 4x or – 4 x Add – 2x to each side. Solve |2x + 12 | = 4x. Check for extraneous solutions. SOLUTION Solve for x. Check for extraneous solutions

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EXAMPLE 3 | 2x + 12 | = 4x | 2(– 2) +12 | = 4(–2) ? |8| = – 8 ? 8 = –8 Check the apparent solutions to see if either is extraneous. Check for extraneous solutions | 2x + 12 | = 4x | 2(6) +12 | = 4(6) ? |24| = 24 ? 24 = 24 The solution is 6. Reject – 2 because it is an extraneous solution. ANSWER CHECK

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GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 1. | x | = 5 SOLUTION | x | = 5 | x | = – 5 or | x | = 5 x = –5 or x = 5 Write original equation. Write equivalent equations. Solve for x. for Examples 1, 2 and 3

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GUIDED PRACTICE for Examples 1, 2 and 3 The solutions are –5 and 5. These are the values of x that are 5 units away from 0 on a number line. The graph is shown below. ANSWER – 3 – 4 – 2 – 1 0 1 234 5 6 7– 5 – 6 – 7 5 5

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GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 2. |x – 3| = 10 SOLUTION | x – 3 | = 10 x – 3 = – 10 or x – 3 = 10 x = 3 – 10 or x = 3 + 10 x = –7 or x = 13 Write original equation. Write equivalent equations. Solve for x. Simplify. for Examples 1, 2 and 3

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GUIDED PRACTICE The solutions are –7 and 13. These are the values of x that are 10 units away from 3 on a number line. The graph is shown below. ANSWER – 3 – 4 – 2 – 1 0 12345 6 7 – 5 – 6 – 78 9 10 11 12 13 10 for Examples 1, 2 and 3

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GUIDED PRACTICE Solve the equation. Check for extraneous solutions. SOLUTION | x + 2 | = 7 x + 2 = – 7 or x + 2 = 7 x = – 7 – 2 or x = 7 – 2 x = –9 or x = 5 Write original equation. Write equivalent equations. Solve for x. Simplify. 3. |x + 2| = 7 for Examples 1, 2 and 3

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GUIDED PRACTICE The solutions are –9 and 5. These are the values of x that are 7 units away from – 2 on a number line. ANSWER for Examples 1, 2 and 3

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GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 4. |3x – 2| = 13 SOLUTION 3x – 2 = 13 or 3x – 2 = – 13 Write original equation. Solve for x. Simplify. |3x – 2| = 13 Write equivalent equations. x = or x = 5 3 –3 2 for Examples 1, 2 and 3 x = –13 + 2 3 or x = 13 + 2 3

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GUIDED PRACTICE The solutions are – and 5. ANSWER 3 3 2 for Examples 1, 2 and 3

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GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 5. |2x + 5| = 3x | 2x + 5 | = 3x 2x + 5 = – 3x or 2x + 5 = 3x x = 1 or x = 5 Write original equation. Write Equivalent equations. Simplify SOLUTION for Examples 1, 2 and 3 2x + 3x = 5 or 2x – 3x = –5

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GUIDED PRACTICE The solution of is 5. Reject 1 because it is an extraneous solution. ANSWER for Examples 1, 2 and 3 | 2x + 12 | = 4x | 2(1) +12 | = 4(1) ? |14| = 4 ? 14 = –8 Check the apparent solutions to see if either is extraneous. | 2x + 5 | = 3x | 2(5) +5 | = 3(5) ? |15| = 15 ? 15 = 15 CHECK

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GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 6. |4x – 1| = 2x + 9 SOLUTION 4x – 1 = – (2x + 9) or 4x – 1 = 2x + 9 Write original equation. Solve For x x = or x = 5 3 1 1 |4x – 1| = 2x + 9 Write equivalent equations. for Examples 1, 2 and 3 4x + 2x = – 9 + 1 or 4x – 2x = 9 + 1 Rewrite equation.

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GUIDED PRACTICE ANSWER The solutions are – and 5. 3 1 1 for Examples 1, 2 and 3 | 4x – 1 | = 2x + 9 | 4(5) – 1 | = 2(5) + 9 ? |19| = 19 ? 19 = 19 Check the apparent solutions to see if either is extraneous. | 4x – 1 | = 2x + 9 | 4( ) – 1 | = 2( ) + 9 3 –1 1 3 1 ? CHECK | | = ? 3 – 19 3 = 3 – 3

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