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EXAMPLE 1

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

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EXAMPLE 1

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

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

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

  4. ? | 5(– 7 )– 10 |= 54 ? | 5(11)– 10 |= 54 ? | –45|= 45 ? 45= 45 45= 45 |45|= 45 EXAMPLE 2 Solve an absolute value equation ANSWER The solutions are 11 and –7. Check these in the original equation. Check: | 5x – 10| = 45 | 5x– 10| = 45

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

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

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

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

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

  10. ANSWER The solutions are –7 and 13. These are the values of xthat are 10units away from 3on a number line. The graph is shown below. 10 10 – 2 0 1 2 3 4 5 6 10 11 – 6 12 – 3 – 1 7 13 – 7 8 9 – 5 – 4 for Examples 1, 2 and 3 GUIDED PRACTICE

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

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

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

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

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

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

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

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

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