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anonymous

  • 4 years ago

can someone check my answer: |2x+5|-1<6 add -1 |2x+5|<7 2x+5>-7 subtract 5 2x>-12

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  1. precal
    • 4 years ago
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    when you drop the asolute value it should be \[-7<2x+5<7\]

  2. anonymous
    • 4 years ago
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    does that mean my answer is wrong?

  3. precal
    • 4 years ago
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    yes because you never switch the inequality from less than to greater than unless you multiply by a negative number or if you divide by a negative number. subtract 5 to all sides and then divide by 2

  4. precal
    • 4 years ago
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    \[-12<2x<2\] \[-6<x<1\] Your solution is from -6 to 1 but not including -6 or 1 just the numbers in between

  5. anonymous
    • 4 years ago
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    why do i subract 5?

  6. precal
    • 4 years ago
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    or add -5 same thing and it is because you want to move it to the other side

  7. anonymous
    • 4 years ago
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    do i subtract if from the -1 or the 6?

  8. precal
    • 4 years ago
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    \[\left| 2x+5 \right|-1<6\] add 1 to both sides \[\left| 2x+5 \right|<7\] drop absolute value \[-7<2x+5<7\] subtract 5 to all 3 sides \[-12<2x<2\] divide by 2 to all 3 sides \[-6<x<1\]

  9. precal
    • 4 years ago
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    hope that helps

  10. anonymous
    • 4 years ago
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    if i were to graph it on a number line what would it look like? as in wouldi t point to the left?

  11. precal
    • 4 years ago
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    put hollow circles on -6 and 1 that shows that they are not included and shade in between those two numbers. -5, -4, -3, -2, -1, 0 are solutions but so are -5.999 to .99999

  12. anonymous
    • 4 years ago
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    |dw:1327885941882:dw|ok so i put the circles on -6 then shade to the right and i put circles on 1 and shade to the left?

  13. Directrix
    • 4 years ago
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    Study up on how to remove absolute value notation in a linear "less than" inequality. This pattern for "less than" absolute-value inequalities always holds: Given the inequality | x | < a, the solution is always of the form –a < x < a. Even when the exercises get more complicated, the pattern still holds. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved Solve | 2x + 3 | < 6. Since this is a "less than" absolute-value inequality, the first step is to clear the absolute value according to the pattern. Then I'll solve the linear inequality. | 2x + 3 | < 6 –6 < 2x + 3 < 6 [this is the pattern for "less than"] –6 – 3 < 2x + 3 – 3 < 6 – 3 –9 < 2x < 3 –9/2 < x < 3/2 Then the solution to | 2x + 3 | < 6 is the interval –9/2 < x < 3/2. This pattern for "less than" absolute-value inequalities always holds: Given the inequality | x | < a, the solution is always of the form –a < x < a. Even when the exercises get more complicated, the pattern still holds. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved Solve | 2x + 3 | < 6. Since this is a "less than" absolute-value inequality, the first step is to clear the absolute value according to the pattern. Then I'll solve the linear inequality. | 2x + 3 | < 6 –6 < 2x + 3 < 6 [this is the pattern for "less than"] –6 – 3 < 2x + 3 – 3 < 6 – 3 –9 < 2x < 3 –9/2 < x < 3/2 Then the solution to | 2x + 3 | < 6 is the interval –9/2 < x < 3/2. http://www.purplemath.com/modules/absineq.htm

  14. precal
    • 4 years ago
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    shade all of it from -6 to 1, think of this as all possible solutions. Any number from -6 to 1 are solutions except for -6 and 1. -5.99 is a solution to .999 is a solution.

  15. anonymous
    • 4 years ago
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    |dw:1327886150381:dw| the squigly lines are the shading lol

  16. precal
    • 4 years ago
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    yes but put hollow circles on -6 and 1 because they are not solutions.

  17. anonymous
    • 4 years ago
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    hollow as in shaded circles?

  18. precal
    • 4 years ago
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    no shaded circles are solid points and represent that the point is a solution. You do not have an equal sign under your inequality therefore it is a hollow circle (not shaded circle)

  19. anonymous
    • 4 years ago
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    thank!

  20. precal
    • 4 years ago
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    You are welcome.

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