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haleyelizabeth2017

  • one year ago

Solve the system \(|y| \ge 2\) and \(|x| \le 1\) by graphing. Please no direct answers :)

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  1. jim_thompson5910
    • one year ago
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    hint: \[\Large |y| \ge 2 \text{ is the same as } -2 \le y \le 2\]

  2. haleyelizabeth2017
    • one year ago
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    Oh!

  3. haleyelizabeth2017
    • one year ago
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    Thank you for clarifying that!

  4. jim_thompson5910
    • one year ago
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    oh sorry, I mixed up the signs I was focusing on x

  5. jim_thompson5910
    • one year ago
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    \[\Large |x| \le 1 \text{ is the same as } -1 \le x \le 1\]

  6. jim_thompson5910
    • one year ago
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    Rule: \[\Large |x| \le k \text{ is equivalent to } -k \le x \le k\] where k is any positive number

  7. haleyelizabeth2017
    • one year ago
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    Ah hah

  8. jim_thompson5910
    • one year ago
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    Rule: \[\Large |x| \ge k \text{ is equivalent to } x \ge k \text{ or } x \le -k\] where k is any positive number

  9. zzr0ck3r
    • one year ago
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    they want you to graph it

  10. jim_thompson5910
    • one year ago
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    using that second rule, \[\Large |y| \ge 2\] breaks down into \[\Large y \ge 2 \text{ or } y \le -2\]

  11. jim_thompson5910
    • one year ago
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    So you need to graph the following \[\Large -2 \le x \le 2, \ y \ge 2, \ y \le -2\]

  12. haleyelizabeth2017
    • one year ago
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    Okay so the solution is where they intersect/overlap?

  13. jim_thompson5910
    • one year ago
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    correct

  14. haleyelizabeth2017
    • one year ago
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  15. haleyelizabeth2017
    • one year ago
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    Thank you so very much!

  16. jim_thompson5910
    • one year ago
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    looks good so far

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