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anonymous

  • one year ago

Solve and check for extraneous solutions 16=3(x-1)-(x-7)

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  1. anonymous
    • one year ago
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    @satellite73

  2. anonymous
    • one year ago
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    not sure what "extraneous" solutions means in this context it is a linear equation right? no radials, no denominators

  3. anonymous
    • one year ago
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    distribute first on the right both the 3 and the minus sign

  4. anonymous
    • one year ago
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    let me know what you get

  5. anonymous
    • one year ago
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    Ok, so I got 2x+6=16

  6. anonymous
    • one year ago
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    think there is a mistake there the \(2x\) is right, but where did the \(6\) come from?

  7. anonymous
    • one year ago
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    -1+7

  8. anonymous
    • one year ago
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    \[3(x-1)-(x-7)=3x-3-x+7\] is a start

  9. anonymous
    • one year ago
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    oops forgot the distributive law didn't you?

  10. anonymous
    • one year ago
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    \[3(x-1)=3x-3\times 1\]

  11. anonymous
    • one year ago
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    Oh okay! So 2x+4?

  12. anonymous
    • one year ago
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    right \[2x+4=16\] takes only 2 steps to solve

  13. anonymous
    • one year ago
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    a) subtract 4 b) divide by 2

  14. anonymous
    • one year ago
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    2x=10 x=2

  15. anonymous
    • one year ago
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    ???

  16. anonymous
    • one year ago
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    first off \(16-4\neq 10\) and secondly \(\frac{10}{2}\neq 2\) so both steps are wrong

  17. anonymous
    • one year ago
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    what is \(16-4\)?

  18. anonymous
    • one year ago
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    16-4=12

  19. anonymous
    • one year ago
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    ok so \[2x+4=16\\ 2x=12\] is a the first step

  20. anonymous
    • one year ago
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    x=6! So sorry, it's late where I am lol so I have a sleepy brain I suppose

  21. anonymous
    • one year ago
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    second step is \(x=\frac{12}{2}\)

  22. anonymous
    • one year ago
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    x=6

  23. anonymous
    • one year ago
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    right

  24. anonymous
    • one year ago
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    How do we check for extraneous solutions?

  25. anonymous
    • one year ago
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    there are no extraneous solutions with linear equations, nothing to check

  26. anonymous
    • one year ago
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    you can replace x by 6 in the original equation and see that it works if you like but it is right, nothing is extraneous here

  27. anonymous
    • one year ago
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    ok:) Can I ask another? I'll be better haha

  28. anonymous
    • one year ago
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    sure

  29. anonymous
    • one year ago
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    x/3=x/2-2 Check for extraneous solutions

  30. anonymous
    • one year ago
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    is this \[\frac{x}{3}=\frac{x}{2}-2\] like that ?

  31. anonymous
    • one year ago
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    yup

  32. anonymous
    • one year ago
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    you have a choice you can work with the fractions or you can get rid of them

  33. anonymous
    • one year ago
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    you pick

  34. anonymous
    • one year ago
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    get rid of them.

  35. anonymous
    • one year ago
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    Actually, which ever is simpler

  36. anonymous
    • one year ago
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    then since the least common multiple of 2 and 3 is 6, if you multiply both sides by 6 they will be gone most people i think like getting rid of fractions

  37. anonymous
    • one year ago
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    \[6\times \frac{x}{3}=6\times (\frac{x}{2}-2)\]

  38. anonymous
    • one year ago
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    I think I did it wrong bc I got 2x/6-3x/6

  39. anonymous
    • one year ago
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    since \(\frac{6}{2}=3\) and \(\frac{x}{3}=2\) you end up with \[2x=3x-18\]

  40. anonymous
    • one year ago
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    you forgot to get rid of the 6 when you cancelled

  41. anonymous
    • one year ago
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    oh okay:)

  42. anonymous
    • one year ago
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    \[6\times \frac{x}{3}=2x\] not \(\frac{2x}{6}\)

  43. anonymous
    • one year ago
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    so now you have \[2x=3x-18\] which takes 2 steps to solve you see where the \(-18\) came from?

  44. anonymous
    • one year ago
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    Yes!

  45. anonymous
    • one year ago
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    x=18

  46. anonymous
    • one year ago
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    yes

  47. anonymous
    • one year ago
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    So it wouldnt be extraneous?

  48. anonymous
    • one year ago
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    no it would not

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