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steve816

  • one year ago

Please help me! How do I convert this to standard form?

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  1. steve816
    • one year ago
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    \[y=\frac{ -11 }{ 9 x}+\frac{ 1 }{ 3 }\]

  2. anonymous
    • one year ago
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    is y the LHS of an equation for a line? If so, the x should be moved to the numerator.

  3. steve816
    • one year ago
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    Sorry, I kind of messed up on writing the equation. It's supposed to be like this:\[y=-\frac{ 11 }{ 9 }x+\frac{ 1 }{ 3 }\]

  4. anonymous
    • one year ago
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    http://jwilson.coe.uga.edu/emt668/EMAT6680.2002/Jackson/Chapter%205%20Lesson%20Plan/Day6.html

  5. steve816
    • one year ago
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    Umm can you please give me a demonstration using this example??

  6. anonymous
    • one year ago
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    Ax+By=C Eliminate the fractions in the problem. Do you know how you can do that?

  7. anonymous
    • one year ago
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    Denominators are 9 & 3 which equal 27. If you multiply the entire equation by the common denominator (27), you can eliminate the fraction. 27 (y=-11x/9 + 1/3) 27y = -33x+9

  8. anonymous
    • one year ago
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    To reach standard form, get the x & y on the same side of the equation.

  9. anonymous
    • one year ago
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    Get ride of the fractions by multiplying each side of the equation by 27. 27 y = -33 x + 9 Then move the x term to the LHS. 33x +27y = 9

  10. steve816
    • one year ago
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    Wait, can't you multiply the equations by 9?

  11. anonymous
    • one year ago
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    Yes.

  12. anonymous
    • one year ago
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    the word ride should have been spelled rid.

  13. anonymous
    • one year ago
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    You could. 9y = -11x+3 That should have been the choice...wasn't thinking...rusty

  14. anonymous
    • one year ago
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    Would give you 11x + 9y = 3

  15. anonymous
    • one year ago
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    Divide each side of the following equation by 3. 33x +27y = 9

  16. steve816
    • one year ago
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    Okay I understand this problem now. Thanks so much @robtobey and @DSS Wish I could give more then 1 medal but really appreciate you guys' help :)

  17. anonymous
    • one year ago
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    Your welcome.

  18. anonymous
    • one year ago
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    You're welcome

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