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Ineedhelpfast12

  • one year ago

How do you solve for this equation?

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  1. Ineedhelpfast12
    • one year ago
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    |dw:1439238581334:dw|

  2. anonymous
    • one year ago
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    does that say\[\frac{ x }{ 3 }+\frac{ 2 }{ 5 }\]?

  3. Ineedhelpfast12
    • one year ago
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    yes

  4. anonymous
    • one year ago
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    is that all of the equation?

  5. Ineedhelpfast12
    • one year ago
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    yes

  6. anonymous
    • one year ago
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    I don't think you can solve, but rather just simplify

  7. Ineedhelpfast12
    • one year ago
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    how though

  8. anonymous
    • one year ago
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    does it say if x equals anything?

  9. Ineedhelpfast12
    • one year ago
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    11\[\frac{ 11 }{ 15}x\]?

  10. Ineedhelpfast12
    • one year ago
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    no it just says solve for x

  11. anonymous
    • one year ago
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    interesting, I'll tag someone @ccieux

  12. phi
    • one year ago
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    ***How do you solve for this equation? *** an equation *always* has an = sign where is your equal sign?

  13. Ineedhelpfast12
    • one year ago
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    Fine how do you solve the PROBLEM?

  14. Ineedhelpfast12
    • one year ago
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    @Nnesha help please

  15. jdoe0001
    • one year ago
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    EQUATions "equate" to something so... you don't really have to "solve" anything per se

  16. jdoe0001
    • one year ago
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    as phi said, there's no = sign

  17. Ineedhelpfast12
    • one year ago
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    anybody out there speak english? I do not understand what jdoe0001 means

  18. anonymous
    • one year ago
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    looks to me like a cross multiplication problem

  19. zzr0ck3r
    • one year ago
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    @jdoe0001 is saying that in order to solve for \(x\) we need an equals sign. You just gave an expression, not an equation.

  20. Ineedhelpfast12
    • one year ago
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    but it is an addition problem

  21. jdoe0001
    • one year ago
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    hmmm

  22. Ineedhelpfast12
    • one year ago
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    I am not trying to solve for x. I just need to simplify x/3+2/5

  23. anonymous
    • one year ago
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    cross multiply then add, but i get what you guys are saying about their not being an equals sign =/

  24. Ineedhelpfast12
    • one year ago
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    This is only Algebra 1 people. What does x6 stand for?

  25. jdoe0001
    • one year ago
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    \(\bf \cfrac{x}{3}+\cfrac{2}{5}\implies \cfrac{}{LCD=15}\implies \cfrac{(15\div 3)x+(15\div5)2}{15}\)

  26. zzr0ck3r
    • one year ago
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    \[\dfrac{x}{3}+\dfrac{2}{5}=\dfrac{5x}{5*3}+\dfrac{2*3}{5*3}=\dfrac{5x+6}{15}\]

  27. Ineedhelpfast12
    • one year ago
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    How about another question?

  28. zzr0ck3r
    • one year ago
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    hopefully it makes sense :)

  29. Ineedhelpfast12
    • one year ago
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    |dw:1439250294613:dw|

  30. Ineedhelpfast12
    • one year ago
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    that is a t

  31. zzr0ck3r
    • one year ago
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    do you see how we simplified before?

  32. Ineedhelpfast12
    • one year ago
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    no

  33. zzr0ck3r
    • one year ago
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    \(\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+cb}{bd}\)

  34. Ineedhelpfast12
    • one year ago
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    ok

  35. zzr0ck3r
    • one year ago
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    for you \[a=x, b=y, c=2, d=t\]

  36. Ineedhelpfast12
    • one year ago
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    so is it|dw:1439250477962:dw|

  37. zzr0ck3r
    • one year ago
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    But you must realize we are not solving anything here, we are just simplifying.

  38. zzr0ck3r
    • one year ago
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    yep

  39. Ineedhelpfast12
    • one year ago
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    ok

  40. Ineedhelpfast12
    • one year ago
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    a couple more

  41. zzr0ck3r
    • one year ago
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    do the same thing...

  42. zzr0ck3r
    • one year ago
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    I will tell you if you have the right answer

  43. Ineedhelpfast12
    • one year ago
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    |dw:1439250550081:dw|

  44. zzr0ck3r
    • one year ago
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    what would be the lcd?

  45. Ineedhelpfast12
    • one year ago
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    80

  46. zzr0ck3r
    • one year ago
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    40

  47. zzr0ck3r
    • one year ago
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    sec

  48. Ineedhelpfast12
    • one year ago
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    oh ya

  49. zzr0ck3r
    • one year ago
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    \[\frac{2t}{8xy}+\dfrac{4tx}{10y}\] We want the denominator to be the same, lcm(8,10)=40 and we are missing an x on the right side \[\dfrac{5*2*t}{4*x*y}+\dfrac{4*4*t*x}{4*10*y*x}\]

  50. Ineedhelpfast12
    • one year ago
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    me no get it

  51. anonymous
    • one year ago
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    did yall solve it?

  52. Ineedhelpfast12
    • one year ago
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    no we did not

  53. Ineedhelpfast12
    • one year ago
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    hey @amsalvad canst thou help?

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