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Destinyyyy

  • one year ago

Need two problems checked...

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  1. Destinyyyy
    • one year ago
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    Solve the absolute value inequality. 36< {-2x+18/5} +9/5 {} absolute value

  2. Destinyyyy
    • one year ago
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    180/5 <{-2x+ 18/5} -2x+18/5> 180/5 -2x+ 18/5< -180/5

  3. Destinyyyy
    • one year ago
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    I got-- x> -81/5 x< 99/5

  4. linn99123
    • one year ago
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    okay so whats ur answer for number 1?

  5. Destinyyyy
    • one year ago
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    I posted both of my answers above.

  6. Destinyyyy
    • one year ago
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    Im not sure what you mean by number one. I only posted one question.

  7. DanJS
    • one year ago
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    looks like you know the way to solve them, not sure about the numbers

  8. dinamix
    • one year ago
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    and i am sure about my answer

  9. Destinyyyy
    • one year ago
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    Okay

  10. xapproachesinfinity
    • one year ago
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    \(|-2x+18/5|+9/5>36 \) \(|-2x+18/5|>189/5 \leftarrow \text{one correction here}\)

  11. Destinyyyy
    • one year ago
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    Where are you getting 189 from???

  12. DanJS
    • one year ago
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    36 - (9/5) = 171/5 for the first thing

  13. xapproachesinfinity
    • one year ago
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    36+9/5=189/5

  14. Destinyyyy
    • one year ago
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    Omg... 36 *5= 180 .. I forgot to minus the 9 -.-

  15. xapproachesinfinity
    • one year ago
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    eh should be 180-9 i added them

  16. xapproachesinfinity
    • one year ago
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    it is ok well make mistakes

  17. xapproachesinfinity
    • one year ago
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    we all*

  18. xapproachesinfinity
    • one year ago
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    but the she got the point

  19. Destinyyyy
    • one year ago
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    Now I have -2x> 153/5 -2x < -153/5

  20. xapproachesinfinity
    • one year ago
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    yah yo got the point of this just made some few error just recheck for certainty

  21. xapproachesinfinity
    • one year ago
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    you know how this works that's what is important

  22. xapproachesinfinity
    • one year ago
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    i got the same result you should be good to go

  23. Destinyyyy
    • one year ago
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    Okay. But now im stuck.. -2 cant be divided into these. And my last problem like this just divided the top number by -2

  24. xapproachesinfinity
    • one year ago
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    multiply by -1/2 and switch the sign > to < and < to >

  25. xapproachesinfinity
    • one year ago
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    if you multiply by negative you should change the inequality

  26. Destinyyyy
    • one year ago
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    Uh? I know what the final answer is because I looked it up on mathway... But dont get how I go from what I have to it.

  27. dinamix
    • one year ago
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    @Destinyyyy u have find it 171/10< x<-153/10 i didnt focus first time

  28. Destinyyyy
    • one year ago
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    The answer is (- infinity sign, -153/10) u (189/10 , infinity sign)

  29. dinamix
    • one year ago
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    yup i was wrong its 189/10 sorry

  30. dinamix
    • one year ago
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    i forget add 18/2 only sorry

  31. Destinyyyy
    • one year ago
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    How do I get from -2x> 153/5 and -2x< -153/5 to the final answer

  32. xapproachesinfinity
    • one year ago
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    i told you to multiply by -1/2 both sides

  33. xapproachesinfinity
    • one year ago
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    first one x<-153/10 see the inequality changed

  34. xapproachesinfinity
    • one year ago
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    second x>153/10

  35. dinamix
    • one year ago
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    not that u have -2x>153/5 ------ 2x<-153/5 -----x<-153/10 @Destinyyyy

  36. xapproachesinfinity
    • one year ago
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    actually you made a mistake somewhere for the scond

  37. dinamix
    • one year ago
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    i fix it now

  38. xapproachesinfinity
    • one year ago
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    it should not be -2x<-153/5

  39. dinamix
    • one year ago
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    this is not - its just Arrow

  40. Destinyyyy
    • one year ago
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    Then Id have to change the denominators to 10... 306/10 5/10 .. Do I minus, divide, multiple?

  41. Destinyyyy
    • one year ago
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    What are you talking about? Its over 5 not 10

  42. dinamix
    • one year ago
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    not that u have (-2x>153/5) ------ (2x<-153/5) -----(x<-153/10 ) @xapproachesinfinity this what i mean

  43. xapproachesinfinity
    • one year ago
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    the second should have been -2x<-189/5 then x>189/10

  44. xapproachesinfinity
    • one year ago
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    that what i meant she made a mistake in calculations

  45. xapproachesinfinity
    • one year ago
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    after that just read what x<-153/10 and x>189/10 mean in interval nottion

  46. xapproachesinfinity
    • one year ago
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    you should get that same answer you have by mathway

  47. xapproachesinfinity
    • one year ago
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    over 10 my bad

  48. Destinyyyy
    • one year ago
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    OMG. Seriously. Im an still at -2x> 153/5 and -2x< -153/5

  49. Destinyyyy
    • one year ago
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    I dont understand your multiple by -1/2.

  50. xapproachesinfinity
    • one year ago
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    you need to isolate x right

  51. xapproachesinfinity
    • one year ago
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    so we multiply by a minus sign which switches the inequality then we multiply by 1/2 to get rid of 2

  52. xapproachesinfinity
    • one year ago
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    it is like you do with equation to islote the unknown

  53. Destinyyyy
    • one year ago
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    ...

  54. Destinyyyy
    • one year ago
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    Still dont understand.

  55. xapproachesinfinity
    • one year ago
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    how would you solve 2x=4

  56. Destinyyyy
    • one year ago
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    Divide both sides by 2.

  57. xapproachesinfinity
    • one year ago
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    so same thing divide the inequality both sides by 2

  58. xapproachesinfinity
    • one year ago
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    that is the same as multiplying out by 1/2 hahaha

  59. Destinyyyy
    • one year ago
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    Okay. Like I said before its not possible. It gives a decimal.... K? So then Id have to change both denominators to 10.

  60. xapproachesinfinity
    • one year ago
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    why decimal ? you get -153/5x2

  61. xapproachesinfinity
    • one year ago
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    i dont see where do you get decimal?

  62. Destinyyyy
    • one year ago
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    What are you talking about? 153 divided by -2= -76.5

  63. xapproachesinfinity
    • one year ago
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    why you need to divide? leave it 153/5x2=153/10

  64. dinamix
    • one year ago
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    \[|-2x+\frac{ 18 }{ 2 }|>\frac{ 171 }{ 5 }\] \[\frac{ -171}{ 5 }>-2x+\frac{ 18 }{ 2 }>\frac{ 171 }{ 5}\] \[\frac{ -171-18 }{ 5 }>-2x>\frac{ 171-18 }{ 5}\] \[\frac{ -189 }{ 5}>-2x>\frac{ 153 }{ 5 }\] \[-\frac{ 153 }{ 10 }>x>\frac{ 189 }{ 10 }\] @xapproachesinfinity could u check my mistake @Destinyyyy this is real solution

  65. xapproachesinfinity
    • one year ago
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    you making this hard on youself

  66. dinamix
    • one year ago
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    cuz i think this is right answer @xapproachesinfinity

  67. Destinyyyy
    • one year ago
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    No im not. You are making this impossible to understand. Ive waited 2 hours for help and now Im just confused. Normally you just divide it. And instead im suppose to multiple it. But all you did was change the denominator. Why???

  68. xapproachesinfinity
    • one year ago
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    yes your answer is good @dinamix

  69. Destinyyyy
    • one year ago
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    You said to multiply by 1/2 so I asked about changing both denominator to 10. And both times you did not respond. So im just suppose to do 5*2= 10 ?

  70. xapproachesinfinity
    • one year ago
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    heheheh thanks i thought i made efforts to show you

  71. Destinyyyy
    • one year ago
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    I understand the first one. Now what about the other one.

  72. dinamix
    • one year ago
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    @xapproachesinfinity ty @Destinyyyy what u didnt understand we can help u

  73. xapproachesinfinity
    • one year ago
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    ok dinamix take over i have to go

  74. dinamix
    • one year ago
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    @xapproachesinfinity plz i dont know how to explain

  75. Destinyyyy
    • one year ago
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    Try showing me?

  76. dinamix
    • one year ago
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    its hard to expalin idea for me @xapproachesinfinity this problem of me

  77. dinamix
    • one year ago
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    @Destinyyyy what i show u

  78. Destinyyyy
    • one year ago
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    Can you just show me how you went from -2x< -153/5 to 189/10

  79. dinamix
    • one year ago
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    -2x>153/5 to x<-153/10 right ?

  80. dinamix
    • one year ago
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    that what u mean or what

  81. Destinyyyy
    • one year ago
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    If I go off from what both of you said before by just multiplying by -1/2 then Ill just get 153/10 which is not the correct answer. Which also makes me think how you said to solve the first one is not the correct way to solve it.

  82. Destinyyyy
    • one year ago
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    The fist sign goes < not >

  83. Destinyyyy
    • one year ago
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    -2x< -153/5 to 189/10

  84. dinamix
    • one year ago
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    \[-2x+\frac{ 18 }{ 5} > 171/5\] and \[-(-2x+\frac{ 18 }{ 5 })>171/5 \]\[\frac{ -171 }{ 5 } >-2x+\frac{ 18 }{ 5 }>\frac{ 171 }{ 5}\]

  85. dinamix
    • one year ago
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    @Destinyyyy its 2 equation i make it only one did u understand now

  86. freckles
    • one year ago
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    36< |-2x+18/5| +9/5 Just because I like to write the absolute value expression first then the inequality thingy.. \[|-2x+\frac{18}{5}|+\frac{9}{5}>36 \\ \text{ so you subtracted } \frac{9}{5} \text{ on both sides } \\ |-2x+\frac{18}{5}|>36-\frac{9}{5} \\ |-2x+\frac{18}{5}|> \frac{180-9}{5} \\ |-2x+\frac{18}{5}| >\frac{171}{5} \\ \text{ now since this is a greater than we have a union } \\ \text{ that is we have } -2x+\frac{18}{5}<-\frac{171}{5} \text{ or } -2x+\frac{18}{5}>\frac{171}{5} \\ \text{ for both inequalities subtract }\frac{18}{5} \text{ on both sides } \\ -2x<\frac{-171}{5}-\frac{18}{5} \text{ or } -2x> \frac{171}{5}-\frac{18}{5} \\ \text{ simplifying right hand sides } \\ -2x <\frac{-189}{5} \text{ or } -2x >\frac{153}{5} \\ \text{ now you can say divide both sides by -2 } \\ \text{ or equivalently say multiply both sides by } \frac{-1}{2} \\ \text{ this is the same meaning just different saying } \] \[\frac{-1}{2} \cdot (-2x)>\frac{-1}{2}(\frac{-189}{5}) \text{ change direction of inequality symbol } \\ \text{ since I multiplied both sides by a negative } \\ x>\frac{189}{10} \\ \text{ now the other inequality } \\ \frac{-1}{2}(-2x)<\frac{-1}{2}(\frac{153}{5}) \\ x<\frac{-153}{10}\] so put the two pieces together and put or between them

  87. dinamix
    • one year ago
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    @freckles like my steps lol

  88. freckles
    • one year ago
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    \[x<\frac{-153}{10} \text{ or } x>\frac{189}{10}\]

  89. freckles
    • one year ago
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    your solution seems to suggest -171 is greater than 171 since it says -171/5>171/5 @dinamix

  90. dinamix
    • one year ago
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    @freckles but why we get same result

  91. freckles
    • one year ago
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    no I didn't get -171>171 because this is not true

  92. Destinyyyy
    • one year ago
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    Omg thank you @freckles I completely understood everything you said. And I saw where I messed up. I did write down the - for -171/5 so when I minus-ed the 18 I had the wrong answer. Now I have the correct one and understand how to solve the ending.

  93. dinamix
    • one year ago
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    but @freckles look the last i get -153/10 >x or x>189/10 look

  94. Destinyyyy
    • one year ago
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    And all that turns into (- infinity sign, -153/10) u (189/10, infinity sign)

  95. Destinyyyy
    • one year ago
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    @freckles Can you check one more? I know that my answer is wrong, but im not sure how it is.

  96. freckles
    • one year ago
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    alright one more and then I eat

  97. Destinyyyy
    • one year ago
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    Okay... Use interval notation to express the solution set and graph the solution set on a number line. x/4 - 1/5 </ x/5 +1

  98. freckles
    • one year ago
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    \[\frac{x}{4}-\frac{1}{5} \le \frac{x}{5}+1\]

  99. freckles
    • one year ago
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    is that correct

  100. Destinyyyy
    • one year ago
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    I got x</ 2 (- infinity sign, 2] But it says im wrong Yes thats correct

  101. freckles
    • one year ago
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    so I retriceumed you subtracted x/5 on both sides and added 1/5 on both sides?

  102. freckles
    • one year ago
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    assumed*

  103. dinamix
    • one year ago
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    x<24 i find it

  104. Destinyyyy
    • one year ago
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    My example had me plus 1/5 to both sides. So I have x/4 </ x/5 +1 +1/5 Then add the +1 and 1/5 together x/4 </ x/5 +6/5

  105. freckles
    • one year ago
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    @Destinyyyy if you don't like the fractions before moving anything you can multiply both sides by 4*5=20

  106. freckles
    • one year ago
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    \[(20)(\frac{x}{4}-\frac{1}{5} ) \le 20(\frac{x}{5}+1) \\ \frac{20}{4}x-\frac{20}{5} \le \frac{20}{5}x+20 \\ 5x-4 \le 4x+20\] this might make it easier for you

  107. Destinyyyy
    • one year ago
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    x</ 24

  108. dinamix
    • one year ago
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    @freckles ty for help her or him anyway , i know the answer but i dont how eplain idea this my problem

  109. Destinyyyy
    • one year ago
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    (- infinity sign, 24]

  110. Destinyyyy
    • one year ago
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    Im a her....

  111. dinamix
    • one year ago
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    yup @Destinyyyy this the answer hreat now u understand

  112. dinamix
    • one year ago
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    ok i was joke only

  113. freckles
    • one year ago
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    yep now if you didn't do the whole canceling fraction thing... \[\frac{x}{4}-\frac{1}{5} \le \frac{x}{5}+1 \\ \text{ subtract } \frac{x}{5} \text{ on both sides } \\ \frac{x}{4} -\frac{x}{5}-\frac{1}{5} \le 1 \\ \text{ now add } \frac{1}{5} \text{ on both sides } \\ \frac{x}{4} -\frac{x}{5} \le 1+\frac{1}{5} \\ \\ \text{ find common denominators } \\ \frac{5x-4x}{20} \le \frac{5+1}{5} \\ \frac{1x}{20} \le \frac{6}{5} \\ \frac{x}{20} \le \frac{6}{5} \\ \text{ multiply both side by } 20 \\ x \le \frac{6(20)}{5} \\ x \le 6 \cdot \frac{20}{5} \\ x \le 6 (4) \\ x \le 24\] just so you know how it looks without removing fractions first

  114. dinamix
    • one year ago
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    @Destinyyyy i find it without use pain last question ;D

  115. dinamix
    • one year ago
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    @Destinyyyy and u can will be like that

  116. Destinyyyy
    • one year ago
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    The example I followed didnt show me like that.. It did x/4 </ x/5 +1 +1/5 x/4 </ x+5 +6/5 x/4 -x/5 </ 6/5 4x-x/5 </ 6/5 4x-x </ 6 3x</ 6 x</2

  117. dinamix
    • one year ago
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    pen*

  118. Destinyyyy
    • one year ago
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    So I would have gotten the right answer if it had told me to do over 20 instead of over 5.. Which I thought was weird. Thank you for helping! @freckles

  119. freckles
    • one year ago
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    np i will eat now

  120. dinamix
    • one year ago
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    u make mistake in line 4

  121. Destinyyyy
    • one year ago
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    Yeah I know that. But that's what the example said to do so I did.

  122. dinamix
    • one year ago
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    \[\frac{ 5x -4x}{ 20 }\]

  123. dinamix
    • one year ago
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    =\[\frac{ x }{ 4}-\frac{ x }{ 5 }\]

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