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anonymous

  • one year ago

What is 81= x-29? Explain, please!

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  1. steve816
    • one year ago
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    Simply add 29 because you want to get the x alone.

  2. anonymous
    • one year ago
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    Explain it word by word, please. My 8th grade sister needs help on this.

  3. steve816
    • one year ago
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    |dw:1443237063291:dw|

  4. anonymous
    • one year ago
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    Can you check her other questions, please? 2\

  5. DanJS
    • one year ago
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    the addition property of equality , haha just had to look that up a min ago

  6. anonymous
    • one year ago
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    4x + 7 - x = 19

  7. anonymous
    • one year ago
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    Help? @steve816

  8. DanJS
    • one year ago
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    you can check by putting in your x value back to the original equation, see if both sides are the same when calculated

  9. Keyoumars
    • one year ago
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    |dw:1443237209574:dw|

  10. anonymous
    • one year ago
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    I already have that one, thank you.

  11. steve816
    • one year ago
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    Great explanation @keyoumars

  12. anonymous
    • one year ago
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    Would you mind helping her with 4x + 7 - x = 19

  13. RagingSquirrel
    • one year ago
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    HI there! This is a matter of getting the variable alone (in this case the variable is x). So with this one: 4x-7-x=19 You want to get your numbers on one side of the equation, and the x's on the other. Start by moving that -7. To do that, just add 7's to both sides. Like this.|dw:1443237639862:dw| So now you have 4x-x=19+7 Simplify the x's here. Since they're both on one side, all you need to do is subtract. 3x=19+7 Add those numbers together. 3x=26 Almost done! However, you still have a three with that x. To get it out, just divide both sides by 3. 3x/3=26/3 Now you have x=8 and 2/3. Don't move on! Check your work by plugging 8 and 2/3 back into the equation. So instead of x, write |dw:1443238066829:dw| 4x-7-x=19 4(26/3)-7-(26/3)=19 4(26/3)-7-(26/3)+7=19+7 4(26/3)-(26/3)=26 3(26/3)=26 26=26 8 and 2/3 is the correct answer!

  14. Keyoumars
    • one year ago
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    thank you@steve816

  15. Keyoumars
    • one year ago
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    thank you @steve816

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