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anonymous

  • 5 years ago

I have a very hard question (for myself...) x-2[x-3(x+4)-5]=3{2x-[x-8(x-4)]}-2 please help :-)

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  1. anonymous
    • 5 years ago
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    ok, let's see

  2. anonymous
    • 5 years ago
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    \[x-2[x-3(x+4)-5] = 3(2x-[x-8(x-4)])-2\] wow, lots of groups!

  3. anonymous
    • 5 years ago
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    ok, so let's think about what the goal is, the goal is unite the groups on each side then solve for x

  4. anonymous
    • 5 years ago
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    yes

  5. anonymous
    • 5 years ago
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    x=6

  6. anonymous
    • 5 years ago
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    ??? that fast?

  7. anonymous
    • 5 years ago
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    adfriedman, nice but so fast!

  8. anonymous
    • 5 years ago
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    it was just expanding brackets

  9. anonymous
    • 5 years ago
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    right, ok fastff, a good approach could be to work from the inside out

  10. anonymous
    • 5 years ago
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    focus on the left side: x−2[x−3(x+4)−5]

  11. anonymous
    • 5 years ago
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    so let's start in the middle of this with 3(x+4)...can you expand this please?

  12. anonymous
    • 5 years ago
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    to 3x+12

  13. anonymous
    • 5 years ago
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    nice. so the equation is now x−2[x−(3x+12)−5] = x-2[x-3x-12-5]

  14. anonymous
    • 5 years ago
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    the really quick approach is to note that only constants multiply the bracketed terms

  15. anonymous
    • 5 years ago
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    ok, now simplify the expression with the [ ... ]

  16. anonymous
    • 5 years ago
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    -3x+12x-5?

  17. anonymous
    • 5 years ago
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    or -3x-12x-5?

  18. anonymous
    • 5 years ago
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    -15x-5?

  19. anonymous
    • 5 years ago
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    almost...forgot about the x

  20. anonymous
    • 5 years ago
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    so you have x-3x-12-5....ok

  21. anonymous
    • 5 years ago
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    simplify this: start with the x-terms

  22. anonymous
    • 5 years ago
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    aren't we suppose to distribute?

  23. anonymous
    • 5 years ago
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    distribute what?

  24. anonymous
    • 5 years ago
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    so inside the [ ] you have x-3x-12-5 which becomes -2x-17, right?

  25. anonymous
    • 5 years ago
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    fastff, how's it going?

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