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? asked in Science & MathematicsMathematics · 8 years ago

Simplifying equations ???? Please help !!?

I'm having a hard time solving these :/ help !!!

1. 1/2h(a+b)

2. 5{-2+3[4-2(3+5)]}

3. [18(x-2)+9x]- {7[3(2y-5)- 8y+7)]+9}

4. -3[9(x-4)+5x] - 8 {3[5(3y+4)] - 12}

4 Answers

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  • None
    Lv 7
    8 years ago
    Favorite Answer

    1. 1/2h(a+b) = 1/(2ha + 2hb)....If you meant (1/2h)(a+b), it's (a + b)/2h

    2. 5{-2+3[4-2(3+5)]} = 5{-2+3[4-2(8)]} = 5{-2+3[4-16]} = (5)( 38) = -190

    3. [18(x-2)+9x]- {7[3(2y-5)- 8y+7)]+9} = 27x - 36 - {7[6y - 15 - 8y+7]+9} =

    27x - 36 - {7[-2y - 8]+9} =27x - 36 - (-14y - 47) = 27x + 14y + 47

    4. -3[9(x-4)+5x] - 8{3[5(3y+4)] - 12} = -3[14x - 36] - 8{3[15y + 20] -12} =

    -42x + 108 - 8(45y + 48) = - 42x - 360y - 276

    Nice work with the parentheses, brackets and braces! Please check what I have done above to find any mistakes.

  • 8 years ago

    1. 1/2h(a + b) is 1/2ah + 1/2bh.

    2. 5{- 2 + 3[4 - 2(3 + 5)]} =

    5{- 2 + 3[4 - 2(8)]} =

    5{- 2 + 3[4 - 16]} =

    5{- 2 + 3[- 12]} =

    5{- 2 - 36} =

    5(- 38) =

    - 190.

    3. [18(x - 2) + 9x] - {7[3(2y - 5) - 8y + 7)] + 9} =

    [18(x - 2) + 9x] - {7[6y - 15 - 8y + 7] + 9} =

    [18x - 36 + 9x] - {7[- 2y - 8] + 9} =

    [27x - 36] - {- 14y - 56 + 9} =

    27x - 36 + 14y + 56 - 9 =

    27x + 14y + 56 - 45 =

    27x + 14y + 11.

    4. - 3[9(x - 4) + 5x] - 8{3[5(3y + 4)] - 12} =

    - 3[9(x - 4) + 5x] - 8{3[15y + 12] - 12} =

    - 3[9x - 36 + 5x] - 8{45y + 36 - 12} =

    - 3(14x - 36) - 8(45y + 24) =

    - 42x + 108 - 360y - 192 =

    - 42x - 360y - 84

    or

    42x + 360y + 84.

  • iceman
    Lv 7
    8 years ago

    You solve the equations and simplify the expressions now in this case we are doing the latter:

    4) -3[9(x-4)+5x] - 8 {3[5(3y+4)] - 12}

    = -3[9x - 36 +5x] - 8{3[15y + 20)] - 12}

    = -3[14x - 36] - 8{45y + 60 - 12}

    = -42x + 108 - 8(45y + 48)

    = -42x + 108 - 360y - 384

    = -42x - 360y -276

  • Anonymous
    8 years ago

    my advise to you try use the MDAS in question #2

    Source(s): using MDAS
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