Section outline

  • Gift Bags

    The Distributive Property 
    ::分配财产

    At the end of the school year, an elementary school teacher makes a little gift bag for each of his students. Each bag contains one class photograph, two party favors, and five pieces of candy. The teacher plans to distribute the bags among each of his 28 students. How many of each item does he need?
    ::学年结束时,小学教师为每个学生制作一个小礼物袋。 每袋都装有一张班级照片、两张政党喜好和五块糖果。 老师计划将袋子分配给28名学生。他需要多少?

    Apply the Distributive Property
    ::应用分配属性

    The Distributive Property can help solve this kind of problem. First, write an expression for the contents of each bag: Items = (photo + 2 favors + 5 candies), or simply I = ( p + 2 f + 5 c ) .
    ::分配属性可以帮助解决这类问题。 首先, 为每个包包的内容写一个表达式: 项目 = (照片 + 2 优惠 + 5 糖果 ) , 或简单的 I = (p+ 2f+5c ) 。

    For all 28 students, the teacher will need 28 times that number of items, so I = 28 ( p + 2 f + 5 c ) .
    ::对于所有28名学生,教师需要的项目数是28倍,所以我=28(p+2f+5c)。

    Next, the Distributive Property states that when a single term is multiplied by a sum of several terms , it may be simplified by multiplying the single term by each of the other terms separately. In other words, 28 ( p + 2 f + 5 c ) = 28 ( p ) + 28 ( 2 f ) + 28 ( 5 c ) ,  which simplifies to 28 p + 56 f + 140 c .
    ::其次,分配财产规定,当一个单一术语乘以几个术语的总和时,可以通过将每个其他术语分别乘以一个术语加以简化。换句话说,28(p+2f+5c)=28(p)+28(2f)+28(5c),简化为28p+56f+140c。

    So the teacher needs 28 class photos, 56 party favors and 140 pieces of candy.
    ::老师需要28张班照 56张派对喜好 140块糖果

    You can see why the Distributive Property works by looking at a simple problem without variables inside the " data-term="Parentheses" role="term" tabindex="0"> parentheses , and considering the .
    ::您可以看到,为什么分配财产通过在括号内不包含变量的简单问题来看待,并且考虑 。

    Using the Order of Operations and the Distributive Property 
    ::使用行动令和分配财产

    Determine the value of 11(2 - 6) using both the Order of Operations and the Distributive Property.
    ::使用《作业命令》和《分配财产》确定11(2-6)的价值。

    Order of Operations tells us to evaluate the amount inside the parentheses first:
    ::行动顺序要求我们首先评估括号内的数额:

    11 ( 2 6 ) = 11 ( 4 ) = 44

    Now let’s try it with the Distributive Property:
    ::现在让我们用分配财产来尝试一下:

    11 ( 2 6 ) = 11 ( 2 ) 11 ( 6 ) = 22 66 = 44

    Note: When applying the Distributive Property you MUST take note of any negative signs!
    ::注:在应用分配财产时,必须注意任何负面迹象!

    Solving for Unknown Values 
    ::解决未知值

    Use the Distributive Property to determine the following.
    ::使用分配财产来确定以下事项。

    a) 11 ( 2 x + 6 )
    :sada) 11(2x+6)

    11 ( 2 x + 6 ) = 11 ( 2 x ) + 11 ( 6 ) = 22 x + 66
    ::11(2xx+6)=11(2x)+11(6)=22x+66

    b) 2 x 7 ( 3 y 2 11 x y )
    :sadb) 2x7(3y2-11xy)

    2 x 7 ( 3 y 2 11 x y ) = 2 x 7 ( 3 y 2 ) + 2 x 7 ( 11 x y ) = 6 x y 2 7 22 x 7 x y
    ::2x7( 3y2- 11xy) = 2x7( 3y2) +2x7( 11xy) = 6xy27- 22x7xy

    We can simplify this answer by canceling the x ’s in the second fraction , so we end up with 6 x y 2 7 22 7 y .
    ::我们可以简化这个答案, 取消 X 在第二个分数中的 X , 所以我们最终会达到 6xy27-227y。

    Identify Expressions That Involve the Distributive Property
    ::说明涉及分配财产的表示

    The Distributive Property can also appear in expressions that don’t include parentheses.
    ::“分配财产”也可以出现在不包括括号的文字中。

    Simplify the following expressions. Even though these expressions aren't written in a form usually associated with the Distributive Property, remember that we treat the numerator of a fraction as if it were in parentheses, and that means we can use the Distributive Property here too. 
    ::简化下面的表达式。 尽管这些表达式并非以通常与分配属性相关的形式写成, 但要记住, 我们对待一个分数的分子, 仿佛它是在括号里, 这意味着我们也可以在这里使用分配属性 。

    a) 2 x + 8 4
    :sada) 2x+84

    2 x + 8 4 can be re-written as 1 4 ( 2 x + 8 ) .  Then we can distribute the 
    ::2x+84 可重写为 14( 2x+8) 。 然后我们可以分发

    1 4 :

    1 4 ( 2 x + 8 ) = 2 x 4 + 8 4 = x 2 + 2

    ::14(2x+8)=2x4+84=x2+2

    b) 9 y 2 3
    :sadb) 9y-23

      9 y 2 3 can be re-written as 1 3 ( 9 y 2 ) ,  and then we can distribute the 
    ::9-23可以改写为13(9-2),然后我们分发

    1 3 :

    1 3 ( 9 y 2 ) = 9 y 3 2 3 = 3 y 2 3

    ::13(9y-2)=9y3-23=3y-23

    Solve Real-World Problems Using the Distributive Property
    ::利用分配财产解决现实世界问题

    The Distributive Property is one of the most common mathematical properties used in everyday life. Any time we have two or more groups of objects, the Distributive Property can help solve for an unknown.
    ::分配财产是日常生活中最常用的数学属性之一。 当我们拥有两组或两组以上的物体时,分配财产可以帮助解决未知的事物。

    Each student on a field trip into a forest is to be given an emergency survival kit. The kit is to contain a flashlight, a first aid kit, and emergency food rations. Flashlights cost $12 each, first aid kits are $7 each and emergency food rations cost $2 per day. There is $500 available for the kits and 17 students to provide for. How many days worth of rations can be provided with each kit?
    ::前往森林实地考察的每个学生应获得紧急生存工具包,其中包括手电筒、急救包和紧急口粮;每个手电筒费用12美元,急救包每套7美元,紧急口粮包每天2美元;可提供500美元,17名学生可提供。每个口粮包可提供多少天的口粮?

    The unknown quantity in this problem is the number of days’ rations. This will be x in our expression.
    ::这个问题的未知数量是口粮的天数。 这将是我们所说的x。

    Each kit will contain one $12 flashlight, one $7 first aid kit, and x times $2 worth of rations, for a total cost of ( 12 + 7 + 2 x ) dollars. With 17 kits, therefore , the total cost will be 17 ( 12 + 7 + 2 x ) dollars.
    ::每个包将包括1个12美元的手电筒、1个7美元的急救包和x值2美元的口粮,总费用为12+7+2x美元,因此,17个包的总费用为17美元(12+7+2x美元)。

    We can use the Distributive Property on this expression:
    ::我们可以使用“分配财产”这一表达方式:

    17 ( 12 + 7 + 2 x ) = 204 + 119 + 34 x

    ::17(12+7+2x)=204+119+34x

    Since the total cost can be at most $500, we set the expression equal to 500 and solve for x .
    ::由于总成本最多500美元,我们设定了等于500美元的表达式,并解决 x。

    204 + 119 + 34 x = 500 323 + 34 x = 500 323 + 34 x 323 = 500 323 34 x = 177 34 x 34 = 177 34 x 5.206

    ::204+119+34x=500323+34x=500323+34x=500323+34x-323=500-32334x=17734x34=17734x=5206

    Since this represents the number of days’ worth of rations that can be bought, we must round to the next lowest whole number . We wouldn’t have enough money to buy a sixth day's worth of supplies.
    ::由于这是可以买到的口粮日数,我们必须四舍五入到下一个最低的口粮日数。 我们没有足够的钱购买第六天的补给品。

    Five days worth of emergency rations can be purchased for each survival kit.
    ::每个生存箱可购买5天的紧急口粮。

    Examples 
    ::实例

    Simplify the following expressions:
    ::简化以下表达式:

    Example 1
    ::例1

    7 ( 3 x 5 )
    ::7(3x-5) 7(3x-5)

    Note the negative sign on the second term. 

    7 ( 3 x 5 ) = 21 x 35

    ::注意第二学期的负符号。 7(3x-5)=21x-35

    Example 2
    ::例2

    2 7 ( 3 y 2 11 )
    ::27(3y2-11)

    2 7 ( 3 y 2 11 ) = 2 7 ( 3 y 2 ) + 2 7 ( 11 ) = 6 y 2 7 22 7
    ::27(3y2-11)=27(3y2)+27(-11)=6y27-227

    Example 3
    ::例3

    z + 6 2
    ::zz+62

    Rewrite z + 6 2 as 1 2 ( z + 6 ) ,  and distribute the 1 2 :
    ::将z+62重写为12(z+6),并分发12:

    1 2 ( z + 6 ) = z 2 + 6 2 = z 2 + 3

    ::12(z+6)=z2+62=z2+3

    Review 
    ::回顾

    For 1-7, use the Distributive Property to simplify the following expressions.
    ::1-7,使用分配财产来简化以下表述。

    1. ( x + 4 ) 2 ( x + 5 )
      :sadx+4)-2(x+5)
    2. 1 2 ( 4 z + 6 )
      ::12(4z+6)
    3. ( 4 + 5 ) ( 5 + 2 )
    4. x ( x + 7 )
      :sadxx+7) (xx+7)
    5. y ( x + 7 )
      ::y(x+7) y(x+7)
    6. x ( 3 x + 5 )
      :sadx3x+5)
    7. x y ( 1 x + 2 y )
      ::xy(1x+2y)

    For 8-15, use the Distributive Property to remove the parentheses from the following expressions.
    ::8-15时,使用分配财产来将括号从以下文字中去掉。

    1. 1 2 ( x y ) 4
      ::12(x-y)- 4
    2. 0.6 ( 0.2 x + 0.7 )
      ::0.60(0.2x+0.7)
    3. 6 + ( x 5 ) + 7
      ::6+(x-5)+7
    4. 6 ( x 5 ) + 7
      ::6-(x-5)+7
    5. 4 ( m + 7 ) 6 ( 4 m )
      ::4(m+7)-6(4-m)
    6. 5 ( y 11 ) + 2 y
      ::-5(y-11)+2y
    7. ( x 3 y ) + 1 2 ( z + 4 )
      :sadx-3y)+12(z+4)
    8. a b ( 2 a + 3 b + b 5 )
      ::a (2a+3b+b5) (2a+3b+b5)

    For 16-23, use the Distributive Property to simplify the following fractions.
    ::对于16-23,使用分配财产来简化下列部分。

    1. 8 x + 12 4
      ::8x+124
    2. 9 x + 12 3
      ::9x+123
    3. 11 x + 12 2
      ::11x+122 11x+122
    4. 3 y + 2 6
      ::3y+26
    5. 6 z 2 3
      ::- 6z-23
    6. 7 6 p 3
      ::7-6p3 7-6p3
    7. 3 d 4 6 d
      ::3d-46d 3d-46d
    8. 12 g + 8 h 4 g h
      ::12克+8h4ghg
    9. A bookcase has five shelves, and each shelf contains seven poetry books and eleven novels. How many of each type of book does the bookcase contain?
      ::书架有5个书架,每个书架都有7本诗歌书和11本小说。书架包含多少书架和11本小说?
    10. Amar is making giant holiday cookies for his friends at school. He makes each cookie with 6 oz of cookie dough and decorates them with macadamia nuts. If Amar has 5 lbs of cookie dough ( 1   l b = 16   o z ) and 60 macadamia nuts, calculate the following.
      1. How many (full) cookies he can make?
        ::他能做多少饼干?
      2. How many macadamia nuts he can put on each cookie, if each is to be identical?
        ::如果每块饼干都一样的话 他能加多少坚果?
      3. If 4 cups of flour and 1 cup of sugar went into each pound of cookie dough, how much of each did Amar use to make the 5 pounds of dough?
        ::如果4杯面粉和1杯糖 进入每磅饼干面团, 每个有多少Amar用 做5磅面粉?

      ::阿玛尔正在为学校的朋友做大型的节日饼干。 他每块饼干加6盎司饼干杜格, 并用坚果装饰它们。 如果阿玛尔有5磅饼干杜格(1磅=16 oz)和60 坚果, 计算如下。 他能做多少( 完整的)饼干? 他能给每块饼干加多少坚果? 如果每块面粉和1杯糖都是一样的, 如果每磅饼干杜格中装了4杯面粉和1杯糖, 阿玛尔能用多少钱做5磅面果?

    Review (Answers)
    ::回顾(答复)

    Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
    ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。