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2011中考模拟分类汇编8.二元一次方程(组)

二元一次方程（组）

一、选择题

1．（2011年浙江省杭州市城南初级中学中考数学模拟试题）若1++y x 与()2

2--y x 互为相

反数，则3)3(y x -的值为（） A.1 B.9 C.–9 D.27 答案：D

2、（2011湖北省崇阳县城关中学模拟）已知()0332

=++++m y x x 中，y 为负数，则m

的取值范围是（）

A. m ＞9

B. m ＜9

C. m ＞-9

D. m ＜-9 答案：A

3、（2011年海宁市盐官片一模）解方程组23739x y x y +=??

+=? ，①－②得（）

A ．32x = B. 32x =- C. 2x = D. 2x =- 答案：D

二、填空题

1、（2011年北京四中五模）方程组???=-=+5

xy y x 的解是 .

答案：?

???

?5y 1x 1y 5x ＝－＝－＝－＝－ 2．（淮安市启明外国语学校2010－2011学年度第二学期初三数学期中试卷）已知x ，y 满足

方程组23,37.x y x y -=??+=?①②

求x +2y 的值为．

答案：4

3、（2011深圳市模四）若关于x ，y 的二元一次方程组?

??=-=+k y x ,

k y x 95的解也是二元一次方程

632=+y x 的解，则k 的值为_______。

答案：

三、解答题

1、(中江县2011年初中毕业生诊断考试)

新年新举措——我县某工艺品销售公司今年一月份调整了职工的月工资分配方案，调整后月工资由基本保障工资和计件奖励工资两部分组成（计件奖励工资＝销售每件的奖励金额×销

②

①

售的件数）. 下表是甲、乙两位职工今年2月份的工资情况信息：（1）试求工资分配方案调整后职工的月基本保障

工资和销售每件产品的奖励金额各是多少？（2）若职工丙今年三月份的工资不低于3000元，

那么丙该月至少应销售多少件产品？

答案：解：（1）设调整后职工的月基本保障工资为x 元，销售每件产品的奖励金额为y 元，

根据题意得方程： ?

?=+=+.2500300,2000200y x y x

解这个方程组得：???==.

5,1000

y x

故调整后职工的基本保障工资为每月1000元，销售每件产品的奖励金额是5元. （2）设职工丙该月至少应销售z 件产品，则1000＋5z ≥3000，解之得：z ≥400.

即职工丙该月至少应销售400件产品.

2、（2011重庆市纂江县赶水镇）解方程组 355223x y x y -=??+=?，

．

答案：解： ①×2+②得，11x=33. 解得，x=3. 把x=3代入①得y=4. ∴??

?==4

y ,

3x 是原方程的解.

3、（重庆一中初2011级10—11学年度下期3月月考）

解方程组?

?=-=+128

53y x y x

答案：．解方程组???=-=+12853y x y x

解：由②得y=2x －3代入①得：

3x+5(2x －1)=8 3x+10x －5=8 13x=13 X=1 代入③得

y=2×1－1=1

∴原方程的解为?

?==11y x

职工甲乙月销售件数（件） 200 300 月工资（元）

2000

2500

① ②

4、（2011年北京四中三模）国庆节，甲、乙两班学生到集市上购买苹果，，苹果的价格如下：

甲班分两次共购买苹果70千克（第二次多于第一次），共付出189元，而乙班则一次购买苹果70千克。

（1）乙班比甲班少付多少元？

（2）甲班第一次、第二次分别购买苹果多少千克？

答案：（1）49元（2）28、42 5、（2011年江苏盐都中考模拟）某汽车制造厂开发了一款新式电动汽车，计划一年生产安装240辆.由于抽调不出足够的熟练工来完成新式电动汽车的安装，工厂决定招聘一些新工人；他们经过培训后上岗，也能独立进行电动汽车的安装.生产开始后，调研部门发现：1名熟练工和2名新工人每月可安装8辆电动汽车；2名熟练工和3名新工人每月可安装14辆电动汽车.

（1）每名熟练工和新工人每月分别可以安装多少辆电动汽车？

（2）如果工厂招聘n （0

成一年的安装任务，那么工厂有哪几种新工人的招聘方案？

（3）在（2）的条件下，工厂给安装电动汽车的每名熟练工每月发2000元的工资，给每

名新工人每月发1200元的工资，那么工厂应招聘多少名新工人，使新工人的数量

多于熟练工，同时工厂每月支出的工资总额W （元）尽可能的少？

解：（1）设每名熟练工和新工人每月分别可以安装x,y 辆电动汽车.

2314x y x y +=??

+=? 解之得

42x y =??

每名熟练工和新工人每月分别可以安装4辆、2辆电动汽车.（3分）（2）设需熟练工m 名，依题意有：2 n ×12+4m ×12=240, n =10-2m

∵0

∴18m n =??

=? 26m n =??=? 34m n =??=? 4

2m n =??=?

（3分）（3）依题意有：W=1200n+（5-12n ）×2000=200 n+10000，要使新工人数量多于熟练工，

满足n=4、6、8，故当n=4时，W 有最小值=10800元（4分

6、（2011年北京四中中考模拟18）某商场按定价销售某种电器时，每台可获利48元；按定价的九折销售该电器6台与将定价降低30元销售该电器9台所获得的利润相等，该电器每台的进价、定价各是多少元？

解：设该电器每台的进价为x 元，定价为y 元。

购苹果数不超过30千克 30千克以上但不超过50千克 50千克以上每千克价格 3元 2.5元 2元

)(小时x

)(吨y

8 2 4 10

)(小时x

)(吨y

8 2 4 10

依题意得???--=-=-)30(9)9.0(648x y x y x y 解得?

??==210162

y x

答：该电器每台的进价、定价各是162元、210元。

7、萧山新星塑料厂有甲、乙、丙三辆运货车，每辆车只负责进货或出货，丙车每小时的运输量最多，乙车每小时的运输量最少，乙车每小时运6吨，下图是甲、乙、丙三辆运输车开始工作后，仓库的库存量y （吨）与工作时间x （小时）之间的函数图像，其中OA 段只有甲、丙两车参与运输，AB 段只有乙、丙两车参与运输，BC 段只有甲、乙两车参与运输。 (1)甲、乙、丙三辆车中，谁是进货车？ (2)甲车和丙车每小时各运输多少吨？

(3)由于仓库接到临时通知，要求三车在8小时后同时开始工作，但丙车在运送

10吨货物后出现故障而退出，问：8小时后，甲、乙两车又工作了几小时，使仓库的库存量为6吨？

（1）乙、丙是进货车，甲是出货车。……………………………………3分（2）设：甲、丙两车每小时运货x 吨和y 吨，

则 ()()()?

??==???-=-++=-108

41065642y x ：x y x y 解得

∴甲车和丙车每小时各运8吨和10吨。…………………………………7分（3）设：经过m 小时后，库存是6吨，

则m(6-8)+10=-4,解得：m=7…………………………………………………9分答：甲、乙两车又工作了7小时，库存是6吨。…………………………10分

8、（2011杭州上城区一模）由于电力紧张，某地决定对工厂实行“峰谷”用电．规定：在每天的8:00至22:00为“峰电”期,电价为a 元/度；每天22:00至8:00为为“谷电”期,电价为b 元/度．下表为某厂4、5月份的用电量和电费的情况统计表：

月份用电量（万度）

电费（万元）

4 12 6.4 5

8.8

(1)若4月份“谷电”的用电量占当月总电量的

，5月份“谷电”的用电量占当月总用电量的

，求a 、b 的值． (2)若6月份该厂预计用电20万度，为将电费控制在10万元至10.6万元之间（不含10万

元和10.6万元），那么该厂6月份在“谷电”的用电量占当月用电量的比例应在什么范围？

答案：(1) 由题意，得

32×12a ＋31

×12b＝6.4 8a ＋4b ＝6.4 43×16a＋4

×16b＝8.8 12a ＋4b ＝8.8 解得 a ＝0.6 b ＝0.4

（2）设6月份“谷电”的用电量占当月总电量的比例为k ．由题意，得10＜20(1－k)×0.6＋20k×0.4＜10.6 解得0.35＜k ＜0.5

答：该厂6月份在平稳期的用电量占当月用电量的比例在35%到50%之间（不含35%和50%）．

9、（北京四中2011中考模拟12）“五一”期间，某商场搞优惠促销，决定由顾客抽奖确定

折扣．某顾客购买甲、乙两种商品，分别抽到七折（按售价的70％销售）和九折（按售价的90％销售），共付款386元，这两种商品原销售价之和为500元．问：这两种商品的原销售价分别为多少元？

答案：解：设甲、乙两种商品的原销售价分别为x ，y 元，根据题意，得：

?=+=+3869.07.0500y x y x ，解得：?

??==180320

y x 答：甲、乙两种商品的原销售价分别为320元,180元。

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