5.4解二元一次方程组——加减消元法 同步练习 2024-2025学年北京版数学七年级下册

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5.4解二元一次方程组——加减消元法 练习

一、单选题

1.用加减消元法解方程组eqIdb5f783050d5c65eeb800cd75d8ac844c下列做法正确的是(   )

A.eqId57bf697265b7ad219f5b3937eef07416 B.eqIda932a606be265439bd13c08ae13fb4a2 C.eqIdd65424f6b1b9ac0c3b98e99631f10ca9 D.eqIdb3426b3d76d00972959e8c4f01cd0662

2.如果二元一次方程组eqId1211aadc54fb5eb7437e52a290e855eb的解是二元一次方程eqIdb82fc6154c5b97fe511816ae1b37040f的一个解,那么a的值是(   )

A.eqId8b2a698891d42c70b597f0da4f215f09 B.eqId354c3a283b2b21cc8ac33995aac20a5c C.eqId4e8b4bbd2f9912adfc9864c0e1e76a9d D.eqId459c84c9addfbd1cdd0a877ba7c584e4

3.已知关于x的不等式组eqIdc1e647bf2c1329c5d5ccbf7194ee354a的解集为eqIde671f8fc15cbd125e689d3bc8caa7e68,则eqId0c6ce02259a85ea191541f4a708738f1的值为(    )

A.eqId274a9dc37509f01c2606fb3086a46f4f B.eqId075dbd0a23a952415f72ea8775934bb7 C.eqId3edbd40e04e2a943051fa83d6e511add D.eqIdb054c58a53818e31842f96a4880c2fc3

4.学科内融合  已知eqId1a10386d1e7469127d99328aa13151f9与eqIdb46dfc588a8937e6f349bdfd1c555993是同类项,那么eqId6f5abf90e8307ef8dcc1030ebe7fd06a的值为(   )

A.0 B.2 C.4 D.6

5.二元一次方程组eqId14a90e09a9305b8dcdf680f205a859f8的解为eqIddfd121f844efd916ef6b66760eb097a7,则eqId72850427e83ff19a24305783e080b280的值为(  )

A.2 B.eqId274a9dc37509f01c2606fb3086a46f4f C.1 D.eqIdacbc6a613224461ade69362d46550474

6.已知二元一次方程组eqIdf4651cadb50b81c5608e67146f2c0f20,那么eqId20d6fc9b90f370fbb27552876b650f8f的值是(    )

A.1 B.0 C.eqId274a9dc37509f01c2606fb3086a46f4f D.eqIdacbc6a613224461ade69362d46550474

7.解方程组eqIddeee7c9cc99f80e2e893f76628ce8de5时,若将①eqId4bfc339cf6dd66599db64fa3fa44e608②,可得(   )

A.eqId8d23961d04951b8b2c5830cb326a6058 B.eqIddb121f493d99aae1574d611d5cd6ec6e C.eqId16cb10cd1f9458649bbf44f093d39fbf D.eqId2f25ba6d3e3a579325ae593896016cb7

8.在解二元一次方程组时,我们的基本思路是“消元”,即通过“代入法”或“加减法”将“二元”化为“一元”,这个过程体现的数学思想是(   )

A.数形结合思想 B.分类讨论思想

C.化归思想 D.类比思想

9.用加减消元法解方程组eqIdbe1e4159c2d6de4f773a79a24d11c5d1时,将eqIdeb79943c3c752cdf4dd1529560c845ac可得(    )page number 0


A.eqId2040b4b2b5d40dcf17c1419a8e261202 B.eqId54533da5f7c7ec315f6b6eb26fde90e3 C.eqId478cac92753442683f550958e0d556ad D.eqIdaad131ed6d3425d6178637e2ce670929

10.已知实数a,b满足eqId1410aa2300567ebd33a06ee36926528d,且eqIda380067a20c25338eb0312e8df6c2760,eqIdd7925a4d2592cb38bd92cd9d6b4e732b,则eqIdc9f2416d1f75a45a314331146550832e的取值范围是(   )

A.eqId21ff3af66f3bca223ef588d5a1328365 B.eqIdc4bc18bdc498539a200114b7e8545a8c C.eqId34e7601d501092ea65275f1461a48017 D.eqIdba5651d0109d99077302ade8859b6757

11.利用加减消元法解方程组eqId45593eaceee62ff8176d8993b3229ccc下列做法正确的是(  )

A.要消去eqId81dea63b8ce3e51adf66cf7b9982a248,可以将①eqIdc02c36e7c8d1e359611ce74b032e960c②eqIdaa6c0fef4112d7a5cfd33e1732c63f7d B.要消去eqId81dea63b8ce3e51adf66cf7b9982a248,可以将①eqId64f8b0ded2b9ee73aaca247e2198a81b②eqId335680d2b7999d6fd74d1ccb7af6d426

C.要消去eqIdd053b14c8588eee2acbbe44fc37a6886,可以将①eqId839a1b08c08f684fbf031c151d0d9712②eqIdf0ae87da0696a2dca811afebaf32e312 D.要消去eqIdd053b14c8588eee2acbbe44fc37a6886,可以将①eqId839a1b08c08f684fbf031c151d0d9712②eqIdaa6c0fef4112d7a5cfd33e1732c63f7d

12.已知关于eqId81dea63b8ce3e51adf66cf7b9982a248,eqIdd053b14c8588eee2acbbe44fc37a6886的二元一次方程组eqId592cc2a3d4d1d2bb5f330c8d7ea657c4,无论eqId294f5ba74cdf695fc9a8a8e52f421328为何值,eqId81dea63b8ce3e51adf66cf7b9982a248与eqIdd053b14c8588eee2acbbe44fc37a6886的值总满足的关系式是(   )

A.eqId5558c083d34cbb0a58d3ce1dc6f5778e B.eqId3732b0e4d40be0f6be43983aad9f6d21 C.eqIdb0f71ed5c062a49e8f575dddac8fedbb D.eqId5fdef24fc79459673e51803d4a3af9b0

二、填空题

13.已知二元一次方程组eqId76aec2dc2b1c93c2385571ff5324780c,则eqId5113986d9507fc5015e71989831c128b        .

14.若关于x,y的二元一次方程组eqIda57041774bc5bf424d508b2762adfccd的解也是二元一次方程eqId6c960fc862fd36db83082f5c50cd604b的解,则eqIddd707b69a11f8de5566f23c1a2a9ff5a        .

15.已知方程组eqIda1874b47dae972fca6f783f921b631e0,则eqId070d1ea22a92808dad7489438c239629        (用只含eqId81dea63b8ce3e51adf66cf7b9982a248的代数式表示).

16.在解关于x、y的二元一次方程组eqId10588f41314cdf7f94dec32c28d1564a时,若eqId30441285f561a58be7871c4b5ba0338d可以直接消去一个未知数,则m、n之间的数量关系可以用等式表示为      .

三、解答题

17.解下列方程组

(1)eqId0e746299312582da98957f81f43a0af4

(2)eqId8dd6a807c1f40b80647f8077b51c3832page number 1


18.若关于eqIda2e647c14561826ba9e396acc5a3792c的方程组eqIdee96867f4caa793f8eda89137348dfb1与方程组eqId101f74536f397678215dca245adc8782的解相同.

(1)求两个方程组的相同解;

(2)求eqId40c209eb477b7739d12bedb2d837fda6的值.

19.已知关于x、y的二元一次方程eqIdc15fb18163df0690365a0d2e7ee88f5a的解如下表

x

…

eqId274a9dc37509f01c2606fb3086a46f4f

eqIdacbc6a613224461ade69362d46550474

0

1

2

…

eqIdc15fb18163df0690365a0d2e7ee88f5a

…

eqId4a94acdfb41489d5694b5a64b9e99754

eqId3edbd40e04e2a943051fa83d6e511add

eqIdacbc6a613224461ade69362d46550474

2

5

…

(1)求k、b的值;

(2)求当eqId914b8992c23d0835e27dced0db075ad0时x的值.

(3)直接写出eqId151aec298a17777331ffe578b5c4dedb的解集.

20.阅读材料“轮换式方程组的解法”,然后解题.

材料:在解方程组eqId2b1dd8b4ea992ce56538b50b1ec73e35时,我们可以先eqId30441285f561a58be7871c4b5ba0338d,得eqId5558c083d34cbb0a58d3ce1dc6f5778e,再eqId58b3617db740a784dc02e35309c9c585,得eqId1e230b64bc83138b314bc1dda2996487,最后重新组成方程组eqId8bdbfbf4f9f80a439222b31fae17dc02解得eqId7f1bfaeb014ce8ce3fcaae3aff1f4754

请你根据材料中的方法解方程组:eqId43ab55b49f62b244a9020d2b992f1c5e

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