一道七年级数学应用题（物资调配问题）求解，必须用一元一次方程解，有详细的过程，越快越好！ 初一数学：实际问题与一元一次方程（人员调配问题及数字问题）
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\u8bbe\u5341\u4f4d\u4e0a\u662fX\uff0c\u4e2a\u4f4d\u4e0a\u662f2X 10X+2X+36=20X+X
\u5f97X=4 \u539f\u6570\u4e3a48
\u8bbe\u5341\u4f4d\u4e0a\u662fX \u5219\u767e\u4f4d\u662fX+7 \u4e2a\u4f4d\u4e0a\u662f3X X+X+7+3X=17 \u5f97X=2 \u4e09\u4f4d\u6570\u4e3a926
\u8bbe\u4e2d\u95f4\u6570\u4e3aX \u5219X-1+X+X+1=18 X=6 \u79ef\u4e3a5*6*7=210
如果从甲书架取50本放到乙书架，这时甲书架有书6X-700-50=(6X-750)本，乙书架有书（X+50)本，两书架的书相等，有方程：
6X-750=X+50
6X-X=50+750
5X=800
X=160
6X-70=6×160-700=960-700=260
答：原来甲书架有书260本，乙书架有书160本
设甲书架有x本书
【如果从甲架上拿100本书放到乙架上，两架上的书就相等】
那么乙书架就有 x-200 本书（x-100=乙+100）
【从乙架上拿100本放到甲架上，那么甲架上的书就是乙架上所剩余的书的5倍】
5*[（x-200）-100] = x+100
甲：x=400
乙：x-200=200
我们根据从甲架上取50本书放到乙架上，两架的书就一样多
可以知道甲比乙多50×2=100本
那么设乙有a本，则甲有a+100本
6×（a-100）=a+100+100
6a-600=a+200
5a=800
a=160本
乙有160本，甲有160+100=260本
解：设原来甲书架上有书x本，那么原来乙书架上有(x-2*50)本，则：
(x-2*50-100)(5+1)=x+100
6x-1200=x+100
5x=1300
x=260
即原来甲书架上有书260本，原来乙书架上有(x-2*50)=260-2*50=160本.
设甲有X本，从甲架上取50本书放到乙架上，两架的书就一样多，则乙有X-100本
X+100=6(X-100-100)
X+100=6X-1200
5X=1300
X=260
260-100=160
原来甲有260本，乙有160本
绛旓細璁惧疄闄呰繍閫乆澶 400梅X=400梅(X+2)+10
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绛旓細鎯婇椈5鏈12鏃ュ洓宸濇倍宸濆彂鐢熷己鐑堝湴闇囧悗锛屾煇鍦版皯鏀垮眬杩呴熷湴缁勭粐浜30鍚ㄩ鐗╁拰13鍚ㄨ。鐗╃殑鏁戠伨鐗╄祫锛屽噯澶囦簬褰撴櫄鐢ㄧ敳銆佷箼涓ょ鍨嬪彿鐨勮揣杞﹀揩閫熻繍寰鐏惧尯銆傚凡鐭ョ敳鍨嬭揣杞︽瘡杈嗗彲瑁呴鐗5鍚ㄥ拰琛ｇ墿1鍚紝涔欏瀷璐ц溅姣忚締鍙椋熺墿3鍚ㄥ拰琛ｇ墿2鍚ㄣ備絾鐢变簬鏃堕棿浠撲績锛屽彧鎷涘嫙鍒9鍚嶉暱閫旈┚椹跺憳蹇楁効鑰呫傚鏋3鍚嶉┚椹跺憳寮鐢茬璐ц溅锛6鍚...
绛旓細瑙ｏ細璁炬湁X杈嗚揣杞 20X-20=15X+10 20X-15X=20+10 5X=30 X=6 鍒欏鏍＄浜6杈嗚揣杞 20*6-20=100锛堝锛夌瓟锛氬鏍℃崘浜100濂楄妗屻
绛旓細8.鏌愪腑瀛︾粍缁囧垵涓瀛︾敓杩涜鏄ユ父锛屽師璁″垝绉熺敤45搴у杞﹁嫢骞茶締锛屼絾鏈15浜烘病鏈夊骇浣嶏紱鑻ョ鐢ㄥ悓鏍锋暟閲忕殑60搴у杞︼紝鍒欏鍑轰竴杈嗚溅锛屼笖鍏朵綑瀹㈣溅鎭板ソ鍧愭弧銆傝瘯闂 锛1锛 鍒濅竴骞寸骇浜烘暟鏄灏戯紵鍘熻鍒掔鐢45搴у杞﹀灏戣締锛熻В锛氱鐢45搴у杞杈嗭紝绉熺敤60搴у杞(x-1)杈,45x+15=60(x-1)瑙ｄ箣寰楋細x=5 45x+15=240...
