三道方程题求数学高手解答!如过程详细完整,高悬赏!!谢谢!!!!

2024-12-26 04:05:09
推荐回答(6个)
回答1:

1 解:设足球每个进价x元,篮球每个进价y元,则∶
60x+50y=5200
{
60﹙x+8%x﹚+50﹙y+12%y﹚-5200=516
解得x=45,y=50
经检验,符合题意
答∶·······················
2 解∶设乙有x本书,甲有﹙52-x﹚本书,丙有﹙68-x﹚本书,则∶
可以看出 丙的比甲的多
∴68-x=3﹙52-x﹚或68-x=3x
∴x=44或x=17
∴x=44,52-x=8,68-x=22 或x=17,52-x=35,68-x=51
答∶甲有8本书,乙有44本书,丙有22本书;
或甲有35本书,乙有17本书,丙有51本书。

3 解∶设男生x人,女生﹙79x-76x﹚/﹙76-71﹚人,则∶
380≤﹙79x-76x﹚/﹙76-71﹚+x≤450
解得∶ 237.5≤x≤281.25﹙男生﹚
142.5≤0.6x≤168.75 ﹙ 女生﹚
又∵x、0.6x必须为整数
∴x最小为238
0.6x最小为143
答:男生至少有238人参加,女生至少有143人参加。

欢迎追问,若满意望采纳O(∩_∩)O

回答2:

1.x,y 60x+50y=5200
60x(1+8%)+50y(1+12%)-5200=516
2.x,y,z(假设x>y>z)
x+y=52
y+z=68
x=3z
3.不会

回答3:

1、足球x篮球y
60x+50y=5200
60*0.08x+50*0.12y=516
2、甲乙丙各有xyz
x+y=52
y+z=68
z=3x
3、男女各有xy
(x+y)*76=79x+71y
380

回答4:

解:设足球原单价X元,篮球Y元。

{4.8x+6y=516
{60x+50y=5200

x=45 y=50
答:足球原价45元,篮球原价50元。
(你有时间限制,所以我只能做一道题。。。)
(如果你提高悬赏分,我会继续做。。。)

回答5:

3.设男生为x,女生为y,则76(x+y)=79x+71y,则:3x=5y,即x:y=5:3,用380除以8,得48,缺4,所以至少需要384人。

回答6:

1.设足球X个,篮球Y个,则由题意得60X+50Y=5200和(108%X*60+112%Y*50)-5200=516
作方程组解。没时间了I'M SORRY

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