標題:
救命啊﹗﹗﹗數學吾識﹗大家救我
發問:
The sum of two numbers is 27 and one of them exceeds twice the other by 3. One number is 5 times another and their difference is 72. 用數學式回答，例如：「1+2-1」 更新: and The sum of two number is 63 and one of them exceeds the other by7
最佳解答:
1) Let x and y be the two numbers x+ y = 27 as per your instruction, x-2y =3 therefore x = 3+2y to substitue: (3+2y) + y = 27 therefore: 3 + 3y = 27 y= 8// x= 19// 2) Let x and y be the two numbers x - y = 72 as per your instruction: x= 5 y to substitute: 5y-y =72 4y= 72 y= 18// therefore: x = 90// 3) Let x and y be the two numbers x + y = 63 as per your instruction: y-x =7, y=7+x to substitute: x+(7+x) =63 2x= 63-7 x=28// therefore: y= 35//
The first question The sum of two numbers is 27 and one of them exceeds twice the other by 3. let X be the first number and Y be the second number X+Y = 27 -->1 Y= 2X+3 -->2 Sub 2 into 1 X+2X+3 = 27 --> 3 3X=24 X=8 Sub X=8 into 2 Y= 16+3 Y=19 Checking 19+8 = 27 Second question One number is 5 times another and their difference is 72. let the first number be X and second number be Y X = 5Y -->1 Y-X=72 -->2 Sub 1 into 2 -4Y= 72 Y = -18 Sub Y=-18 into 2 -18-X=72 -X=90 X= -90 Checking -90-(-18) = -90+18 = -72 third question The sum of two number is 63 and one of them exceeds the other by7 let the first number be X and second number be Y X+Y =63 -->1 X-Y=7 -->2 From 2: X=7+Y -->3 Sub 3 into 1 7+Y+Y =63 2Y= 56 Y= 28 Sub Y=28 into 3 X=7+28 X= 35 Checking 35+28 = 63 2008-07-22 22:53:05 補充： next time give more point...3 question for 5 point...i consider kind doing for u ^^" hehe
救命啊﹗﹗﹗數學吾識﹗大家救我
發問:
The sum of two numbers is 27 and one of them exceeds twice the other by 3. One number is 5 times another and their difference is 72. 用數學式回答，例如：「1+2-1」 更新: and The sum of two number is 63 and one of them exceeds the other by7
最佳解答:
1) Let x and y be the two numbers x+ y = 27 as per your instruction, x-2y =3 therefore x = 3+2y to substitue: (3+2y) + y = 27 therefore: 3 + 3y = 27 y= 8// x= 19// 2) Let x and y be the two numbers x - y = 72 as per your instruction: x= 5 y to substitute: 5y-y =72 4y= 72 y= 18// therefore: x = 90// 3) Let x and y be the two numbers x + y = 63 as per your instruction: y-x =7, y=7+x to substitute: x+(7+x) =63 2x= 63-7 x=28// therefore: y= 35//
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其他解答:The first question The sum of two numbers is 27 and one of them exceeds twice the other by 3. let X be the first number and Y be the second number X+Y = 27 -->1 Y= 2X+3 -->2 Sub 2 into 1 X+2X+3 = 27 --> 3 3X=24 X=8 Sub X=8 into 2 Y= 16+3 Y=19 Checking 19+8 = 27 Second question One number is 5 times another and their difference is 72. let the first number be X and second number be Y X = 5Y -->1 Y-X=72 -->2 Sub 1 into 2 -4Y= 72 Y = -18 Sub Y=-18 into 2 -18-X=72 -X=90 X= -90 Checking -90-(-18) = -90+18 = -72 third question The sum of two number is 63 and one of them exceeds the other by7 let the first number be X and second number be Y X+Y =63 -->1 X-Y=7 -->2 From 2: X=7+Y -->3 Sub 3 into 1 7+Y+Y =63 2Y= 56 Y= 28 Sub Y=28 into 3 X=7+28 X= 35 Checking 35+28 = 63 2008-07-22 22:53:05 補充： next time give more point...3 question for 5 point...i consider kind doing for u ^^" hehe
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