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標題:

maths....二元一次方程

發問:

1)The sum of the ages of Ronald and his mother is 60. Four years ago, his mother's age was 3 times that a Ronald. How old aare Ronald and his mother now?2)Rice A and rice B cost $9 per kg and $5 per kg respectively. For a 30kg mixture of rice A and rice B, the price is $7.4 per kg. What are the weights of... 顯示更多 1)The sum of the ages of Ronald and his mother is 60. Four years ago, his mother's age was 3 times that a Ronald. How old aare Ronald and his mother now? 2)Rice A and rice B cost $9 per kg and $5 per kg respectively. For a 30kg mixture of rice A and rice B, the price is $7.4 per kg. What are the weights of rice A and rice B in the mixture respectively?

最佳解答:

(1) Let the age of Ronald and his mother is x and y years respectively x + y = 60 .... eqn 1 3(x - 4) = y - 4 3x - 12 = y - 4 3x - y = 8 ... eqn 2 adding eqn 1 and eqn 2, we have 4x = 68 x = 17 sub x = 17 into eqn 1, 17 + y = 60 y = 43 Therefore Ronald and his mother are 17 y/o and 43 y/o respectively. (2) Let A and B be the weights of rice A and rice B in the mixture respectively 9A + 5B = 30 (7.4) 9A + 5B = 222 ... eqn 1 A + B = 30 ... eqn 2 eqn 1 - 5 x eqn 2, we have 4A = 72 A = 18 sub A = 18 into eqn 2, we have 18 + B = 30 B = 12 Therefore the weights of rice A and rice B in the mixture are 18kg and 12kg respectively.

其他解答:

1) let x be th age of ronald, then y is the age of his mother x+y =60 (x+4)3 =y+4 x=60-y - (1) (x+4)3 = y+4 - (2) (1) -> (2) (60-y+4)3=y+4 (64-y)3 =y+4 192-3y=y+4 188=4y y=47 -(3) x+47 =60 x=13 (so) tha age of ronald is 13,the age of his mother is 47. 2) let x be the weight of rice A, then y is the weight of riceB x+y=30 9x+5y=7.4(30) x=30-y - (1) 9x+5y=222 - (2) (1)->(2) 9(30-y)+5y=222 270-9y+5y =222 270-4y =222 48=4y y=12- (3) (3) -> (1) 12+x =30 x=18 (so)the weight of rice A is 18kg, the weight of riceB is 12 kg. 希望我冇計錯同埋幫到你啦 =)A9A3995996B435E3