標題:
maths(form 2english)
發問:
solve the following simultaneous equations. 1.p-q=0 p+4q+1=0 2.x+y=3 5x=3y-1 3.2c+5d=27 c+2d=9 4.3y-5x=4 5x-y=2 請詳細回答，，ｔｈｘ～～
最佳解答:
1.p-q=0...(1) p+4q+1=0...(2) in(1)..p=q...(3) sub(3)into(2) q+4q+1=0 5q=-1 q=-0.2 p=-0.2 2.x+y=3...(1) 5x=3y-1...(2) in(1) x=3-y..(3) sub(3)into(2) 5(3-y)=3y-1 15-5y=3y-1 2y=16 y=8...(4) sub(4)into (1) X+8=3 x=-5 3.2c+5d=27...(1) c+2d=9..(2) in(2) c=9-2d...(3) sub(3)into(1) 2(9-2d)+5d=27 18-4d+5d=27 d=9...(4) sub(4)into(2) c+18=9 c=-9 4.3y-5x=4..(1) 5x-y=2...(2) in(2) 5x=2+y..(3) sub(3)into (1) 3y-(2+y)=4 3y-2-y=4 2y=6 y=3...(4) sub(4)into(2) 5x-3=2 5x=5 x=1
其他解答:
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1. p-q=0...1 p+4q+1=0...2 From (1), p-q =0 p=q.....3 substitute (3) into (2), p+4p+1=0 5p=-1 p= -1/5 Sincep=q, q=-1/5 2. x+y=3...1 5x=3y-1...2 From (1), x+y=3 x=3-y.....3 substitute (3) into (2), 5(3-y)=3y-1 15-5y=3y-1 8y=16 y=2 Since x=3-y x=1 ３. 2c+5d=27...1 c+2d=9....2 From (2), c+2d=9 c=9-2d.....3 substitute (3) into (1), 2c+5d=27 2(9-2d)+5d=27 18-4d+5d =27 d =9 Since c=9-2d c=-9 4.3y-5x=4....1 5x-y=2.......2 From (2), 5x-y=2 y=5x-2.........3 substitute (3) into (1), 3(5x-2)-5x=4 15x-6-5x=4 10x=10 x=1 Since y=5x-2 y=3|||||1.p-q=0----(1) p+4q+1=0-----(2) (1): p=q sub into (2) q+4q+1=0 5q=1 q=1/5 p=1/5 2.x+y=3----(1) 5x=3y-1----(2) (1): x=3-y sub into (2) 5(3-y)=3y-1 15-5y=3y-1 16=8y y=2 x=3-y x=3-2 x=1 3.2c+5d=27-----(1) c+2d=9------(2) (2): c=9-2d sub into (1) 2(9-2d)+5d=27 18-4d+5d=27 18-d=27 -d=9 d=-9 c=9-2d c=9-2(-9) c=9-(-18) c=27 4.3y-5x=4----(1) 5x-y=2----(2) (2): 5x-y=2 -y=2-5x y=5x+2 sub into (1) 3(5x+2)-5x=4 15x+6-5x=4 10x+6=4 10x=-2 x=-2/10 x=-1/5 y=5x+2 y=5(-1/5)+2 y=-1+2 y=1
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