用代入法解下列方程以及解釋

3(x1)=4(7-y) <--- 乜野係 "(x1)"? (x+1) or (x-1) ??

2008-05-24 15:16:09 補充：
其實做法全部一樣, 都係先在其中一條方程式中將x (y亦可, 隨你喜歡)化做一個包含y的數值 eg. x = ???y
再將該數值代入所有x內, 就剩下只得y的方程式. 便可求到y. 跟住便可求埋x.
答案:
1. 你應該寫錯咗第 1 part. " 3(x1) = 4(7-y) " . " 3(x1) " 應漏咗d嘢. 請check 番後我再答你.
2. 2(x+4) + 3(y-3) = 12 , (3x+4)-(y-5) = 4
(3x+4)-(y-5) = 4
=> 3x+4-y+5=4
=> 3x+9-4=y
=> 3x+5=y
將(3x+5)代入第一條方程式的y內:
2(x+4)+3(y-3)=12
=>2x+8+3y-9=12
=>3y=13-2x
=>3(3x+5)=13-2x
=>9x+15=13-2x
=>9x+2x=13-15
=>11x=-2
=>x=-2/11 = -0.1813
將x代番入 y=3x+5
=>y=-0.5455+5 = 4.4545
3. (x+4)-(2y-7)=15, 4(2+x)-3(7-5y)=21
(x+4)-(2y-7)=15
=>x+4-2y+7=15
=>x-2y=15-4-7
=>x=4+2y
4(2+x)-3(7-5y)=21
=>8+4x-21+15y=21
=>4x=34-15y
=>4(4+2y)=34-15y
=>16+8y=34-15y
=>23y=18
=>y=18/23 = 0.7826
x= 4+2y = 4+0.7826*2 = 5.5652

4. 1/3x+2/5y=-7, 1/9x+3/5y=7
1/3x+2/5y=-7
=>1/3x=-7-2/5y
=>x=3(-7-2/5y)
=>x=-21-6/5y

1/9x+3/5y=7
=>1/9x=7-3/5y
=>x=63-27/5y
=>21-6/5y = 63-27/5y
=>27/5y-6/5y = 42
=>21/5y=42
=> y=42*5/21
=> y=10
x = -21-6/5y
=>x=-21-12 = -33

5. 0.7x+2/3y=1, 0.6x+1/9y=1
0.7x+2/3y=1
=>0.7x=1-2/3y
=>x= 1.4286-0.4667y
0.6x+1/9y=1
=>0.8572-0.28y+0.1111y=1
=>-0.1689y=0.1428
=>y=-0.8455
x=1.4286-0.4667*-0.8455
=>x=1.8232

樓主問緊既都應係二元一次方程掛

係3(x+1)=4(7-y)