已知方程组x+3y-z=7,2x+y+3z=5,求x+13y-11z的值
解: x+3y-z=7......................(1)
2x+y+3z=5....................(2)
(1)*3+(2),消去z得
3x+9y-3z+(2x+y+3z)=7*3+5
5x+10y=26
x+2y=5.2
x=5.2-2y..........................(3)
将(3)代入(1)中得
5.2-2y+3y-z=7
y=7-5.2+z
y=1.8+z.............................(4)
将(4)代入(3)得
x=5.2-2(1.8+z)
x=5.2-3.6-2z
x=1.6-2z...........................(5)
将(4)(5)代入 x+13y-11z 得
x+13y-11z
=(1.6-2z)+13(1.8+z)-11z
=1.6-2z+13*1.8+13z-11z
=1.6+13*1.8
=1.6+23.4
=25
答:x+13y-11z的值为25
解答:x+3y-z=7 ①
2x+y+3z=5 ②
①*2-②得
5y-5z=9,
y-z=1.8;
①*3+② 得,
5x+10y=26,
x+2y=5.2。
∴原式=x+13y-11z=(x+2y)+11(y-z)
=5.2+11*1.8
=5.2+19.8
=25。
x+3y-z=7 -------- ①
2x+y+3z=5 -------- ②
2①-②
5y-5z=9
y-z=9/5
x+13y-11z
=x+3y-z+10y-10z
=7+10y-10z
=7+10(y-z)
=7+10x9/5
=7+18
=25
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