解三元一次方程组(x-1)/3=(y-3)/4=(z+2)/5 2x-3y+2z=1
问题描述:
解三元一次方程组(x-1)/3=(y-3)/4=(z+2)/5 2x-3y+2z=1
用加减消元法或者代入消元法,答案一定要准确,检验过.(不要设k法)
答
(x-1)/3=(y-3)/44x-4=3y-93y=4x+5y=(4x+5)/3(x-1)/3=(z+2)/55x-5=3z+6z=(5x-11)/3代入2x-3y+2z=12x-(4x+5)+2(5x-11)/3=1-2x-5+10x/3-22/3=14x/3=40/3所以x=10y=(4x+5)/3=15z=(5x-11)/3=13