如果方程组3x+2y=a-1,2x+my=4-3a的解满足x<y,求a的取值范围
问题描述:
如果方程组3x+2y=a-1,2x+my=4-3a的解满足x<y,求a的取值范围
答
3x+2y=a-1①,2x+my=4-3a②①*2-②*3得(4-3m)y=11a-14解得y=(11a-14)/(4-3m)把y=(11a-14)/(4-3m)带入①,得x=(4a-3am+3m-4)/(12+9m)因为x<y,则(4a-3am+3m-4)/(12+9m)<(11a-14)/(4-3m)解得a太麻烦了,多了个m...