直线ax+2y+1+8=0,4x+3y=10,2x-y=10相交一点,
问题描述:
直线ax+2y+1+8=0,4x+3y=10,2x-y=10相交一点,
答
先算4x+3y=10,2x-y=10得值,求得:x=4,y=-2后,代入ax+2y+1+8=0,求得a=-5/4
直线ax+2y+1+8=0,4x+3y=10,2x-y=10相交一点,
先算4x+3y=10,2x-y=10得值,求得:x=4,y=-2后,代入ax+2y+1+8=0,求得a=-5/4