y=3x-5与y=-2x+10的交点为a,联立y=3x-5y=-2x+10得x=3,y=4直线过点（3,4）因为y=-2x+10交x轴于c,则c=5因为s△abc=12而s△abc+s△boc=s△abo+s△aocs△abo+s△aoc=1/2*b*3+1/2*5*4=3b/2+10s△boc=1/2*b*5则有1/2*b*5+1...