方程组﹛y=2x﹢m x/3－y/2＝1的两组解分别为（x1,y1),(x2,y2),且（x1-x2)＋(y1－y2)＝6,求m的值拜托了各

题目

方程组﹛y=2x﹢m x/3－y/2＝1的两组解分别为（x1,y1),(x2,y2),且（x1-x2)＋(y1－y2)＝6,求m的值拜托了各

答案

y1=2x1+m,y2=2x2+m (x1-x2)+(y1－y2)=(x1-x2)+(2x1+m-2x2-m)=5(x1-x2)=6,∴(x1-x2)=6/5 联立y=2x+m,x/3-y/2=1 10x+12mx+3m+6=0 ∴x1x2=(3m+6)/10,x1+x2=-12m/10=-6m/5 ∴(x1-x2)=(x1+x2)-4x1x2=36m^2/25-2(3m+6)/5=(6m-60)/25=6/5 ∴m=15,m=±√15