题目我国明代数学读本《算法统宗》有一道题,其题意为:客人一起分银子,若每人777两,还剩444两;若每人999两,则差888两.若客人为xxx人,银子为yyy两,可列方程组( )A.{7x+4=y9x−8=y\left\{\begin{array}{l}7x+4=y\\ 9x-8=y\end{array}\right.{7x+4=y9x−8=yB.{7x−4=y9x+8=y\left\{\begin{array}{l}7x-4=y\\ 9x+8=y\end{array}\right.{7x−4=y9x+8=yC.{7y+4=x9y−8=x\left\{\begin{array}{c}7y+4=x\\ 9y-8=x\end{array}\right.{7y+4=x9y−8=xD.{7y−4=x9y+8=x\left\{\begin{array}{c}7y-4=x\\ 9y+8=x\end{array}\right.{7y−4=x9y+8=x知识点:由实际问题抽象出一元一次方程章节:方程与不等式 / 一元一次方程