一道数学分式方程式计算题

x^2+7x-8=(x-1)(x+8)
x^2+x+3=（x+2）(x-1)+5
2x^2-x+7=(2x-7)(x+3)+28
原式=（x+8）+(x-1)+[5/(x+2)]+(2x-7)+[28/(x+3)]=4x
化简得[5/(x+2)]+[28/(x+3)]=0
两边同时乘以(x+2)(x+3)得
5(x+3)+28(x+2)=0
x=-71/33