题目解下列方程组：(1){3(x−1)=y+5,5(y−1)=3(x+5).(1)\begin{cases}3(x-1)=y+5,\\5(y-1)=3(x+5).\end{cases}(1){3(x−1)=y+5,5(y−1)=3(x+5). (2){2u3+3v4=124u5+5v6=715.(2)\left\{\begin{array}{l}\dfrac{2u}{3}+\dfrac{3v}{4}=\dfrac{1}{2}\\ \dfrac{4u}{5}+\dfrac{5v}{6}=\dfrac{7}{15}.\\\end{array}\right.(2)⎩⎨⎧32u+43v=2154u+65v=157.知识点：解二元一次方程组——加减消元法章节：方程与不等式 / 二元一次方程组
