题目用加减消元法解方程组{2x+3y=13x−2y=8\left\{\begin{array}{l}2x+3y=1\\ 3x-2y=8\end{array}\right.{2x+3y=13x−2y=8时，对于下列四种变形方法：(1){4x+6y=19x−6y=8(1)\left\{\begin{array}{l}4x+6y=1\\ 9x-6y=8\end{array}\right.(1){4x+6y=19x−6y=8，(2){6x+9y=26x−4y=24(2)\left\{\begin{array}{l}6x+9y=2\\ 6x-4y=24\end{array}\right.(2){6x+9y=26x−4y=24，(3){6x+9y=3−6x+4y=−16(3)\left\{\begin{array}{l}6x+9y=3\\ -6x+4y=-16\end{array}\right.(3){6x+9y=3−6x+4y=−16，(4){4x+6y=29x−6y=24(4)\left\{\begin{array}{l}4x+6y=2\\ 9x-6y=24\end{array}\right.(4){4x+6y=29x−6y=24其中正确的是()(\quad)()A.(1)(2)(1)(2)(1)(2)B.(3)(4)(3)(4)(3)(4)C.(1)(3)(1)(3)(1)(3)D.(4)(4)(4)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
