已知方程组\(\left\{ {\begin{array}{*{20}{l}} {{a_1}x + {b_1}y = {c_1}} \\ {{a_2}x + {b_2}y = {c_2}} \end{array}} \right.\)的解是\(\left\{ {\begin{array}{*{20}{l}} {x = 3} \\ {y = 4} \end{array}} \right.\)，老师让同学们解方程组\(\left\{ {\begin{array}{*{20}{l}} {3{a_1}x + 4{b_1}y = 5{c_1}} \\ {3{a_2}x + 4{b_2}y = 5{c_2}} \end{array}} \right.\)，小聪先觉得这道题好象条件不够，后将方程组中的两个方程两边同除以5，整理得\(\left\{ {\begin{array}{*{20}{l}} {{a_1} \cdot \frac{3}{5}x + {b_1} \cdot \frac{4}{5}y = {c_1}} \\ {{a_2} \cdot \frac{3}{5}x + {b_1} \cdot \frac{4}{5}y = {c_2}} \end{array}} \right.\)，运用换元思想，得\(\left\{ {\begin{array}{*{20}{l}} {\frac{3}{5}x = 3} \\ {\frac{4}{5}y = 4} \end{array}} \right.\)，所以方程组\(\left\{ {\begin{array}{*{20}{l}} {3{a_1}x + 4{b_1}y = 5{c_1}} \\ {3{a_2}x + 4{b_2}y = 5{c_2}} \end{array}} \right.\)的解为\(\left\{ {\begin{array}{*{20}{l}} {x = 5} \\ {y = 5} \end{array}} \right.\)．现给出方程组\(\left\{ {\begin{array}{*{20}{l}} {{a_1}x - {b_1}y = m} \\ {{a_2}x - {b_2}y = n} \end{array}} \right.\)的解是\(\left\{ {\begin{array}{*{20}{l}} {x = 3} \\ {y = 4} \end{array}} \right.\)，请你写出方程组\(\left\{ {\begin{array}{*{20}{l}} {{a_1}x - {b_1}y - 4{a_1} - 2{b_1} = 3m} \\ {{a_2}x - {b_2}y - 4{a_2} - 2{b_2} = 3n} \end{array}} \right.\)的解___．

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