Question

13．三元一次方程組$\left\{\begin{array}{l}{x+y=3}\\{y+z=4}\\{x+z=5}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=1}\\{z=3}\end{array}\right.$．

Answer 1

分析 先將三元一次方程轉(zhuǎn)化為二元一次方程組，再轉(zhuǎn)化為一元一次方程，即可解答本題．

解答 解：$\left\{\begin{array}{l}{x+y=3}&{①}\\{y+z=4}&{②}\\{x+z=5}&{③}\end{array}\right.$
①-②，得
x-z=-1④
③+④，得
x=2，
將x=2代入①，得y=1，
將x=2代入③，得z=3，
故元方程組的解是，$\left\{\begin{array}{l}{x=2}\\{y=1}\\{z=3}\end{array}\right.$，
故答案為：$\left\{\begin{array}{l}{x=2}\\{y=1}\\{z=3}\end{array}\right.$．

點(diǎn)評(píng) 本題考查解三元一次方程組，解題的關(guān)鍵是運(yùn)用消元的思想將方程由三元最終轉(zhuǎn)化為一元一次方程解答．