Question

9．計算：$\left\{\begin{array}{l}{x+y+z=0}\\{4x+2y+z=3}\\{9x+3y+z=6}\end{array}\right.$．

Answer 1

分析 ②-①得出3x+y=3，③-①得出4x+y=3，組成二元一次方程組，求出方程組大的解，再求出z即可．

解答 解：$\left\{\begin{array}{l}{x+y+z=0①}\\{4x+2y+z=3②}\\{9x+3y+z=6③}\end{array}\right.$
②-①得：3x+y=3④，
③-①得：8x+2y=6，
4x+y=3⑤，
由④和⑤組成方程組$\left\{\begin{array}{l}{3x+y=3}\\{4x+y=3}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$，
把$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$代入①得：0+3+z=0，
解得：z=-3，
所以原方程組的解為$\left\{\begin{array}{l}{x=0}\\{y=3}\\{z=-3}\end{array}\right.$．

點評 本題考查了解三元一次方程組，能把三元一次方程組轉(zhuǎn)化成二元一次方程組是解此題的關(guān)鍵．