Question

4．二元一次方程組$\left\{\begin{array}{l}{x+y=5}\\{2x-y=4}\end{array}\right.$的解為（　�。�

• A． $\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$

Answer 1

分析 根據(jù)加減消元法，可得方程組的解．

解答 解：$\left\{\begin{array}{l}{x+y=5①}\\{2x-y=4②}\end{array}\right.$
①+②，得 3x=9，
解得x=3，
把x=3代入①，
得3+y=5，
y=2，
所以原方程組的解為$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$．
故選C．

點(diǎn)評(píng) 本題考查了解二元一次方程組，掌握加減消元法是解題的關(guān)鍵．本題還可以根據(jù)二元一次方程組的解的定義，將四個(gè)選項(xiàng)中每一組未知數(shù)的值代入原方程組進(jìn)行檢驗(yàn)．