Question

11．已知方程組$\left\{\begin{array}{l}{2x-y=1}\\{2x+y=m}\end{array}\right.$的解$\left\{\begin{array}{l}{x≤1}\\{y＞\frac{1}{2}}\end{array}\right.$．求m的取值范圍．

Answer 1

分析 先用m表示出x，y的值，再代入得出m的取值范圍即可．

解答 解：解方程組$\left\{\begin{array}{l}{2x-y=1}\\{2x+y=m}\end{array}\right.$得：$\left\{\begin{array}{l}{x=\frac{m+1}{4}}\\{y=\frac{m-1}{2}}\end{array}\right.$，
因?yàn)榉匠探M$\left\{\begin{array}{l}{2x-y=1}\\{2x+y=m}\end{array}\right.$的解$\left\{\begin{array}{l}{x≤1}\\{y＞\frac{1}{2}}\end{array}\right.$，
可得：$\left\{\begin{array}{l}{\frac{m+1}{4}≤1}\\{\frac{m-1}{2}＞\frac{1}{2}}\end{array}\right.$，
解得：2＜m≤3．

點(diǎn)評 本題考查的是解二元一次方程及解一元一次不等式，根據(jù)題意得出關(guān)于m的不等式是解答此題的關(guān)鍵．