Question

11．已知$\left\{\begin{array}{l}{x=0}\\{y=-\frac{1}{2}}\end{array}\right.$是方程組$\left\{\begin{array}{l}{x-b=y}\\{5x+2a=2y}\end{array}\right.$的解，則代數(shù)式a+b的值為0．

Answer 1

分析 首先把$\left\{\begin{array}{l}{x=0}\\{y=-\frac{1}{2}}\end{array}\right.$代入關(guān)于x、y的方程組$\left\{\begin{array}{l}{x-b=y}\\{5x+2a=2y}\end{array}\right.$可得關(guān)于a、b的方程組$\left\{\begin{array}{l}{0-b=-\frac{1}{2}}\\{0+2a=-1}\end{array}\right.$，解方程組可得a、b的值，進(jìn)而可得代數(shù)式a+b的值．

解答 解：把$\left\{\begin{array}{l}{x=0}\\{y=-\frac{1}{2}}\end{array}\right.$代入關(guān)于x、y的方程組$\left\{\begin{array}{l}{x-b=y}\\{5x+2a=2y}\end{array}\right.$可得：$\left\{\begin{array}{l}{0-b=-\frac{1}{2}}\\{0+2a=-1}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=-\frac{1}{2}}\\{b=\frac{1}{2}}\end{array}\right.$，
把$\left\{\begin{array}{l}{a=-\frac{1}{2}}\\{b=\frac{1}{2}}\end{array}\right.$代入a+b=-$\frac{1}{2}$+$\frac{1}{2}$=0．
故答案為：0．

點(diǎn)評(píng) 此題主要考查了二元一次方程組的解，當(dāng)遇到有關(guān)二元一次方程組的解的問(wèn)題時(shí)，要回到定義中去，通常采用代入法，即將解代入原方程組，這種方法主要用在求方程中的字母系數(shù)．