题目内容
10.若方程组$\left\{\begin{array}{l}{ax+by=1}\\{x+2by=-3c}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$,则a+b+c=( )| A. | $\frac{2}{3}$ | B. | $\frac{1}{2}$ | C. | -$\frac{2}{3}$ | D. | -2 |
分析 把x与y的值代入方程组求出a,b,c的值,即可求出a+b+c的值.
解答 解:把$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$代入方程组得:$\left\{\begin{array}{l}{3a+b=1}\\{3+2b=-3c}\end{array}\right.$,
整理得:$\left\{\begin{array}{l}{3a+b=1①}\\{2b+3c=-3②}\end{array}\right.$,
①+②得:3a+3b+3c=-2,
则a+b+c=-$\frac{2}{3}$,
故选C.
点评 此题考查了二元一次方程组的解,方程组的解即为能使方程组中两方程都成立的未知数的值.
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