初中数学七年级题库5.11：平面直角坐标系与代数综合求值例题解析
例题：已知实数a,b满足\(\displaystyle {{\left( {2a+1} \right)}^{2}}+\left| {a+b+1} \right|=0\)，
且以关于\(\displaystyle x,y\)的方程组\(\displaystyle \left\{ \begin{array}{l}ax+by=m\\2ax-by=m+1\end{array} \right.\)
的解为坐标的点\(\displaystyle P\left( {x,y} \right)\)在x轴上，求m的值。
解析：据题意得\(\displaystyle \left\{ \begin{array}{l}2a+1=0\\a+b+1=0\end{array} \right.\)
求得\(\displaystyle \left\{ \begin{array}{l}a=-\frac{1}{2}\\b=-\frac{1}{2}\end{array} \right.\)
代入方程组中得\(\displaystyle \left\{ \begin{array}{l}-\frac{1}{2}x-\frac{1}{2}y=m\left( 1 \right)\\-x+\frac{1}{2}y=m+1\left( 2 \right)\end{array} \right.\)
所以(1)×2-(2)得\(\displaystyle -\frac{3}{2}y=m-1\)
\(\displaystyle y=-\frac{2}{3}m+\frac{2}{3}\)
因为点P（x,y)在x轴上，得其纵坐标为0
所以\(\displaystyle -\frac{2}{3}m+\frac{2}{3}=0\)
即：\(\displaystyle m=1\)
