Question

6．已知關(guān)于x、y的方程組$\left\{\begin{array}{l}3x-ay=16\\ bx+2y=15\end{array}\right.$的解是$\left\{\begin{array}{l}x=7\\ y=6\end{array}\right.$，求（a+10b）²-（a-10b）²的值．

Answer 1

分析 首先把$\left\{\begin{array}{l}{x=7}\\{y=6}\end{array}\right.$代入關(guān)于x、y的方程組$\left\{\begin{array}{l}{3x-ay=16}\\{bx+2y=15}\end{array}\right.$求得a、b的數(shù)值，再進(jìn)一步化簡(jiǎn)代入即可．

解答 解：把$\left\{\begin{array}{l}{x=7}\\{y=6}\end{array}\right.$代入關(guān)于x、y的方程組$\left\{\begin{array}{l}{3x-ay=16}\\{bx+2y=15}\end{array}\right.$得：$\left\{\begin{array}{l}a=\frac{5}{6}\\ b=\frac{3}{7}\end{array}\right.$，
則（a+10b）²-（a-10b）²=（a+10b+a-10b）（a+10b-a+10b）=40ab，
所以原式=40ab=$\frac{100}{7}$．

點(diǎn)評(píng) 此題考查二元一次方程組的解，解決本題的關(guān)鍵是二元一次方程組的解的概念．