Question

12．已知不等式組$\left\{\begin{array}{l}{x-2a+b＜0}\\{2x+3a-5b＞0}\end{array}\right.$的解集為-1＜x＜6，求a-4b的值．

Answer 1

分析 首先解兩個(gè)不等式可得x＜2a-b，x＞$\frac{5b-3a}{2}$，再由解集為-1＜x＜6可得方程組$\left\{\begin{array}{l}{2a-b=6}\\{\frac{5b-3a}{2}=-1}\end{array}\right.$，再解方程組即可．

解答 解：$\left\{\begin{array}{l}{x-2a+b＜0①}\\{2x+3a-5b＞0②}\end{array}\right.$，
由①得：x＜2a-b，
由②得：x＞$\frac{5b-3a}{2}$，
∵解集為-1＜x＜6，
∴$\left\{\begin{array}{l}{2a-b=6}\\{\frac{5b-3a}{2}=-1}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=4}\\{b=2}\end{array}\right.$，
則a-4b=4-8=-4．

點(diǎn)評(píng) 此題主要考查了解一元一次不等式組，關(guān)鍵是正確計(jì)算出兩個(gè)不等式的解集，得到方程組$\left\{\begin{array}{l}{2a-b=6}\\{\frac{5b-3a}{2}=-1}\end{array}\right.$．