Question

При всех значениях параметра а решите уравнение
1)3ax-2x+1,5a=3x+7
2)3(a-5)x=8-1,1a

Answer 1

$1) 3ax-2x+1,5a=3x+7 \\ 3ax+1,5a=5x+7 \\ 1,5a(2x+1)=5x+7 \\ \frac{3}{2}a=\frac{5x+7}{2x+1} \\ a=\frac{2(5x+7)}{3(2x+1)} \\ (1) a=0, \\ 5x+7=0 \\ 5x=-7 \\ x=-\frac{7}{5} \\ (2) a<0, \\ \frac{2*5(x+\frac{7}{5})}{3*2(x+\frac{1}{2})}<0 \\ x \in (-\frac{7}{5};-\frac{1}{2}) \\ (3) a>0, \\ \frac{2*5(x+\frac{7}{5})}{3*2(x+\frac{1}{2})}>0 \\ x \in (-\infty ; -\frac{7}{5}) \cup (-\frac{1}{2};+\infty ) \\ \\ \\ 2) 3(a-5)x=8-1,1a \\ 3ax-15x=8-1,1a \\ 3ax+1,1a=15x+8 \\ a(3x+1,1)=15x+8 \\ a=\frac{15x+8}{3x+1,1} \\ (1) a=0, \\ 15x+8=0 \\ 15x=-8 \\ x=-\frac{8}{15} \\ (2) a<0, \\ \frac{15(x+\frac{8}{15})}{3(x+\frac{11}{30})}<0 \\ x \in (-\frac{8}{15};-\frac{11}{30}) \\ (3) a>0, \\ \frac{15(x+\frac{8}{15})}{3(x+\frac{11}{30})}>0 \\ x \in (-\infty; -\frac{8}{15}) \cup (-\frac{11}{30};+\infty)$