X is a real no. and -5< x ≤ 3 is the solution to the two simulyaneous inequalities ax+15>3x and 8-x ≥ bx-10 where a and b are positive contants such that a>3 and b>0, find the possible values of a and b
Hello !
$\displaystyle ax+15>3x \implies x(a-3)>-15$
Since a>3, a-3>0, then we can divise by (a-3) without changing the direction of the inequation :
$\displaystyle x>\tfrac{-15}{a-3}$
But $\displaystyle x \le 3$
$\displaystyle \implies \boxed{3} \ge x > \boxed{\tfrac{-15}{a-3}}$
$\displaystyle 3>\frac{-15}{a-3} \implies a-3>-5 \implies a>-2$, i.e. $\displaystyle a \in (-2; +\infty)$
But we wanted $\displaystyle a \in (3; +\infty)$
Thus the solution is $\displaystyle a \in (-2; +\infty) \cap (3; +\infty) \Leftrightarrow a \in (3; +\infty)$
Solution is a>3
I think what the question wants you to do is find values for a and b so that solving those inequalities would get you -5< x ≤ 3.X is a real no. and -5< x ≤ 3 is the solution to the two simulyaneous inequalities ax+15>3x and 8-x ≥ bx-10 where a and b are positive contants such that a>3 and b>0, find the possible values of a and b
So after you do this part You just have to solve $\displaystyle \frac{-15}{a-3} = -5$ (because x>-5) and then repeat the process for b.