# Thread: Help!very eager to know the way to solve it!

1. ## Help!very eager to know the way to solve it!

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

2. Hello !
Originally Posted by helloying
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
$\displaystyle ax+15>3x \implies x(a-3)>-15$
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

3. Originally Posted by Moo
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 $\displaystyle x \ge 3$

$\displaystyle \implies \boxed{3} \ge x > \boxed{\tfrac{15}{a-3}}$

$\displaystyle 3>\frac{15}{a-3} \implies a-3>5 \implies a>8$

The only condition for a was that a>3, which is the case here. So the solution is $\displaystyle \boxed{a>8}$

Can you do it for y ? It's exactly the same reasoning.

Shouldnt it be x(a-3)>-15 because it have been brought over? so the ans will be different but I couldnt get the ans if I made it to -15.

4. Originally Posted by helloying
Shouldnt it be x(a-3)>-15 because it have been brought over? so the ans will be different but I couldnt get the ans if I made it to -15.
I guess it wasn't my day at all
I edited it

Ok, it looks like there are problems with your exercises, can't you check if you've copied it right ?

5. 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
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.
So after you do this part

Since a>3, a-3>0, then we can divise by (a-3) without changing the direction of the inequation :
You just have to solve $\displaystyle \frac{-15}{a-3} = -5$ (because x>-5) and then repeat the process for b.