# How do you solve for x in 3x-5 < x + 9 \le 5x + 13 ?

Oct 31, 2014

By separating into two inequalities,

$\left\{\begin{matrix}3 x - 5 < x + 9 \\ x + 9 \le 5 x + 13\end{matrix}\right.$

Let us work on the first inequality.

$3 x - 5 < x + 9$

by subtracting $x$,

$\implies 2 x - 5 < 9$

$\implies 2 x < 14$

by dividing by $2$,

$\implies x < 7$

Let us work on the second inequality.

$x + 9 \le 5 x + 13$

by subtracting $x$,

$\implies 9 \le 4 x + 13$

by subtracting $13$,

$\implies - 4 \le 4 x$

by dividing by $4$,

$\implies - 1 \le x$

By combining the two inequalities, we have

$- 1 \le x < 7$,

or in interval notation, the solution set is

$\left[- 1 , 7\right)$.

I hope that this was helpful.