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Answer 1 · 2017-09-17T14:07:49

$[- 3, 9)$ $[- 3, 9)$

We have: $0 \leq \frac{1}{3} x + 1 < 4$ $0 le frac(1)(3) x + 1 < 4$

Let's subtract $1$ $1$ from the inequality:

$\Rightarrow 0 - 1 \leq \frac{1}{3} x + 1 - 1 < 4 - 1$ $Rightarrow 0 - 1 le frac(1)(3) x + 1 - 1 < 4 - 1$

$\Rightarrow - 1 \leq \frac{1}{3} x < 3$ $Rightarrow - 1 le frac(1)(3) x < 3$

Then, let's multiply by $3$ $3$ :

$\Rightarrow - 1 \times 3 \leq \frac{1}{3} x \times 3 < 3 \times 3$ $Rightarrow - 1 times 3 le frac(1)(3) x times 3 < 3 times 3$

$therefore - 3 le x < 9$

Therefore, the solution interval to the inequality is $[- 3, 9)$ .