B1. For which values of the real number $a$ does the expression $(a-0.\overline{3})(-3)+(-4a)$ have a value of at least -6? Present the solution graphically as well.
B1. From the problem statement, we write the inequality: $(a-0, \overline{3})(-3)+(-4 a) \geq-6$. We replace the periodic decimal number $0, \overline{3}$ with the fraction $\frac{1}{3}$. Then we remove the parentheses: $-3 a+1-4 a \geq-6$ and simplify the inequality to $-7 a \geq-7$. We divide by $-7$ and get the solution $a \leq 1$.

Writing the inequality: $(a-0, \overline{3})(-3)+(-4 a) \geq-6 \ldots$ 1 point

Writing the periodic number as a fraction $0, \overline{3}=\frac{1}{3}$ 1 point Simplifying the equation to the form $-7 a \geq-7$ 1 point

Solution $a \leq 1$

Graphical representation 2 points