## Problema 3

Notăm $a \odot b=2 a+3 b-5$ unde $a, b \in \mathbb{N}^{*}$.

a) Calculaţi 5 ○ 7

b) Rezolvaţi ecuaţia : $13 \cdot(2 \odot 2) \cdot(x \odot 8)=2015$


## Problema 3

a) $\boldsymbol{a} \odot \boldsymbol{b}=\mathbf{2 a}+\mathbf{3} \boldsymbol{b}-\mathbf{5}=>5 \bigcirc 7=2 \cdot 5+3 \cdot 7-5=\mathbf{2 6}$

b) $13 \cdot(2 \odot 2) \cdot(x \odot 8)=2015=>13 \cdot(2 \cdot 2+3 \cdot 2-5)(2 x+3 \cdot 8-5)=2015$

![](https://cdn.mathpix.com/cropped/2024_06_07_dfc9162f0525ae4a7f10g-2.jpg?height=66&width=1337&top_left_y=2011&top_left_x=363)

