Răspuns: [tex]\bf x\in (1, + \infty )[/tex]
Explicație pas cu pas:
Salutare!
[tex]\bf \dfrac{2}{3}x- \dfrac{1}{5} > \dfrac{7}{15} \:\:\:\Big|\cdot15 \:\:\:\text{\it(inmultim toata relatia cu 15)}[/tex]
[tex]\bf 15\cdot \dfrac{2}{3}x- 15\cdot\dfrac{1}{5} >7[/tex]
[tex]\bf \not15\cdot \dfrac{2}{\not3}x- \not15\cdot\dfrac{1}{\not5} >7[/tex]
[tex]\bf 5 \cdot \dfrac{2}{1}x- 3\cdot\dfrac{1}{1} >7[/tex]
[tex]\bf 10x- 3 > 7[/tex]
[tex]\bf 10x > 7+3[/tex]
[tex]\bf 10x > 10\:\:\:\Big|:10[/tex]
[tex]\bf x > 1\implies \boxed{\bf x\in (1, + \infty )}[/tex]
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