Respuesta :

well, the idea is, first off, we move the "recurring" or repetitive digits to the left-side of the decimal point.  So, in this case is just one number, is 3, so, we'll move it to the left by simply multiplying the number by a power of 10, so just 10 in this case then.

now, let's make x = 4.3333333....

[tex]\bf x=4.3333\overline{3}\qquad then\qquad \begin{array}{lclll} 10\cdot x\implies &43.3333\overline{3}\\ &\downarrow \\ &39+4.3333\overline{3}\\ &\downarrow \\ &39+x \end{array} \\\\\\ thus\qquad 10x=39+x\implies 9x=39\implies x=\cfrac{39}{9}[/tex]

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