Esercizio. 
Risolvere la seguente equazione
![\[\left(\dfrac{1}{5}\right)^{-1} \left[x+\dfrac{x-1}{2}-\dfrac{x-3}{3}-\left(x-\dfrac{x+1}{2}+\dfrac{x+3}{3}\right)\right] = \dfrac{x+1}{6} - (-5)^{-1} \left(\dfrac{x-2}{3}+\dfrac{x+3}{2}\right)\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-bb14fe85d8d187f0712e6523bd7c8b71_l3.png "Rendered by QuickLaTeX.com")
Soluzione.
\textbf{Soluzione.}\\
\textbf{Soluzione.}\\
![\[\begin{aligned} & \left(\dfrac{1}{5}\right)^{-1} \left[x+\dfrac{x-1}{2}-\dfrac{x-3}{3}-\left(x-\dfrac{x+1}{2}+\dfrac{x+3}{3}\right)\right] = \dfrac{x+1}{6} - (-5)^{-1} \left(\dfrac{x-2}{3}+\dfrac{x+3}{2}\right) \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad 5 \left[\dfrac{7x+3}{6}-\dfrac{5x+3}{6}\right] = \dfrac{x+1}{6}+\dfrac{1}{5} \; \dfrac{5(x+1)}{6} \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad \dfrac{5}{6} \cdot 2x = \dfrac{x+1}{6}+ \dfrac{x+1}{6} \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad \dfrac{5}{6} \cdot 2x = 2 \cdot \dfrac{x+1}{6}\quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad \dfrac{4}{3}x = \dfrac{1}{3} \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad x= \dfrac{1}{4} \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b87260ee01d2f4e57fabbbb8b0aa43c3_l3.png "Rendered by QuickLaTeX.com")
Fonte: Moduli di lineamenti di matematica – N.Dodero, P.Baroncini e R.Manfredi
Esercizio 36 – Equazione di primo grado
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Fonte: Moduli di lineamenti di matematica – N.Dodero, P.Baroncini e R.Manfredi