Esercizio. 
Risolvere la seguente equazione
![\[\dfrac{1}{2}\left[\dfrac{x+3}{2}-\left(x-\dfrac{1}{2}\right)+\dfrac{x-2}{2}\right]+\dfrac{1}{4}x=\dfrac{x-2}{4}-\left(x-\dfrac{x-3}{2}\right) \, \dfrac{3}{4}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-f18997663fe6739a723d2e112497febc_l3.png "Rendered by QuickLaTeX.com")
Soluzione.
\textbf{Soluzione.}\\
\textbf{Soluzione.}\\
![\[\begin{aligned} & \dfrac{1}{2}\left[\dfrac{x+3}{2}-\left(x-\dfrac{1}{2}\right)+\dfrac{x-2}{2}\right]+\dfrac{1}{4}x=\dfrac{x-2}{4}-\left(x-\dfrac{x-3}{2}\right) \, \dfrac{3}{4} \quad \Leftrightarrow \quad\\ \\ & \quad \Leftrightarrow \quad \dfrac{1}{2}\left[\dfrac{x+3}{2}-\dfrac{2x-1}{2}+\dfrac{x-2}{2}\right]+\dfrac{1}{4}x=\dfrac{x-2}{4}-\dfrac{x+3}{2} \, \dfrac{3}{4} \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad \dfrac{1}{2} + \dfrac{1}{4}x = \dfrac{x-2}{4} - \dfrac{3}{8} (x+3) \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad \dfrac{4+2x}{8} = \dfrac{2x-4-3(x+3)}{8} \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad 3x = -17 \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad x=-\dfrac{17}{3} \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-14e13405981fb65327662110ed00f860_l3.png "Rendered by QuickLaTeX.com")
Fonte: Moduli di lineamenti di matematica – N.Dodero, P.Baroncini e R.Manfredi
Esercizio 33 – Equazione di primo grado
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Fonte: Moduli di lineamenti di matematica – N.Dodero, P.Baroncini e R.Manfredi