Esercizio 37 – Equazione di primo grado

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Esercizio. (\bigstar\bigstar\bigstar)

Risolvere la seguente equazione

    \[\left[x-6\left(x-\dfrac{1}{3}\right)\right]:\left(\dfrac{2}{3}-\dfrac{1}{2}\right)+\left(x-\dfrac{4x+1}{3}\right):\left(-\dfrac{1}{6}\right) = \left(2x+\dfrac{1-2x}{2}\right):\left(\dfrac{1}{3}-\dfrac{1}{2}\right)-5x\]

 

Soluzione.
\textbf{Soluzione.}\\

    \[\begin{aligned} & \left[x-6\left(x-\dfrac{1}{3}\right)\right]:\left(\dfrac{2}{3}-\dfrac{1}{2}\right)+\left(x-\dfrac{4x+1}{3}\right):\left(-\dfrac{1}{6}\right) = \left(2x+\dfrac{1-2x}{2}\right):\left(\dfrac{1}{3}-\dfrac{1}{2}\right)-5x \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad \left(x-6x+2\right):\left(\dfrac{4-3}{6}\right)+\left(\dfrac{3x-4x-1}{3}\right) \cdot \left(-6\right) = \left(\dfrac{4x+1-2x}{2}\right):\left(-\dfrac{1}{6}\right)-5x \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad \left(2-5x\right) \cdot 6 + 2(x+1) = -3(2x+1)-5x \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad 12-30x+2x+2=-6x-3-5x \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad -17x=-17 \quad \Leftrightarrow \quad \\\\ & \quad \Leftrightarrow \quad x=1 \end{aligned}\]

 


Fonte: Moduli di lineamenti di matematica – N.Dodero, P.Baroncini e R.Manfredi