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Esercizio 35 – Equazione di primo grado

Equazioni di primo grado: equazioni

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Esercizio. (\bigstar\bigstar\bigstar)

Risolvere la seguente equazione

    \[-3^2 \left[x-\left(\dfrac{x-1}{2}-\dfrac{x-3}{3}\right)\right]+\left[2x-\left(\dfrac{x+3}{2}-\dfrac{x-1}{3}\right)\right]:\left(\dfrac{3}{2}+\dfrac{1}{3}\right)^2=\dfrac{7x-3}{22}\]

 

Soluzione.
\textbf{Soluzione.}\\

    \[\begin{aligned} & -3^2 \left[x-\left(\dfrac{x-1}{2}-\dfrac{x-3}{3}\right)\right]+\left[2x-\left(\dfrac{x+3}{2}-\dfrac{x-1}{3}\right)\right]:\left(\dfrac{3}{2}+\dfrac{1}{3}\right)^2=\dfrac{7x-3}{22} \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad -9 \left[x-\dfrac{3x-3-2x+6}{6}\right]+\left[2x-\left(\dfrac{3x+9-2x+2}{6}\right)\right]:\left(\dfrac{11}{6}\right)^2=\dfrac{7x-3}{22} \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad -9 \left[x-\dfrac{x+3}{6}\right]+\left[2x-\left(\dfrac{x+11}{6}\right)\right]\cdot \dfrac{6^2}{11^2}=\dfrac{7x-3}{22} \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad -9 \left[\dfrac{6x-x-3}{6}\right]+\left[\dfrac{12x-x-11}{6}\right]\cdot \dfrac{6^2}{11^2}=\dfrac{7x-3}{22} \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad -\dfrac{3}{2} \left( 5x-3\right)+ \dfrac{6}{11}(x-1)=\dfrac{7x-3}{22} \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad \dfrac{-15x+9}{2} + \dfrac{6x-6}{11}-\dfrac{7x-3}{22}=0 \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad \dfrac{-165x+99+12x-12-7x+3}{22}=0 \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad -160x + 90=0 \quad \Leftrightarrow \quad\\\\ & \quad \Leftrightarrow \quad x = \dfrac{9}{16} \end{aligned}\]

 


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