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lica la regola del trasporto e separa i termi- ti e i termini incogniti nelle seguenti equa- Attenzione ai segni! \( 9 x+3+x=15+7 x \) \( 1+12 x-19=9 x-10 \) \( 20-7 x-11=-4 x+2 \) \( 14-10 x-8=6-14 x \)

Ask by Rodriquez Hodgson. in Italy
Mar 13,2025

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**Risultati delle equazioni:** 1. **Equazione 1:** \( x = 4 \) 2. **Equazione 2:** \( x = \frac{8}{3} \) 3. **Equazione 3:** \( x = \frac{7}{3} \) 4. **Equazione 4:** \( x = 0 \)

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Iniziamo a risolvere la prima equazione: \( 9x + 3 + x = 15 + 7x \). Combinando i termini simili, abbiamo \( 10x + 3 = 15 + 7x \). Ora, si spostano i termini con \( x \): \( 10x - 7x = 15 - 3 \), che porta a \( 3x = 12 \) e, infine, \( x = 4 \). Passiamo alla seconda: \( 1 + 12x - 19 = 9x - 10 \). Riscriviamo l'equazione totalizzando i termini noti: \( 12x - 18 = 9x - 10 \). Isoliamo \( x \): \( 12x - 9x = -10 + 18 \) si riduce a \( 3x = 8 \), quindi \( x = \frac{8}{3} \). Adesso affrontiamo la terza equazione: \( 20 - 7x - 11 = -4x + 2 \). Riorganizzando otteniamo \( 9 - 7x = -4x + 2 \), quindi \( -7x + 4x = 2 - 9 \). Questo si semplifica in \( -3x = -7 \), che quindi ci dà \( x = \frac{7}{3} \). Infine, lavoriamo sull'ultima: \( 14 - 10x - 8 = 6 - 14x \). Portiamo tutto a sinistra per ottenere \( 6 - 10x = 6 - 14x \). Spostiamo nuovamente i termini: \( -10x + 14x = 6 - 6 \), che diventa \( 4x = 0 \) quindi \( x = 0 \). Ecco i valori di \( x \) per le quattro equazioni!

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