[tex]\it 4x+(x+1)+(x+2)+(x+3)+\ ...\ +(x+10)+120=203|_{-120}\ \Rightarrow\\ \\ \\\Rightarrow4x+\underbrace{x+x+x+\ ...\ +x}_{10\ termeni}+1+2+3+\ ...\ +10=83\ \Rightarrow\\ \\ \\ \Rightarrow14x+\dfrac{10\cdot11}{2}=83 \Rightarrow\ 14x+55=83|_{-55}\ \Rightarrow 14x=28|_{:14}\ \Rightarrow\ x=2[/tex]
