[tex]\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2 \qquad \begin{cases} c=\stackrel{hypotenuse}{11 + x}\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{17}\\ \end{cases}\implies (11+x)^2=x^2+17^2 \\\\\\ (11+x)(11+x)=x^2+17^2\implies \stackrel{F~O~I~L}{121+22x+x^2}~~ = ~~x^2+289 \\\\\\ 121+22x=289\implies 22x=168\implies x=\cfrac{168}{22}\implies x=\cfrac{84}{11}\implies x=7\frac{7}{11}[/tex]
