## College Algebra (10th Edition)

$\displaystyle \frac{13}{11}$
2 by 2 determinant: $D=\left|\begin{array}{ll} a & b\\ c & d \end{array}\right|=ad-bc$ 3 by 3 determinant: $\left|\begin{array}{lll} {a_{11}}&{a_{12}}&{a_{13}}\\ {a_{21}}&{a_{22}}&{a_{23}}\\ {a_{31}}&{a_{32}}&{a_{33}}\end{array}\right|=\\\\=a_{11}\left|\begin{array}{cc} {a_{22}}&{a_{23}}\\ {a_{32}}&{a_{33}}\end{array}\right|-a_{12}\left|\begin{array}{cc} {a_{21}}&{a_{23}}\\ {a_{31}}&{a_{33}}\end{array}\right|+a_{13}\left|\begin{array}{cc} {a_{21}}&{a_{22}}\\{a_{31}}&{a_{32}}\end{array}\right|$ --- $\left|\begin{array}{rrr}{x}&{1}&{1}\\{4}&{3}&{2}\\{-1}&{2}&{5}\end{array}\right|=x\left|\begin{array}{cc} {3}&{2}\\{2}&{5}\end{array}\right|-1\left|\begin{array}{cc}{4}&{2}\\{-1}&{5}\end{array}\right|+1\left|\begin{array}{cc}{4}&{3}\\{-1}&{2}\end{array}\right|$ $=x(15-4)-(20+2)+(8+3)$ $=11x-22+11$ $= 11x -11$ So, we are solving $11x -11=2$ $11x=13$ $x=\displaystyle \frac{13}{11}$