Accepted Answer
[tex]Chair=x[/tex][tex]Table=y[/tex][tex]Desk=z[/tex][tex]\begin{Bmatrix}x+y&=&40\\y+z&=&83\\x+z&=&77\end{matrix}[/tex]keep the first row as normal, then in the other ones, we can isolate Y and X[tex]\begin{Bmatrix}x+y&=&40\\y&=&83-z\\x&=&77-z\end{matrix}[/tex]now we can replace at first row...[tex]\begin{Bmatrix}(77-z)+(83-z)&=&40\\y&=&83-z\\x&=&77-z\end{matrix}[/tex][tex]\begin{Bmatrix}160-2z&=&40\\y&=&83-z\\x&=&77-z\end{matrix}[/tex][tex]\begin{Bmatrix}-2z&=&40-160\\y&=&83-z\\x&=&77-z\end{matrix}[/tex][tex]\begin{Bmatrix}-2z&=&-120\\y&=&83-z\\x&=&77-z\end{matrix}[/tex][tex]\begin{Bmatrix}z&=&60\\y&=&83-z\\x&=&77-z\end{matrix}[/tex]now we can replace the Z to discovery the other value[tex]\begin{Bmatrix}z&=&60\\y&=&83-60\\x&=&77-60\end{matrix}[/tex][tex]\boxed{\boxed{\boxed{\begin{Bmatrix}Chair&=&17\\Table&=&23\\Desk&=&60\end{matrix}}}}[/tex]