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