Kelcilanereis
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# A dealer sells a certain type of chair and a table for $40. He also sells the same sort of table and a desk for$83 or a chair and a desk for \$77. Find the price of a chair, table, and of a desk.

$Chair=x$ $Table=y$ $Desk=z$ $\begin{Bmatrix}x+y&=&40\\y+z&=&83\\x+z&=&77\end{matrix}$ 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 $\begin{Bmatrix}z&=&60\\y&=&83-60\\x&=&77-60\end{matrix}$ $\boxed{\boxed{\boxed{\begin{Bmatrix}Chair&=&17\\Table&=&23\\Desk&=&60\end{matrix}}}}$