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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.

Answer :

D3xt3R
[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]

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