IDNLearn.com is the place where your questions are met with thoughtful and precise answers. Get accurate and detailed answers to your questions from our dedicated community members who are always ready to help.
Sagot :
[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]
[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]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.